11-0895 (AR) Revision 1 Structural CalcsAi�
RrAT3C"1'
r'J1`-"� `I' ` £rl_s
78080 Calle Amigo, Suite 102 phone: (760)771-9993
La Quinta, CA 92253 Fax: (760)771-9998
Cell: (760)808-9146
Structural Calculation
For Page Residence
At 48-801 San Lucas Dr.
La Quinta, CA.
Type Of Project: Residential
Remodeling+Addition
Designer: Michael Kiner Architect
Date: September 30, 2011
Design by: R.A.
JN: 11028
Revision#1
9/30/2011
RECEIVED
OCT 10 2011
BY:1
— ipt C
CITY OF LA QUINTA
BUILDING & SAFETY DEPT.
APPROVED
FOR CONSTRUCTION
tj —oq5
�PPgGHq� o�!
W G
tr NO. C 67613
EXP, 6/_3& /�
t _ ,
• DESIGN CRITERIA
2009 INTERNATIONAL BUILDING CODE
2010 CALIFORNIA BUILDING CODE
SOILS BEARING PRESSURE = 1500 PSF (PER SOILS REPORT)
EXTERIOR WALL = 25.00 PSF
INTERIOR WALL = 10.00 PSF
ROOF LOAD
ROOF DEAD LOAD
TILE ROOFING
= 10.00 PSF
SHEATHING
= 2.00 PSF
FRAMING
= 2.50 PSF
DRYWALL
= 2.00 PSF
INSULATION
= 1.50 PSF
MISCELLENUOUS
= 7.00 PSF
TOTAL DEAD LOAD
= 25.00 PSF
TOTAL LIVE LOAD
= 20.00 PSF
TOTAL LOAD
= 45.00 PSF
L'
El
0
a
•
•
RA PROJECT Page Res
PAGE.
CLIENT beam If1 hdr..left of rii. bdrm
Structural JOB NO DESIGN BY:
DATE: REVIEW BY:
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
4%5.e
&K to
0 No
No. 1, Douglas Fir -Larch
L = 11.5
It
Wo = 165
lbs / It
WL = "60.._
. lbs / It
Poo = 0
lbs
Lo = 0
ft
Poe = 6 _:
lbs
L2 = 0 '
It
L
Purr 1 + P02
Wo
d L -
&-I 360 Camber => 0.33 inch
dKaD,L=L/240.
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no)
0 No
Code Duration Factor. Cn Condition
1 0.90
Dead Load
2 1.00
Occupancy Live Load
3 1.15
Snow Load
4 1.25
Construction Load
5 1.60
Wind/Earthquake Load
6 2.00
Impact Load
Choice =>
4 Construction Load
ANALYSIS
51.6
DETERMINE REACTIONS, MOMENT, SHEAR
Wserrwt = 9
lbs / It RLeR = 1.35 kips
Vhu. = 1.20
kips, at 7.5 inch from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'min =
580
ksi
d =
7.50
in
FbE =
11345
psi
A =
41.3
int
1 =
193
in
S,r =
51.6
in
Re =
7.833
< 50
l E =
20.6
(ft, Tab 3.3.3 footnote 1)
. =
THE BEAM DESIGN IS ADEQUATE.
Cc) CM Ct Ci CL CF
Designation
1.25 1.00 1.00 1.00 0.99 1.00
1
CHECK BENDING AND SHEAR CAPACITIES r
fb = MMa. / S. = 900 psi < F -
b -
No. 1, Douglas Fir -Larch
f, = 1.5 VMa. / A = 44 psi <
3
CHECK DEFLECTIONS
d (L, M.) = 0.08 in, at 5.750 ft from left end,
Select Structural, Southern Pine
d (Ku D + L , M.) = 0.30 in, at 5.750 ft from left end
5
Where KQ = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
No. 2, Southern Pine
d 0.50, M.) = 0.33 in, at 5.750 It from left end
THE BEAM DESIGN IS ADEQUATE.
Cv Cc Cr
1.00 1.00 1.00
1673 psi
F, [Satisfactory]
[Satisfactory]
d L = L / 360 [Satisfactory]
d Kco.L = L/ 240 [Satisfactory]
0
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
RRigtt =
1.35 kips
. =
3.87
ft -kips, at 5.75 It from left end
E = E. =
1600
ksi Fy = 1687.5 psi
Fb =
1,350
psi F = FbE / Fe = 6.72
Fv =
170
psi Fe = 1,673 psi
E' =
1,600
ksi Fv' = 213 psi
Cv Cc Cr
1.00 1.00 1.00
1673 psi
F, [Satisfactory]
[Satisfactory]
d L = L / 360 [Satisfactory]
d Kco.L = L/ 240 [Satisfactory]
0
0
0
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
AXIAL LOAD F = `,.0: --kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F� = F� CD CP CF = 654 psi
Where F, = 925 psi
CD = 1.50
CF = 1.00 (Lumber only)
Cp = (1+F) / 2c - [(1+F) / 2c)2 - F / c]'.5 =
F-' = F. CD CF = 1480 Psi
LQ = Ke L = 1.0 L = 138 in
b = 5.5 in
SF = stendemess ratio = 25.1 <
F,E = 0.822 E'mn / SF2 = 757 psi
E'min = 580 ksi
F = F,E / F,* = 0.512
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < Fc'
THE ALLOWABLE FLEXURAL STRESS IS
Fo = 2141 psi, [ for CD = 1.6 ]
THE ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 900 psi < F;
CHECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(fe / Fe )2 + fb / fFe (1 - fc / FCE)] = 0.420
F F
1 /
0.442
50 [Satisfies NDS 2005 Sec. 3.7.1.41
[Satisfactory]
[Satisfactory]
< 1 [Satisfactory]
L
•
•
RA
PROJECT: 'Page Res:' PAGE
+ Structural CLIENT beam 42 p An beam atm bdrrn DESIGN BY:
JOB NO.: DATE: REVIEW BY:
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
:,ONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
,GW -'5 1/2-x 13-1/2' 4:
Glulam24F-1.8E
L =.;:•18:51
'ft
WD =--- 300
lbs / ft
WL = 240:
lbs / ft
PD1 _ .' . 0-
lbs
L1= -0_
ft
PD2 = U
Ibs
L2 = 0.
'ft
AL=L/360.
Wind/Earthquake Load
AKaD.L=L/240
2.00
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 0 No
Code
Duration
Factor Co Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
33.5
(ft, Tab 3.3.3 footnote 1)
DETERMINE REACTIONS,
MOMENT, SHEAR
WscrHe =
18
lbs / ft RL,.n = 5.16 kips
Vm. =
4.53
kips, at 13.5 inch from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'min =
930
ksi
d =
13.50
in
FbE =
6215
psi
A =
74.3
int
I =
1,128
in'
Sx =
167.1
in3
RB =
13.400
< 50
lE =
33.5
(ft, Tab 3.3.3 footnote 1)
CD CM G
Ci CL C.
1.25 1.00 1.00
1.00 0.96 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 1714
psi < Fb =
fv' = 1.5 VMax / A =
92 psi <
HECK DEFLECTIONS
2,400
A (L. Max) = 0.31
in, at 9.250 ft from left end,
4 (Ka D . L. Mm) = 0.72
in, at 9.250 ft from left end
Where Ku = 1.00
, (NDS 3.5.2)
rERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
Id (1.5D• Mm) = 0.62 in, at 9.250 ft from left end
L
F�P-1
C�,
1 Pos
1
W` F
23.86
ft-ldps, at 9.25 ft from left end
E = E. =
1800
ksi Fb* = 3000 psi
Fb =
2,400
psi F = FbE / Fb' = 2.07
Fv =
265
psi Fe = 2,876 psi
Camber => 0.62 inch
THE BEAM DESIGN IS ADEQUATE.
Cv CC C,
0.99 1.00 1.00
2876 psi [Satisfactory]
F, [Satisfactory]
< dl=L/360
< 'dl(iD.L=L/240
[Satisfactory]
[Satisfactory]
0
R" =
5.16 kips
Mr. =
23.86
ft-ldps, at 9.25 ft from left end
E = E. =
1800
ksi Fb* = 3000 psi
Fb =
2,400
psi F = FbE / Fb' = 2.07
Fv =
265
psi Fe = 2,876 psi
E' =
1,800
ksi F,; = 331 psi
Cv CC C,
0.99 1.00 1.00
2876 psi [Satisfactory]
F, [Satisfactory]
< dl=L/360
< 'dl(iD.L=L/240
[Satisfactory]
[Satisfactory]
0
N
U
•
CHECK THE BEAM CAPACITY WITH AXIAL LOA[
AXIAL LOAD F
THE ALLOWABLE COMPRESSIVE STRESS IS
F.' = F- Co CP CF = 411 psi
Where F� = 1600 psi
Co = 1.60
CF = 1.00 (Lumber only)
Cp = (1+F) / 2c - [(1+F) / 2c)Z - F / c]°.5 =
F.* = F. CD CF = 2560 psi
Le = Ke L = 1.0 L = 222 in
b = 5.5 in
SF = slenderness ratio = 40.4 <
FSE = 0.822 E'er;„ / SFZ = 419 psi
E'min = 830 ksi
F = F.E / F.' = 0.164
C = 0.9
THE ACTUAL COMPRESSIVE STRESS IS
% = F / A = 0 psi < F.'
ALLOWABLE FLEXURAL STRESS IS
Fb' = 3681 psi, [ for CD = 1.6 j
19
F F
0.161 '
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1714 psi < Fe [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fc / Fo )z + fb / [Fb' (1 - f. / F.E)] = 0.466 < 1 [Satisfactory]
O
•
C]
RA PROJECT Page Res PAGE:
1 ' Structural CLIENT beam ft3 beam at m'bathroom DESIGN BY
JOB NO f DATE REVIEW BY
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
'GL8 5:1/2 x 13 1/2 - Glutam 24F -1.8E
L=',..15.5 -'ft
Wo = -330..' lbs / ft
WL = 240'.'' lbs / ft
Poe =' ; 0....' lbs
Lt = 0 .`ft
PDz = 0" `! lbs
L2=-' 0. 'ft
A = L/ 360;.
dKcD.L=L/240
Does member have continuous lateral support by top diaphragm 7
(1= yes, 0= no) 0 No
Code
Duration Factor, Factor Cn Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
kLYSIS
1.E =
28.6
'ERMINE REACTIONS, MOMENT, SHEAR
Wseirwe =
18
lbs / ft Rt�ft = 4.55 kips
Vm. =
3.89
kips, at 13.5 inch from left end
DETERMINE SECTION PROPERTIESB ALLOWABLE STRESSES
b =
5.50
in
E'min =
930
ksi
d =
13.50
in
FbE =
7276
psi
A =
74.3
int
1 =
1,128
in'
Sx =
167.1
in
RB =
12.385
< 50
1.E =
28.6
(ft, Tab 3.3.3 footnote 1)
CO ' CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 0.97 1.00
IECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 1268
psi < Fb =
f,'= 1.5 VMa, / A =
79 psi <
ECK DEFLECTIONS
2,400
d (L. F&x) = 0.15
in, at 7.750 ft from left end,
d (Ka D . L. M.* = 0.38
in, at 7.750 ft from left end
Where K, = 1.00
, (NDS 3.5.2)
(ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
Id 0.50. M�t = 0.33
in, at 7.750 ft from left end
L
R" =
4.55 kips
_
.r L
ft4dps, at 7.75 It from left end
PDi�P.
/
1/
WL
2,400
psi F = FbE / Fe = 2.43
Fv =
265
WD
E' =
1,800
ksi 17; = 331 psi
}
Camber => 0.33 inch
THE BEAM DESIGN IS ADEQUATE.
Cv C, C,
1.00 1.00 1.00
2904 psi
F,; [Satisfactory]
[Satisfactory]
d L = L / 360 [Satisfactory]
dKv0+L = L / 240 [Satisfactory]
11
R" =
4.55 kips
Mneme =
17.65
ft4dps, at 7.75 It from left end
E = Ex =
1800
ksi Fb* = 3000 psi
Fb =
2,400
psi F = FbE / Fe = 2.43
Fv =
265
psi Fb' = 2,904 psi
E' =
1,800
ksi 17; = 331 psi
Cv C, C,
1.00 1.00 1.00
2904 psi
F,; [Satisfactory]
[Satisfactory]
d L = L / 360 [Satisfactory]
dKv0+L = L / 240 [Satisfactory]
11
0
•
•
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
AXIAL LOAD F = >=p: bps
THE ALLOWABLE COMPRESSIVE STRESS IS
Fr-= Fc Co CP CF = 580 psi
Where Fe = 1600 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0-5 = 0.226
F. = Fc Co CF = 2560 psi
Le = Ke L = 1.0 L = 186 in
b = 5.5 in
SF = slenderness ratio = 33.8 < 50
FCE = 0.822 E',,;,,/ SF = 597 psi
E'min = 830 ksi
F = F.E / Fc' = 0.233
C = 0.9
'HE ACTUAL COMPRESSIVE STRESS IS
f = Cl -
[Satisfies NDS 2005 Sec. 3.7.1.41
A - 0 psi < Fc' [Satisfactory]
ALLOWABLE FLEXURAL STRESS IS
Fe' = 3717 Psi, [ for Co = 1.6 ]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1268 psi < F;
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fe/Fc )2+fb/[Fe (1-fe/Fer)] = 0.341
[Satisfactory]
< 1 [Satisfactory]
0
1 1
ITTM
F
F
1
[Satisfies NDS 2005 Sec. 3.7.1.41
A - 0 psi < Fc' [Satisfactory]
ALLOWABLE FLEXURAL STRESS IS
Fe' = 3717 Psi, [ for Co = 1.6 ]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1268 psi < F;
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fe/Fc )2+fb/[Fe (1-fe/Fer)] = 0.341
[Satisfactory]
< 1 [Satisfactory]
0
11
•
RA PROJECT Page Res
PAGE:
t CLIENT beam !f4 tidr at rearof m bathroom DESIGN BY
Structural JOB NO
V►li,.,:�':Ct�a1.: n..:ac�.-` �==- -- .......-,�•-_:.. � _ DATE REVIEW BY: -
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
t3 x 6 No. 1, Douglas Fir4arch
L =
:6:25'
It
WD =
' 235:
lbs / ft
WL _-'
140-
`' lbs / ft
Pot =
O:,
r: lbs
L1 =
0:
ft
Poz =
`0
.. lbs
L2 = 0
_L
ft
L
Duration Factor C Condition
1
0.90
L
2
_
Occupancy Live Load
3
FD jj
Snow Load
Poz
+
7
W` ` 8 + #
-
5
8
Wind/Earthquake Load
wD
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
I E =
11.6
DETERMINE REACTIONS, MOMENT, SHEAR
WsenW =
dL - /. Camber=> 0.10 inch
d r.D.L=L/240,
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 0 No
Code
Duration Factor C Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Constnxtion Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
I E =
11.6
DETERMINE REACTIONS, MOMENT, SHEAR
WsenW =
7
lbs / ft RLan = 1.19 kips
VMax =
1.02
kips, at 5.5 inch from left end
MINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'min =
560
ksi
d =
5.50
in
FbE =
27589
psi
A =
30.3
int
I =
76
in°
Sx =
27.7
in
RB =
5.023
< 50
I E =
11.6
(ft, Tab 3.3.3 footnote 1)
CO CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 1.00 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 806
psi < Fb =
f,,'= 1.5 VMax / A =
50 psi <
ECK DEFLECTIONS
5
d (L. Mix) = 0.04
in, at 3.125 ft from left end,
d (Ka D . L. Marl = 0. 11
in, at 3.125 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
FERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.5D. Max) = 0.10 in, at 3.125 ft from left end
THE BEAM DESIGN IS ADEQUATE.
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
R" = 1.19 kips
M'. = 1.86 ft4dps, at 3.13 ft from left end
E = Ex =
1600
ksi
Fb =
1,350
psi
Fv =
170
psi
E' =
1,600
ksi
Cv Cc Cr
1.00 1.00 1.00
1682 psi
F,' [Satisfactory]
Fb = 1687.5 psi
F = FbE / Fb. = 16.35
Fb = 1,682 psi
Fv = 213 psi
[Satisfactory]
d L = L/ 360 [Satisfactory]
d Kc o . L = L/ 240 [Satisfactory]
01
•
•
0
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
r AXIAL LOAD F = ;:.0... kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F.' = Fc CD Cp CF = 1245 psi
Where Fc = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
Cp = (1+F) / 2c - [(1+F) / 2c)Z - F / c]°,5 = 0.841
Fc = Fc CD CF = 1480 psi
Le = KL = 1.0 L = 75 in
b = 5.5 in
SF = slenderness ratio = 13.6 < 50
FcE = 0.822 E'm„ / SF = 2564 psi
E'min = 580 ksi
F = FcE / Fc = 1.732
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
f = F/A =
FTTT ��
F F
}
[Satisfies NDS 2005 Sec. 3.7.1.4]
c 0 psi < Fc' [Satisfactory]
ALLOWABLE FLEXURAL STRESS IS
Fe = 2153 psi, [ for CD = 1.6 ]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 806 psi < Fo
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fc/FC, )2+tb/[Fe (1-fc/Fcr)I= 0.374
[Satisfactory]
< 1 [Satisfactory]
x
•
0
•
RA PROJECT Page R�
CLIENT beam #5 hdr at rear.',of exercise room PAGE.
Structural DESIGN BY:
JOB NO f DATE
►IVnnA`:Ro�en n'e�s..� n ._ ._:...� _ :.; ' REVIEW BY.
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
:;ONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
6'X:8",, No. 1, Douglas Fir -larch
L ="`_:7'25; . ft
Wo =
,235":.; Ibs / ft
WL =
140 ` lbs / ft
PD1 ='
. 0.<: lbs
Occupancy Live Load
ft
PD2 =
::>0 Ibs
L2
0 ft
L,
Duration Factor C Condition
1
FL
Dead Load
�
Po j
Occupancy Live Load
{ Po2
WL
Snow Load
4
1.25
'
'D
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
I E =
d L = L / 360 Camber => 0.07 inch
dKoD.L=Ll240 :.
Does member have continuous lateral support by top diaphragm ?
0= yes, 0= no) 0 No
Code
Duration Factor C Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
I E =
13.7
DETERMINE REACTIONS, MOMENT, SHEAR
Wsetrw =
9
lbs / ft RLe11 = 1.39 kips
VM. =
1.15
kips, at 7.5 inch from left end
MINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'min =
560
ksi
d =
7.50
in
FbE =
17085
psi
A =
41.3
int
I =
193
in'
Sx =
51.6
in3
RB =
6.383
< 50
I E =
13.7
(ft, Tab 3.3.3 footnote 1)
CD CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 0.99 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 587
psi < Fb =
fv = 1.5 VMax / A =
42 psi <
ECK DEFLECTIONS
5
d (L. Max) = 0.03
in, at 3.625 ft from left end,
d (Kn D + L. Max) = 0.08
in, at 3.625 ft from left end
Where Ku = 1.00
, (NDS 3.5.2)
FERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A(1sD. Max) = 0.07 in, at 3.625 ft from left end
THE BEAM DESIGN IS ADEQUATE.
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -Larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Rwgr = 1.39 kips
Mm. = 2.52 ft -kips, at 3.63 ft from left end
E = Ex =
1600
ksi
Fb =
1,350
psi
Fv =
170
psi
E' =
1,600
ksi
Cv Cc Cr
1.00 1.00 1.00
1678 psi
F,; [Satisfactory]
Fb = 1687.5 psi
F = FbE / Fb* = 10.12
Fb' = 1,678 psi
Fv = 213 psi
[Satisfactory]
d L = L/ 360 [Satisfactory]
d rOD. L = L/ 240 [Satisfactory]
N
d
•
•
•
HECK THE BEAM CAPACITY WITH AXIAL LOA[
KIAL LOAD F = ,:0 : kips
iE ALLOWABLE COMPRESSIVE STRESS IS
F.' = F. Co CP CF = 1140 psi
Where F. = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cju.5 =
F. = F. Co CF = 1480 Psi
Le = Ke L = 1.01- = 87 in
b = 5.5 in
SF = slenderness ratio = 15.8 <
FIE = 0.822E',,„/SF 2 = 1905 Psi
E'min = 580 ksi
F = F.E / F.* = 1.287
C = 0.8
E ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < Fc'
ALLOWABLE FLEXURAL STRESS IS
Fe = 2148 psi, [ for Co = 1.6 j
1
F F
0.770
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe)/ S = 587 psi < Fe [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(f. / F�' )2 + fb / (Fe' (1 - fc / FCE)j = 0.273 < i (Satisfactory]
8
•
•
•
RA PROJECT PAGE RES PAGE:
i
CLIENT.$EAM#9 FL H AT FRONTOFGREATROO DESIGN BY
STRUCTURAL JOB NO . DATE :%'.� �/ : REVIEW BY.
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
LIMIT OF LIVE LOAD
LIMIT OF LONG-TERM
`6X`10
(1= Yes, 0= no)
`No.1,Douglas Fir -Larch
L
8
It
Wo = :'-
455;;
lbs / ft
WL = .;
340,.
; Ibs / ft
Po, _::
,.. �;.:
.: lbs
1.25
0
It
PD2=.::0:,
Wind/Earthquake Load
lbs
Lz= ;,: 0
:ft
L
(1= Yes, 0= no)
1 Yes
Code
L Z_�
1
Po 1
Dead Load
PDz
w�
Occupancy Live Load
3
WD
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
d - L / :360 Camber => 0.10 inch
dV.D.L=L/240
Does member have continuous lateral support by top diaphragm ?
(1= Yes, 0= no)
1 Yes
Code
Duration Factor, Cn Codes
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
2 Occupancy Live Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
Wsenw, =
11
lbs / It Rye = 3.23 kips
Vwzx =
2.59
kips, at 9.5 inch from left end
NINE SECTION PROPERTIES& ALLOWABLE STRESSES
THE BEAM DESIGN IS ADEQUATE.
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -Larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Rray,r = 3.23 kips
M' = 6.45 ft4dps, at 4.00 ft from left end
b =
5.50 in Emir, = N/A
E = E.=
1600 kei
Fb = N/A
d =
9.50 in FbE = N/A
Fb =
1,350 psi
F = FbE / Fb* = N/A
A =
52.3 int 1 = 393 in
Fv =
170 psi
Fb' = 1,350 psi
Sx =
82.7 in RB = N/A
E' =
1,600. ksi
Fv' = 170 psi
lE =
N/A
CD CM Ct
Ci CL CF
1.00 1.00 1.00
1.00 1.00 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 936
psi < Fb =
f,;= 1.5 VMax / A =
74 psi <
CHECK DEFLECTIONS
d (L. M.) = 0.05
in, at 4.000 ft from left end,
d (Ka o . L . M.) = 0.12
in, at 4.000 ft from left end
Where K= = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.50, Max) = 0.10
in, at 4.000 ft from left end
Cv C, Cr
1.00 1.00 1.00
1350 psi
F,; [Satisfactory]
[Satisfactory]
dL=L/360
40-L=L/240
[Satisfactory]
[Satisfactory]
6
I
•
•
C,
HECK THE BEAM CAPACITY WITH AXIAL LOA[
;IAL LOAD F kips
IE ALLOWABLE COMPRESSIVE STRESS IS
F.* = Fo Co CP CF = 1367 psi
Where F. = 925 psi
CD = 1.80
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - 1(1+F) / 2c)2 - F / c]o.e =
Fo = F. CD CF = 1480 Psi
Lo = K, L = 1.01- = 96 in
d = 9.5 in
SF = slenderness ratio = 10.1 <
FoE = 0.822 E',a„ / SF = 4669 psi
E'min = 580 ksi
F = Foe / Fe = 3.155
C = 0.8
ACTUAL COMPRESSIVE STRESS IS
fo = F / A = 0 psi < Fc'
ALLOWABLE FLEXURAL STRESS )S
Fe = 2160 psi, [ for CD = 1.6 ]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 936 psi < FD
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(fc / FC,
)2 + fb / [Fe (1 - fo / Foe)] = 0.433
F F
0.924
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
[Satisfactory]
< 1 [Satisfactory]
9
11
•
RA PROJECT PAGE RES =: PAGE:
' CLIENT 'BEAM #11 FL HDR:A71"RQNT'F GREAT Ra DESIGN BY :
STRUCTURAL
JOB NO DATE
..._ ., .... ,.. .u,_..,.,. .._-, .._.. � ,oir �' REVIEW BY:
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 1 Yes
L
Glulam 24F -1.8E
L= .'1%5*..
R
Wo = ;580 ?..
lbs / ft
WL = ' 440;
, lbs / ft
Pao = :0 :5:
lbs
2 1.00 Occupancy Live Load
PD2 = - ::0. ' s
lbs
L2=_ 0
ft
d L = L / 360'
dK D-L=L/240
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 1 Yes
L
C�,Po,
Poe
WL
1 0.90 Dead Load
0
2 1.00 Occupancy Live Load
4
Camber => 0.34 inch
THE BEAM DESIGN IS ADEQUATE.
Code Duration Factor. C Condition
1 0.90 Dead Load
2 1.00 Occupancy Live Load
3 1.15 Snow Load
4 1.25 Construction Load
5 1.60 Wind/Earthquake Load
6 2.00 Impact Load
Choice => 2 Occupancy Live Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
Wsar o rt = 22 lbs / ft RLQ = 8.60 kips
Rw yx = 8.60 kips
Vm. = 7.03 kips, at 18 inch from left end
Mm. =
35.46
ft -kips, at 8.25 ft
from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b = 5.13 in E'min = N/A
E = Ex =
1800
ksi Fb =
N/A
d = 18.00 in FbE = N/A
Fb =
2,400
psi F = FbE / Fe = N/A
A = 92.3 int 1 = 2,491 in
Fv =
265
psi Fb' =
2,361 psi
Sx = 276.8 in 3 RB = N/A
E' =
1,800
ksi F,; =
265 psi
lE = N/A
Cc CM Ct Ci CL CF Cv
Cp
Cr
1.00 1.00 1.00 1.00 1.00 1.00 0.98
1.00
1.00
:HECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 1538 psi < Fb = 2361
psi
[Satisfactory]
fv' = 1.5 VMax / A = 114 psi < F,
[Satisfactory]
HECK DEFLECTIONS
d iL, M.0 = 0.16 in, at 8.250 It from left end,
<
d L = L / 360
[Satisfactory]
d (Ka D . L. Mui = 0.39 in, at 8.250 ft from left end
<
d K� D , L = L / 240
[Satisfactory]
Where Ka = 1.00 , (NDS 3.5.2)
(ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
d (1.50. Max) = 0.34 in, at 8.250 It from left end
e fil ff / 0 00 1 r-r�kD
•
•
i
HECK THE BEAM CAPACITY WITH AXIAL LOA[
CAL LOAD F = 'V 0' i : kips
-IE ALLOWABLE COMPRESSIVE STRESS IS
Fe = F. CD CP CF = 2411 psi
Where F. = 1600 psi
Co = 1.50
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 -F / c]0.5
F-' = F. Co CF = 2560 psi
Le = K� L = 1.01- = 198 in
d = 18 in
SF =slenderness ratio = 11.0 <
FIE = 0.822 E' --n /SF2 = 5318 psi
E'min = 930 ksi
F = F,E / F�* = 2.468
C = 0.9
E ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < F.'
ALLOWABLE FLEXURAL STRESS 1S
Fo = 3777 psi, [ for Co = 1.6 ]
/ 1
F F
0.942
50 [Satisfies NDS 2005 Sec. 3.7.1.41
[Satisfactory]
-- ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1538 psi < Fti [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(f. / F.' )2 + fp / (Fb, (1 - fc / FcE)j = 0.407 < 1 [Satisfactory]
•
•
0
RA PROJECT PAGE RES s
STRUCTURAL JCLIENT - OB NO.:$ #13 HpR A DAN�OF
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
6X;92 No. I. Douglas Fir -Larch
L
8:5
;ft
WD ='
:580;
lbs / ft
WL
Dead Load
lbs ft
1.00
Occupancy Live Load
3
1.15
Snow Load
ft
PD2 =
Construction Load
5
L2 =
; 0.
, ft
PAGE:
)M DESIGN BY:
REVIEW BY: - y
L
Po, I
(1= yes, 0= no)
1Pos
/
wl
'D
Duration Factor Co Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
Camber => 0.09 inch
d K� D. L = L/ 246
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no)
1 Yes
Code
Duration Factor Co Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
2 Occupancy Live Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
Wstlrwt =
14
lbs / ft RLe = 4.39 kips
VMax =
3.40
kips, at 11.5 inch from left end M'x =
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'min =
N/A
d =
11.50
in
FbE =
N/A
A =
63.3
int
1 =
697 in
Sx =
121.2
in'
RB =
N/A
LE =
N/A
CD CM G
Ci CL CF
1.00 1.00 1.00
1.00 1.00 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 924
psi < Fb =
f,'= 1.5 VMax / A =
81 psi <
ECK DEFLECTIONS
5
Id (L. Max) = 0.05
in, at 4.250 ft from left end,
'd (Kar o . L. Max) = 0.11
in, at 4.250 ft from left end
Where Kn = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
d (1.5D, Max) = 0.09 in, at 4.250 ft from left end
THE BEAM DESIGN IS ADEQUATE.
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fr -larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
RMON = 4.39 kips
9.34 ft -kips, at 4.25 ft from left end
E = E. =
1600
ksi
Fb =
1,350
psi
Fv =
170
psi
E' =
1,600
ksi
Fb = N/A
F = FbE / Fb = WA
Fb = 1,350 psi
F,, = 170 psi
C„ C' C,
1.00 1.00 1.00
1350 Psi [Satisfactory]
Fv' [Satisfactory]
< dL=L/360
< dK«D.L=L/240
f,m . 4 l2 t /rli���
[Satisfactory]
[Satisfactory]
1ECK THE BEAM CAPACITY WITH AXIAL LOAC
.IAL LOAD F
E ALLOWABLE COMPRESSIVE STRESS IS
F-' = FI CD CP CF = 1396 psi
Where F. = 925 psi
CO = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cf.5 =
Fc = Fc CD CF = 1480 Psi
Le = Ko L = 1.01- = 102 in
d = 11.5 in
SF = slenderness ratio = 8.9 <
FIE = 0.822 E;,. /SF2 = 6060 psi
E'min = 580 ksi
F = Foe / F.* = 4.095
C = 0.8
ACTUAL COMPRESSIVE STRESS IS
fI = F / A = 0 psi < Fc'
E ALLOWABLE FLEXURAL STRESS 1S
1 1
F F
0.944
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
Fn' = 2160 psi, [ for CD = 1.6 )
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 924 psi < Fti [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(fI / FI. )2 + fb / [Fb' (1 - fI / F.01 = 0.428 < 1 (Satisfactory]
•
0
101
C]
•
•
Raymond
B. Franqie
PROJECT: THE PAGE RESIDENCE
CLIENT: BEAM # 14 cant bm at.patio
JOB NO.: DATE:
INPUT DATA $ DESIGN SUMMARY
BEAM SECTION GLB 51/8 x 18 Glulam 24F -1.8E
BEAM SPAN L t = . 30 ft
CANTILEVER L2= 3 ft, (0 for no cantilever)
DEAD LOADS
PROJECTED LIVE LOADS
wDL,l =
0.205
kips/ ft
wDL.2 =
0.14
kips / ft
wLL.i =
0.205
kips / ft
wLL,2 =
0.14
kips / ft
INCENTRATED LOADS
P,
P2 I
P3
Dead ki s
0
0 1
0
Live kis
0
0
0
Location from Left End ft
0
0
33
LOPE 4 :12 (0 = 18.43 ° )
DEFLECTION LIMIT OF LIVE LOAD
ALL =L/360
Code
Duration Factor. Cn
Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=> 4
Construction Load
YSIS
1
REACTIONS, MOMENTS & SHEARS
Rz=051 cosB+w )Li+(cos0+w,, l(Li+0.5Lz)Lz+EP d =
l 1111/ Li Lt
w-,wn.., l
R� cosB+w,•.,)L,+(cos9+w,,, ILz+EP–Rz= 6.27 kips
PAGE:
DESIGN BY:
REVIEW BY:
w—
_ wu.,
j_Z_0 =iT
R,f
'N
Slope
t2
I R
I
t L L,
MMm
L�-2
A-
THE BEAM DESIGN IS ADEQUATE.
7.22 kips Xt = 15.00 ft
X2 = 15.00 ft
X3= 0.00 ft
M = 46.7 ft -kips M trn = 0.5 + wuz) Li
wd, + P3L2 = 1.3 ft -kips
Ya = 6.36 kips, at R2 left. (coso i
2MINE SECTION PROPERTIES AND DESIGN FACTORS
L„ = M -(X3 . L2 ) = 3.0 ft, (NDS 2005 Table 3.3.3)
GLB 5 1/8 x 18 Properties
b = 5.13 in Fb = 2,400
psi LE = 6.2 ft, (fab 3.3.3 footnote 1)
d = 18.00 in Fv = 265 psi, (NDS 97 CM included) R 7.1 < 50
A = 92.3 int 6=
E' = 1,800 ksi E'y= 1,600 ksi
M A
•
•
•
Sx =
276.8
in
Fp =
2,753 psi
I =
2,491
in
F, =
331 psi
E = E.=
1800
ksi
E'f in =
930 ksi
CD
CM
Ct
Ci CL
CF CV Cc C1
1.25
1.00
1.00
1.00 0.99
1.00 0.92 1.00 1.00
CHECK BENDING AND
SHEAR CAPACITIES
Cantilever. Jr. =Mmin lSX = 56 psi < Fb =
Middle Span: fb = MMS IS,, = 2026 psi < Fb =
Shear. f„' = 1.5 VMa, lA = 103 psi <
(neglected d offset conservatively)
CHECK DEFLECTION AT LIVE LOAD CONDITION
L = L 1loos B = 31.62 ft, beam sloped span
Put Put Pua
a or b, ft 31.62 31.62 3.16
P, (k) 1 0.00 1 0.00 1 0.00
W1 =war COS2 0= 0.18 klf, perpendicular to beam
W2 = WU 2 OOS2 B = 0.13 ktf, perpendicular to beam
P3a3(L+a3) w2a3(4L+3a3)
FbE = 21959 psi
Fb = 3000 psi
F = FbE / Fb. = 7.32
2753 psi
2775 psi
F„' [Satisfactory]
[Satisfactory]
[Satisfactory]
3E/ + 24E/ )°OSB 0.01 in, downward to vertical direction.
< 2 L2 / 360 = 0.20 in [Satisfactory]
_ 0.0 2 z`t.s SwtL4l __
Oneia — E ElL L —b + 384E1 J cosB 0.88 in, downward to vertical direction.
< L, / 360 = 1.00 in [Satisfactory]
N)
•
•
r\ • - Ac V\ , , ` \I I\ )
Reza PROJECT: BEAM#'15, CANT BM. AT PATIO -
A$ har our CLIENT: ;PAGERESIDEBCE ,-
9 p JOB NO.: 11.028,- 4 __ ..� ' DATF719/28/20M
Wood Beam.Dn nSAse on NDS:2005 ' d W iP
INPUT DATA & DESIGN SUMMARY
BEAM SECTION GLB-5,1/2 x 19 1/2 Glulam
BEAM SPAN L =i 11.5 `.ft yl
CANTILEVER L2 2 5 'ft. (0 for no cantil
SLOPED DEAD LOADS WDLA =, 0.4 2Jsoce 14,
WDL.2-� 0
PROJECTED LIVE LOADS WLL•1 = '0
• N—
WLL.2 = 0
CONCENTRATED LOADS PDL = 4.22
PLL = 3
SLOPE 0 :12 (0 = 0.00 °) '`'-i
DEFLECTION LIMIT OF LIVE LOAD I x. I x I x,
d LL = L /360y
LONG T
- ERM DEFLECTION (NDS 3.5.2)
a_
4 Kcr D . L = L / -180
Code
Duration Factor C„
Condit
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
R:=0.5 =+,,,_, L,*
�coso
-=+,,,_, (r•,+osc.)r=°+rr"r--=
�wsn
Wind/Earthquake Load
r•, r•,
Impact Load
\
-' " .' Construction Load
1
(W50
C050 +
(cos A
ALYSIS
TERMINE REACTIONS, MOMENTS & SHEARS
0.73 kips
A', = 1.83 ft
X2= 1.83 ft
X3 = 7.85 ft
PAGE:
DESIGN BY: RA
REVIEW BY: �RA
41 11
Acr- 0s: 5' 9 OV PSL
THE BEAM DESIGN IS ADEQUATE.
11.09 kips
18.1 ft -kips
0.7 ft -kips
Amax = 11.09 kips, at R2 left.
TERMINE SECTION PROPERTIES AND DESIGN FACTORS
L u = M -(X3 , L2 ) = 7.8 ft, (NDS 2005 Table 3.3.3)
GLB 5 1/2 x 10 1/2 Properties
_,Ejt] .,n'(a L.-3,,)1 0 = 2,400 psi
_ p /E = 15.4 ft, (Tab 3.3.3 footnote 1)
3rar 2asr 24 /-'/
U ry = 265 psi, (NDS 97 CM included) R 8.0 < 50
E' = 1,800 ksi E'y= 1,600 ksi
16L'r 384EI 32 El
oao
•
C]
•
SX = 101.1 in Fe = 2,969 psi
i"bE = 17379 psi
I = 531
in' _
F,; - 331 psi
Fb' = 3000 psi
E = Ex = 1800 ksi E'min = 930 ksi
F = FbE / Fb* = 5.79
3/i/
� C C'
?4
/:/ 24 // C
f
1.40 I.Uu I.vu 1.uv v.99 1.00 1.00 1.00
1.00
/, `L�
CIAmwd Pcos
fl=
" /="2) OPACITIES
161.7 3846/
32/:7
Cantilever:
fb' = MM;r, /SX = 2143 psi < Fb =
2969 psi [Satisfactory]
Middle Span:
fb = MMax IS. = 79 psi < Fb =
3000 psi [Satisfactory]
Shear:
f„' = 1.5 VMax /A = 288 psi <
F„' [Satisfactory]
(neglected d offset conservatively)
CHECK DEFLECTION AT LIVE LOAD CONDITION
L = L , /cos 0 =
11.50 ft, beam sloped span
a = L 2/cos 0 =
2.50 ft, beam sloped cantilever length
P = PLL cos 0 =
3.00 kips, perpendicular to beam
W, = W LL.1 cos 2
0 = 0.00 klf, perpendicular to beam
W2 = W LL 2 cos 2
0 = 0.00 klf, perpendicular to beam
0.16 in, downward to vertical direction.
< 2 L2 / 360 = 0.17 in [Satisfactory]
-0.11 in, uplift to vertical direction.
< L, / 360 = 0.38 in ,[Satisfactory]
:_CK DEFLECTION AT LONG-TERM LOAD, Kr DL + LL, CONDITION
P = PKcroL+LL cos 0 = 9.33 kips, perpendicular to beam Ku = 1.50 , (NDS 3.5.2)
w, = Kcr W DL., cos 0 + W LL.1 cost 0 = 0.60 klf, perpendicular to beam
W2 = Kcr W DL.2 cos 0 + W LL,2 cos 2 0 = 0.00 klf, perpendicular to beam
0.32 in, downward to vertical direction.
< 2 L2 / 180 = 0.33 in [Satisfactory]
-0.10 in, uplift to vertical direction.
< L, / 180 = 0.77 in [Satisfactory]
D
(confd)
•
•
RA
'STRUCTURAL
PROJECT PAGE RES: PAGE:
CLIENT BEAM #18 fYP HDR'AT F OF 4ALLWAY DESIGN BY:
JOB NO. DATE
1�. REVIEW BY
INPUT DATA 8 DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
LIMIT OF LIVE LOAD
LIMIT OF LONG-TERM
6 ic;8 ','No. 1, Douglas Fir -Larch
L = .
6;25',:1
ft
WD =
: '5W:'
IbS / ft
WL = -..
400
lbs / ft
PD1 = .
0,
lbs
1.25
Construction Load
5
PD2 =
Q:.
IbS
L2 =
Q -
ft
L
Poi
111111
L
1
wL
WD
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snowy Load
4
1.25
Construction Load
5
d L = L / 360 Camber => 0.10 inch
dreD.L=L/240:
Does member have continuous lateral support by top diaphragm ?
(1= Yes, 0= no) 1 Yes
Code
Duration Factor, C. Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snowy Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
2 Occupancy Live Load
kLYSIS
Rwyt =
FERMINE REACTIONS,
MOMENT, SHEAR
Wse f m =
9
lbs / ft RLen = 3.09 kips
VM„ =
2.47
kips, at 7.5 inch from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'mi„ =
N/A
d =
7.50
in
FbE =
N/A
A =
41.3
int
1 =
193 in
S, =
51.6
in'
RB =
WA
lE =
N/A
Choice
=> 2
CD CM Ct
Ci CL CF
1.00 1.00 1.00
1.00 1.00 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 1124
psi < Fb =
f,; = 1.5 VM,, / A =
90 psi <
CHECK DEFLECTIONS
3
d (L, Max) = 0.04
in, at 3.125 ft from left end,
d (Kv D. L , Max) = 0.11
in, at 3.125 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD
+ SELF WEIGHT)
A (11.50, Max) = 0.10
in, at 3.125 ft from left end
THE BEAM DESIGN IS ADEQUATE.
Cv C, Cr
1.00 1.00 1.00
1350 Psi [Satisfactory]
Fy' [Satisfactory]
< dL=L/ 360
< dKWD+L=L/240
dmf 16f li 010r6p
[Satisfactory]
[Satisfactory]
9
Code
Designation
1
Select Structural, Douglas Fr -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -Larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Rwyt =
3.09 kips
MM„ =
4.83
ft-Idps, at 3.13 It from left end
E = E, =
1600
ksi Fb* = N/A
Fb =
1,350
Psi F = FbE / Fb' = N/A
F„ =
170
psi Fe = 1,350 psi
E' =
1,600
ksi F, = 170 psi
Cv C, Cr
1.00 1.00 1.00
1350 Psi [Satisfactory]
Fy' [Satisfactory]
< dL=L/ 360
< dKWD+L=L/240
dmf 16f li 010r6p
[Satisfactory]
[Satisfactory]
9
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
AXIAL LOAD F = 0,`>''< kips
THE ALLOWABLE COMPRESSIVE STRESS IS
• F� = Fc Co CP CF = 1370 psi
Where Fc = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.e =
Fc` = Fc CO CF = 1480 psi
Le = K, L = 1.0 L = 75 in
d = 7.5 in
SF = slenderness ratio = 10.0 <
FcE = 0.822 E'„j„ / SF2 = 4766 psi
E'min = 580 ksi
F = Foe / Ft* = 3.221
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < F.'
E ALLOWABLE FLEXURAL STRESS IS
F F
0.925
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
Fe = 2160 psi, [ for CD = 1.6 J
E ACTUAL FLEXURAL STRESS IS
fD = (M + Fe) / S = 1124 psi < F; [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fo / Fc. )2 + fD / [FD (1 - fo / FoE)l = 0.520 < 1 [Satisfactory]
r:
�J
•
RA PROJECT
:Pa*R- PAGE:
CLIENT beam #20 drop beam`s! {sundry DESIGN BY
Structural JOB NO.: . DATE::. REVIEW BY:
Wood;BeamDesign:8ase an -NDS 2005 . -'; '; ::
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
A L = L /:36D Camber => 0.18 inch
d Ka D . L = L / :240..
Does member have continuous lateral support by top diaphragm ?
(1= yes,
No. 1, Douglas Fir -Larch
L=`
'10:'
ft
wo =
`:330
lbs / ft
WL =
::240.
` lbs / ft
PD, _
0' .
lbs
Snow Load
.0'
It
PD2 =
_0 :
lbs
LZ =
Q
ft
A L = L /:36D Camber => 0.18 inch
d Ka D . L = L / :240..
Does member have continuous lateral support by top diaphragm ?
(1= yes,
0= no) 0
-F
Code
Duration Factor, C„
Po,
w,.
C
+ PD
/
'D
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
4
Construction Load
4
A L = L /:36D Camber => 0.18 inch
d Ka D . L = L / :240..
Does member have continuous lateral support by top diaphragm ?
(1= yes,
0= no) 0
No
Code
Duration Factor, C„
Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=> 4
Construction Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
THE BEAM DESIGN IS ADEQUATE.
Code
Designation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -Larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Wsenvuc =
11
lbs / ft RLQ = 2.91 kips
1.00 0.99 1.00
RRkft =
2.91
Id ps
f,'= 1.5 VM,, / A =
Vm„ =
2.45
kips, at 9.5 inch from left end
M" =
7.27
ft4dps, at 5.00 ft from left end
in, at 5.000 ft from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
, (NDS 3.5.2)
b =
5.50
in E'min = 580 ksi
E = E,=
1600
ksi
Fb* = 1687.5
psi
d =
9.50
in FbE = 9889 psi
Fb =
1,350
psi
F = FbE / Fe =
5.86
A =
52.3
int 1 = 393 in"
F, =
170
psi
Fe = 1,671
psi
SX =
82.7
in' RB= 8.389 <50
E' =
1,600
ksi
F„ = 213
psi
1E=
18.7
(ft, Tab 3.3.3 footnote 1)
CD CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 0.99 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMa. / S. = 1054
psi < Fb =
f,'= 1.5 VM,, / A =
70 psi <
ECK DEFLECTIONS
d (L, Max) = 0.09
in, at 5.000 ft from left end,
d (Ka D . L, M�) = 0.21
in, at 5.000 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
(ERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
d 0.50, MW = 0.18 in, at 5.000 ft from left end
Cv C, Cr
1.00 1.00 1.00
1671 psi
F� [Satisfactory]
[Satisfactory]
d L = L/ 360 [Satisfactory]
d Ker D. L = L/ 240 [Satisfactory]
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
AXIAL LOAD F 0 =:' Idps
THE ALLOWABLE COMPRESSIVE STRESS IS
Fe = Fc Co CP CF = 808 psi
Where F. = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.e =
Fc = F. Co CF = 1480 psi
Le = KB L = 1.01- = 120 in
b = 5.5 in
SF = slenderness ratio = 21.8 <
FCE = 0.822 E'ffd„ / SF = 1002 Psi
E'min = 580 ksi
F = FIE / Fc* = 0.677
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
f, = F / A = 0 psi < F,'
E ALLOWABLE FLEXURAL STRESS IS
m � Z
F F
0.546 T
50 [Satisfies NDS 2005 Sec. 3.7.1.41
[Satisfactory]
Fb' = 2138 psi, [ for Co = 1.6 ]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1054 psi < Fp [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(fo / F., )Z + fb / [Fe (1 - f. / FCE)] = 0.493 < 1 [Satisfactory]
Cl
(0
•
•
0
RA
STRUCTURAL
PROJECT PAGE RES
CLIENT BEAM 022
JOB NO.:
)S�2005
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
S DR ATfNTRY OO
DATE: . -ham pai
PAGE:
DESIGN BY:
REVIEW BY:
L
PD+ 1
L
No. 1, Douglas Fir -Larch
L=
5.
ft
Wo =
:520
lbs / ft
WL =
320
lbs / ft
Poo =
0>
lbs
4
1.25
ft
P02 = ..
0 .
lbs
L2 =
0 ,,
ft
PAGE:
DESIGN BY:
REVIEW BY:
L
PD+ 1
L
+ PD2
WL
wo
Duration
Factor CD Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
d L = L / 360 Camber => 0.04 inch
d KaD«L = L/240
Does member have continuous lateral support by top diaphragm ?
(1= Yes, 0= no)
1 Yes
Code
Duration
Factor CD Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
2 Occupancy Live Load
ANALYSIS
2.65
ft -kips, at 2.50 ft from left end
DETERMINE REACTIONS, MOMENT, SHEAR
Wsulwr =
9
Itis / It RLeft = 2.12 kips
Vm. =
1.59
kips, at 7.5 inch from left end
NINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in
E'fT11„ =
N/A
d =
7.50
in
FbE =
N/A
A =
41.3
int
I =
193 in'
Sx =
51.6
in'
RB =
N/A
1E=
N/A
Choice
=> 2
CD CM Ct Ci
1.00 1.00 1.00 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 617 psi <
f,'= 1.5 Vma, / A 56 ps
HECK DEFLECTIONS
THE BEAM DESIGN IS ADEQUATE.
CL CF Cv CD Cr
1.00 1.00 1.00 1.00 1.00
Fb = 1350 psi
< F, [Satisfactory]
d (L, Max) = 0.01 in, at 2.500 ft from left end, <
d (Ka D . L. Ma* = 0.04 in, at 2.500 ft from left end <
Where Ka = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.51), mu) = 0.04 in, at 2.500 It from left end
[Satisfactory]
dL=L/360
dKaD.L=L/240
[Satisfactory]
[Satisfactory]
Code
Designation
1
Select Structural, Douglas Fr -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -Larch
4
Select Structural, Southern Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Rwpt =
2.12 kips
M'. =
2.65
ft -kips, at 2.50 ft from left end
E = E. =
1600
ksi Fb = N/A
Fb =
1,350
psi F = FbE / Fb* = N/A
Fv =
170
psi Fe = 1,350 psi
E' =
1,600
ksi F, = 170 psi
CL CF Cv CD Cr
1.00 1.00 1.00 1.00 1.00
Fb = 1350 psi
< F, [Satisfactory]
d (L, Max) = 0.01 in, at 2.500 ft from left end, <
d (Ka D . L. Ma* = 0.04 in, at 2.500 ft from left end <
Where Ka = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.51), mu) = 0.04 in, at 2.500 It from left end
[Satisfactory]
dL=L/360
dKaD.L=L/240
[Satisfactory]
[Satisfactory]
•
•
0
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
AXIAL LOAD F = 0` kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F.' = F. Co CP CF = 1414 psi
Where F� = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 =
F-' = Fc Co CF = 1480 Psi
La = K8 L = 1.0 L = 60 in
d = 7.5 in
SF = slenderness ratio = 8.0 <
FcE = 0.822 E'„d„ / SF2 = 7449 psi
E'min = 580 ksi
F = Fee / Fc' = 5.033
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < F.'
ALLOWABLE FLEXURAL STRESS 1S
Fti = 2160 psi, [ for Co = 1.6 J
T
1 1
F F
0.955
4
50 [Satisfies NDS 2005 Sec. 3.7.1.4)
[Satisfactory]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 617 psi < Fti [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(f. / FC' )Z + fb / (Fb, (1 - f. / F..), = 0.286 < f (Satisfactorvl
'UT DATA & DESIGN SUMMARY
MBER SIZE
MBER SPAN
IFORMLY DISTRIBUTED DEAD LOAD
IFORMLY DISTRIBUTED LIVE LOAD
NCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
GLB 5 4/8 x 131/2
RA
PROJECT
PAGE RES
PAGE:
lbs / ft
STRUCTURAL
CLIENT
JOB NO.
SEAM #23 HDR AT ENTRY TOWER
DATE , �?L�7/
DESIGN BY:
REVIEW BY:
'UT DATA & DESIGN SUMMARY
MBER SIZE
MBER SPAN
IFORMLY DISTRIBUTED DEAD LOAD
IFORMLY DISTRIBUTED LIVE LOAD
NCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
GLB 5 4/8 x 131/2
Glulam 24F -I ZE
L 14
ft
Wo = _ 470
lbs / ft
WL = 280.`,
lbs ft
Pot = 2500.
lbs
9:5
ft
PD2 = ; ' p
Itis
L2
ft
A L = L /:360,
1.25
d 1c« o . L = L / •240;'
5
Does member have continuous lateral support by top diaphragm ?
(1= yes,
0= no)
1 Yes
Code
Duration
Factor. Co Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
2 Occupancy Live Load
_YSIS
ERMINE REACTIONS, MOMENT, SHEAR
L
Po,
LZ
,r
# P.z
WL
Wp
CHECK BENDING AND SHEAR CAPACITIES
Rwo„ =
7.06
kips
•
Vm. =
6.20
kips, at 13.5 inch from right end
MMS, =
24.82
ft4dps, at 8.00 ft from left end
Camber => 0.50 inch
THE BEAM DESIGN IS ADEQUATE.
CD CM Cr
WSeMm =
16
lbs / It RLC„ = 6.17 kips
CHECK BENDING AND SHEAR CAPACITIES
Rwo„ =
7.06
kips
•
Vm. =
6.20
kips, at 13.5 inch from right end
MMS, =
24.82
ft4dps, at 8.00 ft from left end
Where Ke = 1.00
DETERMINE SECTION
DETERMINE CAMBER AT 1.5 (DEAD
PROPERTIES& ALLOWABLE STRESSES
A (1.50. Me ) = 0.50
in, at 7.250 ft from left end
b =
5.13
in E'mi„ = N/A
E = E,=
1800
ksi
Fb = N/A
d =
13.50
in FbE = N/A
Fb =
2,400
psi
F = FbE / Fb* = N/A
A =
69.2
int 1 = 1,051 in
Fv =
265
psi
Fe _ - 2,400 psi
Sx =
155.7
in RB = NIA
E' =
1,800
ksi
F, = 265 psi
lE =
N/A
CD CM Cr
Ci CL CF
1.00 1.00 1.00
1.00 1.00 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 1914
psi < Fb =
f,; = 1.5 VMax / A =
134 psi <
CHECK DEFLECTIONS
d (L, Max) = 0.13
in, at 7.000 ft from left end,
d rue. D . L, M.) = 0.46
in, at 7.250 ft from left end
Where Ke = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD
+ SELF WEIGHT)
A (1.50. Me ) = 0.50
in, at 7.250 ft from left end
C]
Cv Cr Cr
1.00 1.00 1.00
2400 psi
F,; [Satisfactory]
[Satisfactory]
d L = L/ 360 [Satisfactory]
d Rv D . L = L / 240 [Satisfactory]
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
'AXIAL LOAD F = :r,.;.:0,; ?: kips
• THE ALLOWABLE COMPRESSIVE STRESS IS
F-' = Fc CD CP CF = 2347 psi
Where F. = 16W psi
F F
CD = 1.50 --�
CF = 1.00 (Lumber only) ` 1
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.5 = 0.917 T T
Fj = F. CO CF = 2560 psi
Le = Ke L = 1.01- = 168 in
d = 13.5 in
SF = slenderness ratio = 12.4 < 50 [Satisfies NDS 2005 Sec. 3.7.1.4]
FeE = 0.822 E'ntn / SF = 4936 psi
E'min = 930 ksi
F = FoE / F.' = 1.928
C = 0.9
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < F.' [Satisfactory]
ALLOWABLE FLEXURAL STRESS 1S
Fe = 3840 psi, [ for CD = 1.6 ]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1914 psi < F; [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.21
(f. / FC' )Z + fb / (Fb' (1 - f. / F.E)1 = 0.498 < 1 (Satisfactory]
•
•
•
•
•
RA PROJECT 06ge-Res . PAGE
Structural CLIENT beam #24 hdrnghtofk,tchen DESIGN BY:
JOB NO. DATE. REVIEW BY:
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
PD1
Duration Factor. C., Condition
PD2
6 x%6
Dead Load
No. 1, Douglas Fr -Larch
L
6 25 ft
Wo =
:275:.
lbs / ft
WL =;
180-
lbs/ft
Por =
, .0':
: lbs
6
,5
ft
PD2 =
' 0:
: lbs
L2=
A
ft
PD1
Duration Factor. C., Condition
PD2
wL
Dead Load
2
wDI
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
Id L = L / 3W:; Camber => 0.12 inch
d KaD.L = L/'240
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 0 No
Code
Duration Factor. C., Condition
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 'Construction Load
',LYSIS
6
-ERMINE REACTIONS, MOMENT, SHEAR
Wseffwt =
7
lbs / ft RLee = 1.44 kips
V,,,,,, =
1.23
Idps, at 5.5 inch from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.50
in E'min = 580
ksi
d =
5.50
in FbE = 27589
Psi
A =
30.3
int 1 = 76
in
S, =
27.7
in' RB = 5.023
< 50
l E =
11.6
(ft, Tab 3.3.3 footnote 1)
in, at 3.125 ft from left end
CD CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 1.00 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / S. = 975
psi < Fb =
fv' = 1.5 VMax / A =
61 psi <
CHECK DEFLECTIONS
No. 1, Douglas Fir -Larch
A (L, W4 = 0.05
in, at 3.125 ft from left end,
d (Kcr D + L. Max) = 0.13
in, at 3.125 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
Id (1.50, Maxi = 0.12
in, at 3.125 ft from left end
THE BEAM DESIGN IS ADEQUATE.
Cv Cc Cr
1.00 1.00 1.00
1682 psi
F,; [Satisfactory]
[Satisfactory]
AL = L/ 360 [Satisfactory]
d Kao . L = Ll 240 [Satisfactory]
300
Code
Desianation
1
Select Structural, Douglas Fir -Larch
2
No. 1, Douglas Fir -Larch
3
No. 2, Douglas Fir -larch
4
Select Structural, Southem Pine
5
No. 1, Southern Pine
6
No. 2, Southern Pine
Choice
=> 2
Ra;ot =
1.44 kips
MM„ =
2.25
ft -kips, at 3.13 ft from left end
E = E, =
1600
ksi Fb = 1687.5
psi
Fb =
1,350
psi F = FbE / Fb =
16.35
Fv =
170
psi Fb' = 1,682
psi
F =
1,600
ksi F� = 213
psi
Cv Cc Cr
1.00 1.00 1.00
1682 psi
F,; [Satisfactory]
[Satisfactory]
AL = L/ 360 [Satisfactory]
d Kao . L = Ll 240 [Satisfactory]
300
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
AXIAL LOAD F = `_`0"•':: kips
• THE ALLOWABLE COMPRESSIVE STRESS IS
F� = F� Co CP CF = 1245 psi
Where F, = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
Cp = (1+F) / 2c - [(1+F) / 2c)' - F / c]o.s =
F,' = F� Co CF = 1480 psi
Le = Ke L = 1.0 L = 75 in
b = 5.5 in
SF = slenderness ratio = 13.6 <
FSE = 0.822 E'm„ / SFZ = 2564 psi
E'min = 580 ksi
F = F.E / F.' = 1.732
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
% = F / A = 0 psi < Fc'
: ALLOWABLE FLEXURAL STRESS IS
Fe' = 2153 psi, [ for Co = 1.6 ]
� 1
F F
T
0.841
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 975 psi < Fe [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(f. / F�, )Z + fb / [FD (1 - f. / FcE)] = 0.453 < 1 [Satisfactory]
(31
•
RA PROJECT Page3Res PAGE
Structural JOB
beam#25 ridge beam:a4latcheri20 DESIGN BY
JOS NO. DATE ;• REVIEW BY
Woad --Beam Design Base-,oti NDS 200.5
INPUT DATA & DESIGN SUMMARY
MEMBER SIZE
MEMBER SPAN
UNIFORMLY DISTRIBUTED DEAD LOAD
UNIFORMLY DISTRIBUTED LIVE LOAD
CONCENTRATED DEAD LOADS
(0 for no concentrated load)
DEFLECTION LIMIT OF LIVE LOAD
DEFLECTION LIMIT OF LONG-TERM
.GLB 5:1/8 x 18•.: `
`; Glulam 24F -1.8E
L = : 20:5.,
ft
wo = : 386.',
lbs / ft
WL = 2fIO
lbs / ft
Por = 0.:
lbs
L, = 0:.
ft
PD2 = : O. "
lbs
L2 = . '-0 ;
. ft
A L = L / ,360,
5
dKaD.L=L/24t)
Wind/Earthquake Load
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no)
0 No
Code
Duration Factor, C Coen
1
0.90
Dead Load
2
1.00
Occupancy Live Load
3
1.15
Snow Load
4
1.25
Construction Load
5
1.60
Wind/Earthquake Load
6
2.00
Impact Load
Choice
=>
4 Construction Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
WseKHn =
22
lbs / ft RL�ft = 6.99 kips
VMa■ =
5.97
kips, at 18 inch from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b =
5.13
in
E'miD =
930
ksi
d =
18.00
in
FbE =
3579
psi
A =
92.3
int
I =
2,491
ins
S,r =
276.8
in3
RB =
17.658
< 50
LE =
37.9
(ft, Tab 3.3.3 footnote 1)
L,
RR44 =
6.99 kips
Mtmr =
L
ft-Idps, at 10.25 ft from left end
PD,
1800
P.
11
W`
2,400
psi F = FbE / Fb. = 1.19
wD
265
psi Fb' = 2,633 psi
E' =
1,800
ksi F,; = 331 psi
Camber => 0.53 inch
THE BEAM DESIGN IS ADEQUATE.
CD CM Ct Ci CL CF Cv C,: Cr
1.25 1.00 1.00 1.00 0.88 1.00 0.96 1.00 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMax / SX = 1553 psi < Fb = 2633 psi [Satisfactory]
f, = 1.5 VMax / A = 97 psi < F„ [Satisfactory]
ECK DEFLECTIONS
d (L, MU) = 0.25 in, at 10.250 ft from left end, <
d (Ka D . L. WW* = 0.60 in, at 10.250 ft from left end <
Where Ka = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
Id 0.5D. Wax) = 0.53 in, at 10.250 ft from left end
A L = L / 360 [Satisfactory]
A r, D+L = L / 240 [Satisfactory]
RR44 =
6.99 kips
Mtmr =
35.83
ft-Idps, at 10.25 ft from left end
E = Ex =
1800
ksi Fb = 3000 psi
Fb =
2,400
psi F = FbE / Fb. = 1.19
Fv =
265
psi Fb' = 2,633 psi
E' =
1,800
ksi F,; = 331 psi
CD CM Ct Ci CL CF Cv C,: Cr
1.25 1.00 1.00 1.00 0.88 1.00 0.96 1.00 1.00
ECK BENDING AND SHEAR CAPACITIES
fb = MMax / SX = 1553 psi < Fb = 2633 psi [Satisfactory]
f, = 1.5 VMax / A = 97 psi < F„ [Satisfactory]
ECK DEFLECTIONS
d (L, MU) = 0.25 in, at 10.250 ft from left end, <
d (Ka D . L. WW* = 0.60 in, at 10.250 ft from left end <
Where Ka = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
Id 0.5D. Wax) = 0.53 in, at 10.250 ft from left end
A L = L / 360 [Satisfactory]
A r, D+L = L / 240 [Satisfactory]
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
• AXIAL LOAD F = 0:_ii:> kips
• THE ALLOWABLE COMPRESSIVE STRESS IS
Fc = Fc CD CP CF = 292 psi
Where F. = 1600 psi
CO = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.5 =
F. = F. CD CF = 2560 psi
Le = Ke L = 1.0 L = 246 in
b = 5.125 in
SF = slenderness ratio = 48.0 <
FIE = 0.822 E',- / SF = 296 psi
E'min = 830 ksi
F = F.E / F.* = 0.116
C = 0.9
THE ACTUAL COMPRESSIVE STRESS IS
fc = F / A = 0 psi < F.'
E ALLOWABLE FLEXURAL STRESS IS
Fp = 3371 psi, [ for CD = 1.6 ]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1553 psi < F;
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fo / FC' )2 + fb / [Fe (1 - f. / FcE)l = 0.461
•
0 -
i �
F ]j IF
0.114 T
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
[Satisfactory]
< f [Satisfactory]
•
•
C7
RA PROJECT PB" Res a PAGE
Structural CLIENT beam #26 ntige beam,at outdoorpatto DESIGN BY: r s
J08 NO. �: << �: DATE REVIEW BY:
INPUT DATA & DESIGN SUMMARY
Ci CL CF
MEMBER SIZE
; 1 ' 5;1/8 x 15;. ' Glulam 24F -1.8E
MEMBER SPAN
L = `,x.:19;° ft
UNIFORMLY DISTRIBUTED DEAD LOAD
w =
o -,;.'330 � lbs / ft
UNIFORMLY DISTRIBUTED LIVE LOAD
WL= 0; 240 lbs / ft
CONCENTRATED DEAD LOADS
P t' 'i:;. :`
iii -U lbs
(0 for no concentrated load)
L, _ ;0;•`" f ft
in, at 9.500 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
L2 = ;,0: ^ ft
DEFLECTION LIMIT OF LIVE LOAD
AL = L/ 360" .=
DEFLECTION LIMIT OF LONG-TERM
AKaD~L=L/''240`'',:"
L
Ci CL CF
THE BEAM DESIGN IS ADEQUATE.
Does member have continuous lateral support by top diaphragm ?
L
fb = MMD„ / S. = 1658
Po, j
f,'= 1.5 VMa„ / A =
1 P.
1
wL
d (L• Max) = 0.27
WD
Code Duration Factor. C, Condition
in, at 9.500 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
1 0.90 Dead Load
in, at 9.500 ft from left end
Camber => 0.59 inch
CD CM Ct
Ci CL CF
THE BEAM DESIGN IS ADEQUATE.
Does member have continuous lateral support by top diaphragm ?
CHECK BENDING AND SHEAR CAPACITIES
fb = MMD„ / S. = 1658
psi < Fb =
f,'= 1.5 VMa„ / A =
(1= yes, 0= no) 0 No
CHECK DEFLECTIONS
d (L• Max) = 0.27
in, at 9.500 ft from left end,
Code Duration Factor. C, Condition
in, at 9.500 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
1 0.90 Dead Load
in, at 9.500 ft from left end
2 1.00 Occupancy Live Load
3 1.15 Snow Load
4 1.25 Construction Load
5 1.60 Wind/Earthquake Load
6 2.00 Impact Load
Choice => 4 Construction Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
wS.ffw = 18 lbs / ft RL�e = 5.59 kips
R" =
5.59 kips
Vmr = 4.85 kips, at 15 inch from left end
Mme, =
26.55
ft -kips, at 9.50 ft from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b = 5.13 in E'min = 930 ksi
E= E.=
1800
ksi Fb = 3000
psi
d = 15.00 in FbE = 4690 psi
Fb =
2,400
psi F = FbE / Fb' =
1.56
A = 76.9 int 1 = 1,441 in°
F,,. =
265
psi Fe = 2,794
psi
S,, = 192.2 in3 RB = 15.425 < 50
E' =
1,800
ksi F, = 331
psi
l E = 34.7 (ft, Tab 3.3.3 footnote 1)
CD CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 0.93 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMD„ / S. = 1658
psi < Fb =
f,'= 1.5 VMa„ / A =
95 psi <
CHECK DEFLECTIONS
d (L• Max) = 0.27
in, at 9.500 ft from left end,
A (Ka D . L, MO = 0.66
in, at 9.500 ft from left end
Where Ka = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.50. tax) = 0.59
in, at 9.500 ft from left end
D
Cv C'. C,
0.99 1.00 1.00
2794 psi .
F� . ' (Satisfactory]
(Satisfactory]
d L = L/ 360 [Satisfactory]
A Ka D. L = L/ 240 [Satisfactory]
r
•
CHECK THE BEAM CAPACITY WITH AXIAL LOAD
'AXIAL LOAD F = `::Os:. ', kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F.' = F. Co CP CF = 340 psi
Where F, = 1600 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / cf.5 =
F. = F. Co CF = 2560 psi
Le = K, L = 1.01- = 228 in
b = 5.125 in
SF = slenderness ratio = 44.5 <
FIE = 0.822C,,„/S172 = 345 psi
E'min = 830 ksi
F = FcE / Fc' = 0.135
C = 0.9
THE ACTUAL COMPRESSIVE STRESS IS
f� = F / A = 0 psi < Fc'
ALLOWABLE FLEXURAL STRESS IS
Fe' = 3576 psi, [ for CD = 1.6 J
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1658 psi < F,
:CK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fc/Fc )2+fb/[Fb (1-f./F.01 = 0.463
a
i i
F F
0.133
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
[Satisfactory]
< i [Satisfactory]
•
•
•
RA
Camber => 0.12 inch
DEFLECTION LIMIT OF LONG-TERM L / 246:„>•;'
PROJECT
PageRes °� r, L
�k
PAGE
THE BEAM DESIGN IS ADEQUATE.
CLIENT
beam #27 tidr at front of kttchen
DESIGN BY:
(1= yes, 0= no) 0 No
Structural JOB NO:
;
x ,:::-.,� _ :.',•> ,.> DATE s ..';::: .�...,
REVIEW BY:
Code
INPUT DATA & DESIGN SUMMARY '
MEMBER SIZE "_ 6 z 12 7 No. 1, Douglas Fir-Larch
�d`
MEMBER SPAN L = + 8 25.`'* ft 'jI�
UNIFORMLY DISTRIBUTED DEAD LOAD wo ,. ,.60;, lbs / ft P., j + P.
UNIFORMLY DISTRIBUTED LIVE LOAD WL = r>;=`0 v lbs / ft W`
CONCENTRATED DEAD LOADS PD, _ 'r.:5600S lbs "o
y.
(0 for no concentrated load) L, _ It
P02 t) lbs
L2 ft
DEFLECTION LIMIT OF LIVE LOAD' d L = L /.360 ,:_,
Camber => 0.12 inch
DEFLECTION LIMIT OF LONG-TERM L / 246:„>•;'
b = 5.50 in E'min = 580 ksi
E = E,=
THE BEAM DESIGN IS ADEQUATE.
Does member have continuous lateral support by top diaphragm ?
1687.5 psi
d = 11.50 in FbE = 9347 psi
(1= yes, 0= no) 0 No
1,350
psi F = FbE / Fe = 5.54
Code Duration Factor, Co Condition
Code
Designation
1 0.90 Dead Load
1
Select Structural, Douglas Fit -Larch
2 1.00- Occupancy Live Load
2
No. 1, Douglas Fir -Larch
3 1.15 Snow Load
3
No. 2, Douglas Fr -Larch
4 1.25 Construction Load
4
Select Structural, Southern Pine
5 1.60 Wind/Earthquake Load
5
No. 1, Southern Pine
6 2.00 Impact Load
6
No. 2, Southern Pine
Choice => - 4 Construction Load
Choice
=> 2
ANALYSIS
CHECK BENDING AND SHEAR CAPACITIES
DETERMINE REACTIONS, MOMENT, SHEAR -
wseavt = 14 lbs / ft R,e, = 1.83 kips
Ra;a„ =
4.38 kips
VM,z = 4.31 kips, at 11.5 inch from right end
M = 9.66
ft -kips, at 6.00 ft from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b = 5.50 in E'min = 580 ksi
E = E,=
1600
ksi Fb =
1687.5 psi
d = 11.50 in FbE = 9347 psi
Fb =
1,350
psi F = FbE / Fe = 5.54
A = 63.3 int I = • 697 in°
F„ _
170
psi Fe _ -
1,669 psi
s
SK = 121.2 in RB = 8.629 < 50
E' =
1,600
ksi F,; =
213 psi
/E = 16.3 (ft, Tab 3.3.3 footnote 1)
CD CM Ct Ci CL CF Cv
Cr
Cr
1.25 1.00 1.00 1.00 0.99 1.00 1.00
1.00
1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / SK = 956 psi < Fb = 1669
psi
[Satisfactory]
f,'= 1.5 VMax / A = 102 psi < F,
[Satisfactory]
CHECK DEFLECTIONS
d (L, Max) = 0.00 in, at 4.125 It from left end,
<
d L = L / 360
[Satisfactory]
d (Ka D + L , Max) = 0.08 in, at 4.500 ft from left end
<
It Ko D « L = L 1240
[Satisfactory]
Where Kp = 1.00 , (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
A (1.50, Mu) = . 0.12 - in, at 4.500 ft from left end
-
•
•
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
'AXIAL LOAD F = ::;0 ';` kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F� = F. Co CP CF = 1020 psi
Where F. = 925 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]o.s =
Fc' = Fc CD CF = 1480 Psi
Le = Ke L = 1.0 L = 99 in
b = 5.5 in
SF = slenderness ratio = 18.0 <
FSE = 0.822 E'„i„ / SF2 = 1471 psi
E'min = 580 ksi
F = FSE / F,' = 0.994
C = 0.8
THE ACTUAL COMPRESSIVE STRESS IS
f, = F / A = 0 psi < Fc'
E ALLOWABLE FLEXURAL STRESS IS
Fe = 2137 psi, [ for CD = 1.6 j
i i
F F
0.689
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 9% psi < F; [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
(fc / F., )2 + fb / [Fp (i - fe / FCE)] = 0.448 < 1 [Satisfactory]
•
•
•
-- RA /�
PROJECT
Page;Res -. `
PAGE
Structural
CLIENT
beam ff2a hdr left of[outdoa..
DESIGN BY:
MEMBER SPAN
JOB NO.
.: 7..,.: DATE
REVIEW BY: '
INPUT DATA & DESIGN SUMMARY
Ci CL CF
1.25 1.00 1.00
1.00 0.97 1.00
MEMBER SIZE
GL'B 5 1/ISx 12: ; :.
`' Glulam 24F -1.8E
f,; = 1.5 VM. / A =
MEMBER SPAN
L = 15:25
ft
d (L. M.W = 0.15
UNIFORMLY DISTRIBUTED DEAD LOAD
wD = 290.
lbs / ft
Po, j + PD1
UNIFORMLY DISTRIBUTED LIVE LOAD
WL = 160.::-
lbs / ft
W`
CONCENTRATED DEAD LOADS
PD1 = : 0
lbs
-D
(0 for no concentrated load)
Lt = 0 . ,
ft
PD2 = 0:
lbs
L2= 0,,
ft
DEFLECTION LIMIT OF LIVE LOAD
d L = L / 360
Camber => 0.42 inch
DEFLECTION LIMIT OF LONG-TERM
d Ka D. L = L / 240
6 2.00 Impact Load
THE BEAM DESIGN IS ADEQUATE.
Does member have continuous lateral support by top diaphragm ?
(1= yes, 0= no) 0 No
Ci CL CF
1.25 1.00 1.00
1.00 0.97 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 1318
Code Duration Factor. Co Condition
f,; = 1.5 VM. / A =
75 psi <
CHECK DEFLECTIONS
d (L. M.W = 0.15
1 0.90 Dead Load
d (Kar D . L. M.0 = 0.43
in, at 7.625 ft from left end
Where K= = 1.00
, (NDS 3.5.2)
2 1.00 Occupancy Live Load
3 1.15 Snow Load
4 1.25 Construction Load
5 1.60 Wind/Earthquake Load
6 2.00 Impact Load
Choice => 4 Construction Load
ANALYSIS
DETERMINE REACTIONS, MOMENT, SHEAR
wsdrve = 15 lbs / ft RLm = 3.54 kips
Re;g =
3.54
kips
Vm. = 3.08 kips, at 12 inch from left end
Mm =
13.51
ft4dps, at 7.63 ft from left end
DETERMINE SECTION PROPERTIES& ALLOWABLE STRESSES
b = 5.13 in E'miD = 930 ksi
E = Ex=
1800
ksi
Fb = 3000
psi
d = 12.00 in FbE = 7307 psi
Fb =
2,400
psi
F = FbE / Fb* =
2.44
A = 61.5 int 1 = 738 in
Fv =
265
psi
Fb' = 2,904
psi
3
Sx = 123.0 in RB = 12.358 < 50
E' =
1,800
ksi
F� = 331
psi
/E = 27.9 (ft, Tab 3.3.3 footnote 1)
CD CM Ct
Ci CL CF
1.25 1.00 1.00
1.00 0.97 1.00
CHECK BENDING AND SHEAR CAPACITIES
fb = MMax / Sx = 1318
psi < Fb =
f,; = 1.5 VM. / A =
75 psi <
CHECK DEFLECTIONS
d (L. M.W = 0.15
in, at 7.625 ft from left end,
d (Kar D . L. M.0 = 0.43
in, at 7.625 ft from left end
Where K= = 1.00
, (NDS 3.5.2)
DETERMINE CAMBER AT 1.5 (DEAD + SELF WEIGHT)
d n.so. F&x) = 0.42 in, at 7.625 ft from left end
Cv Cr Cr
1.00 1.00 1.00
2904 psi
F,; [Satisfactory]
[Satisfactory]
d L = L/ 360 [Satisfactory]
d Kar D . L - L / 240 [Satisfactory]
f 3f)
0
•
Cl
CHECK THE BEAM CAPACITY WITH AXIAL LOAC
AXIAL LOAD F = 0..: ; kips
THE ALLOWABLE COMPRESSIVE STRESS IS
F.' = F� Co CP CF = 522 psi
Where F. = 1600 psi
Co = 1.60
CF = 1.00 (Lumber only)
CP = (1+F) / 2c - [(1+F) / 2c)2 - F / c]0.e =
F. = F. Co CF = 2560 psi
1, = Ka L = 1.01- = 183 in
b = 5.125 in
SF = slenderness ratio = 35.7 <
FCE = 0.822 E',,;,, / SFZ = 535 psi
E'min = 830 ksi
F = FcE / Fc' = 0.209
C = 0.9
THE ACTUAL COMPRESSIVE STRESS IS
f, = F / A = 0 psi < Fc'
ALLOWABLE FLEXURAL STRESS 1S
Fe = 3717 psi, [ for CD = 1.6 ]
� 1
F F
0.204
50 [Satisfies NDS 2005 Sec. 3.7.1.4]
[Satisfactory]
E ACTUAL FLEXURAL STRESS IS
fb = (M + Fe) / S = 1318 psi < Fti [Satisfactory]
ECK COMBINED STRESS [NDS 2005 Sec. 3.9.2]
qC / F., )2 + fb / [Fti (1 - f. /'F.E)] = 0.335 < 1 [Satisfactory]
-...............
.. J ...........
-
0
DESIGN CRUERIA-SEISMIC .LOADING
LOCATION: LA QUINTA, CA (ZIPCODE: 92253)
Ss =1.515 S, = 0.600 SDs =1.010 SBI = 0.600
SEISMIC DESIGN CATEGORY D (IBC SECTION 1613.5.6)
SITE CLASS D (ASSUMED) (IBC SECTION 1613.5.2)
OCCUPANCY CATEGORY H (IBC TABLE 16045)
SEISMIC AVORTANCE FACTOR 1 1.00 (ASCE 7-05, TABLE11.5-1)
RESPONSE MODIFICATION COEFFICIENT R =6.5 (ASCE 7-05, TABLE 12.2-1)
ATHARDYFRAA& PANEL R = 6.5 w (BF CATALOG 6-08)
AT CAMMEVER STEEL- COL UMV R = L5 j (ASCE 7-05, TABLE 122-1)
REDUNDANCY FACTOR p =1.30 (ASCE 7-05, SECT. '12.3.4.2)
BASE.SBEAR V = p x'[SwV(1.4 x R)] x W i (ASCE 7-05, EQU. 12.8-1 & 12.8-2)
BASE SHEAR V =1.30 x [1-01.0 x 1.0/(1-4 x 6.5)] = 0.145W
DESIGN =WIND LOADING
ASSUMED MOST CONSERVATIVE ROOF ANGLE2:00
• ps3o =17.8 PSF (ASCE 7-05, FIGURE 6-2)
WIND E)CPOSURE"CATEGORY C ----(ASCF-1-,05S;ECT. 6.5.6.3)
WIND HAPORTANCE FACTOR I= 1.00 (ASCE 7-05, TABLE 6-1)
BASIC WUMD SPEED 90 MPH (ASCE 7-05, FIGURE 6-1)
TOPOGRAPHIC FACTOR Kzr 1.00 (ASCE 7-05, SECT. 6.5-7.2)
VVEND FACTOR pi = X x Kir x I x ps3o (ASCE 7-05; EQU. 6-1)
11 21.54 psf
MAM MUM HEIGHT 15 ft {1L 1-21) WIM FACTOR ,!)9
MA,!MvfUMEEIGHT.20ft:::(?L=1-29)VMWFACTOR 1
.pli = 22.97 psf
Nf A3CDvlU- M HEIGHT 25 ft ::(). =1.35) ViM FACTOR pg = 24.03 psf
MA)CMUM HEIGHT 30 ft (X =1.40) VAM FACTOR p,§ = 2492 psf
9 LATERAL ANALYSIS
SECTION 1 LONGITUDINAL
ROOF AVERAGE HEIGHT = 20.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT
20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF
WIND LOAD = 22.97 x ( 20.00 - 10.00/2 ) = 345.. 00 PLF
SEISMIC LOAD = 0.145 x ( 25.00x66.00 + 15.00x2x5 + I O.00x2x5) = 276.00 PLF
WIND GOVERNS = 345.00 PLF
MAX. SHEAR = 345.00 x 28.00 / 2 x 66.00 = 74.00 PLF
CHORD FORCE = 345.00 x 28.00 x 28.00 / 8 x 65.00 = 513.00 LBS
USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C
USE (6) 16D' S PER TOP PLATE SPLICE
SECTION 2 LONGITUDINAL
ROOF AVERAGE HEIGHT = 15.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT
15 FT. MAX ROOF HEIGHT WIND FORCE = 21.54 PSF
WIND LOAD = 21.54 x ( 15.00 - 10.00/2 ) = 216.00 PLF
SEISMIC LOAD = 0.145 x ( 25.00x40.00 + 15.00x2x5 + I0.00xlx5) = 174.00 PLF
SEISMIC GOVERNS = 216.00 PLF
MAX. SHEAR = 216.00 x 28.00 / 2 x 40.00 = 76.00 PLF
CHORD FORCE = 216.00 x 28.00 x 28.00 / 8 x 40.00 = 530.00 LBS
USE/2" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C
• USE (6) 16D' S PER TOP PLATE SPLICE
• LATERAL ANALYSIS
SECTION 1 TRANSVERSE
ROOF AVERAGE HEIGHT = 17.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT
20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF
WIND LOAD = 22.97 x ( 17.00 - 10.00/2 ) = 276.00 PLF
SEISMIC LOAD = 0.145 x ( 25.00x110.00 + 15.00x2x5 + 10.00x6x5) = 464.00 PLF
SEISMIC GOVERNS = 464.00 PLF
MAX. SHEAR = 464.00 x 28.00 / 2 x 110.00 = 59.00 PLF
CHORD FORCE = 464.00 x 28.00 x 28.00 / 8 x 110.00 = 414.00 LBS
USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C
USE (6) 16D'S PER TOP PLATE SPLICE
SECTION 2 TRANSVERSE
ROOF AVERAGE HEIGHT = 17.00 FT TOP PLATE AVERAGE HEIGHT = 10.00 FT
20 FT. MAX ROOF HEIGHT WIND FORCE = 22.97 PSF
WIND LOAD = 22.97 x ( 17.00 - 10.00/2 ) = 276.00 PLF
SEISMIC LOAD = 0.145 x ( 25.0003.00 + 15.00x2x5) = 118.00 PLF
WIND GOVERNS = 276.00 PLF
MAX. SHEAR = 276.00 x 20.00 / 2 x 33.00 = 84.00 PLF
CHORD FORCE = 276.00 x 20.00 x 20.00 / 8 x 33.00 = 419.00 LBS
• USE '/z" CDX UNBLOCKED WITH 8D' S AT 6", 12" O/C
USE (6) 16D'S PER TOP PLATE SPLICE
•
0
/4?,-pv/54 (glZg/11)
• SHEAR WALL DESIGN
SW #I LEFT ELEVATION
TOTAL LOAD = 345.00 x 20.00/2 = 3450.00 LBS.
SHEAR WALL LENGTH = 4.00 + 11.00 (7.00 AROUND WINDOWS) = 11.00 FT.
SHEAR WALL = 3450.00 / 11.00 = 314.00 PLF
USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0
MAX. DRAG = 1088.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 2948.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #2 RIGHT OF M. BDRM AND OFFICE
TOTAL LOAD = 345.00 x 35.00/2 = 6038.00 LBS.
SHEAR WALL LENGTH = 7.00 + 7.00 = 14.00 FT.
SHEAR WALL = 6038.00 / 14.00 = 431.00 PLF
USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0
MAX. DRAG = 3782.00 LBS. USE (28) 16D'S PER TOP PLATE SPLICE OR ST6236
MAX. UPLIFT = 3962.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #3 LEFT OF HER CLOSET
TOTAL LOAD = 345.00 x 35.00/2 = 6038.00 LBS.
SHEAR WALL LENGTH= 10.00 FT.
• SHEAR WALL = 6038.00 / 1.00 = 604.00 PLF
USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0
MAX. DRAG = 3194.00 LBS. USE (28) 16D'S PER TOP PLATE SPLICE OR ST6236
MAX. UPLIFT = 5539.00 LBS USE SIMPSON HDU8 HOLDOWN EACH END
SW #4 LEFT OF BATH 2
TOTAL LOAD = 345.00 x 47.00/2 = 8108.00 LBS.
SHEAR WALL LENGTH = 20.00 FT.
SHEAR WALL = 8108.00 / 20.00 = 405.00 PLF
USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.0
MAX. DRAG = 4065.00 LBS. USE (30) 16D'S PER TOP PLATE SPLICE OR ST6236
MAX. UPLIFT = 3055.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #5 LEFT OF GARAGE
TOTAL LOAD = 345.00 x 55.00/2 = 9488.00 LBS.
SHEAR WALL LENGTH = 16.00 FT.
SHEAR WALL = 9488.00 / 16.00 = 593.00 PLF
USE SHEAR WALL TYPE 13 WITH 5/8" x 12" AB. AT 8" O.0
MAX. DRAG = 3225.00 LBS. USE (24) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 5250.00 LBS USE SIMPSON HDU8 HOLDOWN EACH END
ftl.56 e., b A(. 06'elft 51s7/AA JPU f7= X� U°S
• 4-x4x1,Sa150 = cp otos
C]
•
RA PROJECT: :PAGE RES. — PAGE
CLIENT: SHEAR WALLS 1:. DESIGN BY
JOB NO :
DATE: REVIEW BY:
INPUT DATA
LATERAL FORCE ON DIAPHRAGM: vy. WPM _
: 344
plf,for wind
(SERVICE LOADS) vcftSEISWC=,
.:250:
ptf,forseismic, ASO
DIMENSIONS: L,= 3S, :' ft, L2 =
4.
:ft, L3 = 3.5 ft
H, 3 ft, H2
4.
ft. H3 = ft
KING STUD SECTION 1 pcs, b
4
in, h 8 in
SPECIES (1 = DFL, 2 = SP) Y
DOUGLAS FIR -LARCH
GRADE (1, 2, 3 4 5, or 6) 4
No. 2
EDGE STUD SECTION TPCs. b -
4
4n, h = 's6 in
SPECIES 0 = DFL, 2 = SP) 1
DOUGLAS FIR -LARCH
GRADE (1,2,3,4,5,or(l) 4'.
No.2
PANEL GRADE (0 or 1) = 4 " r_ Sheathing
and Single Floor
MINIMUM NOMINAL PANEL THICKNESS =
4t2' -
in
COMMON NAIL SIZE (0=6d, 1=8d, 2=10d)
2
10d
SPECIFIC GRAVITY OF FRAMING MEMBERS
O:S"
STORY OPTION ( 1=ground level, 2 --upper levet)
t.
ground level shear wall
DESIGN SUMMARY
'%LOCKED 15132 SHEATHING WITH 10d COMMON NAILS
3 in O.C. BOUNDARY 8 ALL EDGES / 12 in O.C. FIELD,
5/8 in DIA. x 10 in LONG ANCHOR BOLTS @ 30 in O.C.
SIR
V dio
T�40� W'��
HOLD-DOWN FORCES: TL = 3.14 k , TR = 3.14 k (USE PHD2--DW SIMPSON HOLD-DOWN)
MAX STRAP FORCE: F = 1.05 k (USE SIMPSON CS18 OVER WALL SHEATHING WITH FLAT BLOCKING)
KING STUD: 1 - 4" x 6' DOUGLAS FIR -LARCH No. 2. CONTINUOUS FULL HEIGHT,
EDGE STUD: 1 - 4"x 8" DOUGLAS FIR -LARCH No. 2, CONTINUOUS FULL HEIGHT.
SHEAR WALL DEFLECTION: e = 0.53 in
V
712
t L2 3
L
TL TR
ASSUME INFLECTION POINT AT MIDDLE OF WINDOW
Lt 1.2/2
F1 I F2
= 1 F4 + 2 F4
I <+ I
F5 F6
F9 F10
F5
5
F5
F13
F5
7
F5
L1 + 0.5 L2
n
o j
1-2/2 L3
F2 I F3
J 3 F4 + 4
F7 + F8
F11 F12
FB
6
FS
F14
F8
8
F8
F15 Fib F17 F18
—F5 - F19 — — F20
---' FS
s 9 F21 1 0 � F21 1 1 F2, I 1 2
F22 F23 F23 1 F24
TL FREE -BODY INDIVID lA PANFIS OF WA TR
M-1
0
0
ANALYSIS
Panel GladeCommon
Nail
Min.
Penetration'dmess
(m)
_ Min.
cul)
Blocked Nail Spacing
Boundary R A8 Edges
CHECK MAX SHEAR WALL DIMENSION RATIO
h / w = 1.1 <
3* .z ;_; ,.
[Satisfactory]
2
Sheathing and Single-Fkwr
DETERMINE FORCES 8 SHEAR STRESS OF FREE -BODY INDIVIDUAL PANELS
OF WALL
15/32
310
460
INDMDLIAL PANEL W (h) H (h) MAX SHEAR STRESS (plp
NO.
FORCE (M
NO.
FORCE (mf)
1 3.50 3.00
194
F1
680
F13
1570
2 200 3.00
523
F2
1047
F14
1570
3 2.00 3.00
523
F3
680
F15
2557
4 3.50 3.00
194
F4
1570
F16
987
5 3.50 2.00
493
F5
1727
F17
987
6 3.50 2.00
493
F6
1047
F18
2557
7 3.50 2.00
493
F7
1047
F19
1047
8 3.50 2.00
493
F8
1727
F20
1047
9 3.50 3.00
194
F9
583
F21
1570
10 200 3.00
523
F10
987
F22
880
11 2.00 3.00
523
F11
987
F23
1047
12 3.50 3.00
194
F12
583
F24
680
DETERMINE REQUIRED CAPACITY vb =
523 pB, ( 1
Side Panel Required, the Max. Nail Spacing =
THE SHFAR C_APA('MF=A PFR
..---....- ......- �,..�,...�.� ..o.� �o...,�.c., uy ormulo Uia my Iautw per n5k. note a.
JE MAX SPACING OF 5V DIA ANCHOR BOLT (NDS 2005, Tab. 11E)
518 in DIA, x 10 in LONG ANCHOR BOLTS @ 30 in O.C.
TNF Nr)l IlMV1ru c^or
3 in)
Panel GladeCommon
Nail
Min.
Penetration'dmess
(m)
_ Min.
cul)
Blocked Nail Spacing
Boundary R A8 Edges
SEISMIC
6
4
3
2
Sheathing and Single-Fkwr
10d
1 518
15/32
310
460
600
770
..---....- ......- �,..�,...�.� ..o.� �o...,�.c., uy ormulo Uia my Iautw per n5k. note a.
JE MAX SPACING OF 5V DIA ANCHOR BOLT (NDS 2005, Tab. 11E)
518 in DIA, x 10 in LONG ANCHOR BOLTS @ 30 in O.C.
TNF Nr)l IlMV1ru c^or
3 in)
KING STUD CAPACITY
van
(P)
Wall Seismic
at mid -story pbs)
Overturning
Moments (ft4bs)
Resisting Safety Net Uplift
Moments (ft4bs) Factors cabs)
Hdddorn
SIMPSON
SEISMIC
250
176
28380
left 0 0.9 TL = 2580
O4,
Right 0 0.9 TR = 2580
WIND
314
EAL„
34540
Left 0 2/3 TL = 3140
ti
Q�
Right 0 2/3 TR = 3140
KING STUD CAPACITY
( I L 6 I R le
values should Include upper vel UPLIFT forces if applicabh
ECK MAXIMUM SHEAR WALL DEFLECTION: (IBC Section 2305.33
0.99
kips
F, =
39?
=A&,dmg+Aslmf+6Nall :to+OCto,d .rpae do=8y
+V
�
+0.75;henxf=
in, ASD
1600
EAL„
Gt
00
ce/hl�0.525
e,n= 0.429 in
b,albwadsD
Where: vi,= 523 plf . ASD L„ =
11 ft
E = 1.7E+06 psi
a I
(ASCE 7-0512.8.E
A = 16.50 in` h =
10 It
G = 9.0E+04 psi
Ca 4 I 1
t= 0.298 in G, =
0.024 in
do = 0.15 in
,(ASCE 7-05 Tab 12.2-1 8 lab 11.51
Cp = 0.38 A = 19.25 in'
E =
1600
A. = 0.02 ho
CF = 1.10
Fc = 894 psi > fc = 163 psi
(ASCE 7-05 Tab 12.12-1)
KING STUD CAPACITY
P. =
0.99
kips
F, =
1350
psi
Co = 1.60
C, = 0.38 A = 19.25 in'
E =
1600
ksi
CF= 1.10
F,'= 894 psi > fc = 51 psi
[Satisfactory)
EDGE STUD CAPACITY
P.=
3.14
kips, (this
value should include upper level DOWNWARD loads H applicable)
F, =
1350
psi
Co = 1.60
Cp = 0.38 A = 19.25 in'
E =
1600
ksi
CF = 1.10
Fc = 894 psi > fc = 163 psi
[Satisfactory]
0
tl
•
•
•
Re ✓ V—o( ( 9 2 q / 11
Reza PROJECT: Shear Wal['#3 J �` PAGE:
As harnnQur CLIENT 'Page Residence DESIGN BY: A.A.
M JOB NO : 11028 fT T DATE# REVIEW BY: ',R.A.
INPUT DATA
WALL LENGTH
WALL HEIGHT
WALL THICKNESS
FOOTING LENGTH
FOOTING WIDTH
FOOTING THICKNESS
FOOTING EMBEDMENT DEPTH
ALLOWABLE SOIL PRESSURE
DEAD LOAD AT TOP WALL
LIVE LOAD AT TOP WALL
TOP LOAD LOCATION
WALL SELF WEIGHT
LATERAL LOAD TYPE (0--wind,1=seismic)
WIND LOADS AT WALL TOP
Lw = ,
1:0
It
h =
10
,ft
t =°
6
in
L =
22
ft
L, =
2
ft z
B =.
2
;ft
T=.
24
in
D =
2
r ft
qa =;
1.5-
ksf D
Pr,DL =
0.1
kips
Pr.LL = r
0.1
kips i— L 1
a -;
2.25
ft
Pw=e
1
akips
0
wind
F = 6 ! kips THE FOOTING DESIGN IS ADEQUATE.
M = 0 f ft -kips
CONCRETE STRENGTH fc. =f 2.5 lksi
REBAR YIELD STRESS fy =} 60 ksi
TOP BARS, LONGITUDINAL 4 # 5,
BOTTOM BARS, LONGITUDINAL 4 3 # 1 5
BOTTOM BARS, TRANSVERSE # 3 @ .12. ' in o.c. < _= Not Required
ANALYSIS
CHECK OVERTURNING FACTOR (IBC 06 1605.2.1, 1801.2.1, & ASCE 7-05 12.13.4)
F=MR/Mo= 2.05 > 1.6/0.9 for wind
Where Pf = 12.76 kips (footing self weight)
Mo = F (h + D) + M = 72 ft -kips (overturning moment)
MR = (Pr.00 (Lt + a) + Pf (0.5 L) + Pw (L, + 0.51-w) = 148
CHECK SOIL CAPACITY (ALLOWABLE STRESS DESIGN)
Ps = 8.8 kips (soil weight in footing size)
P = (Pr.DL + Pr,LL) + Pw + (Pf - Ps) = 5.16 kips (total vertical net load)
MR = (Pr,DL + Pr. LL) (1-1 + a) + Pf (0.5 L) + Pw (L, + 0.51-w) = 148
e = 0.5 L - (MR - Mo) / P = -3.77 It (eccentricity from middle of footing)
P(1+ Le) L
\\ for e < —
9Arnx = BL 6
2P ,for e>L = 0.00 ksf
—
3B(0.5L—e)' 6
Where e = -3.77 It, < (L/6)
CHECK FOOTING CAPACITY (STRENGTH DESIGN)
MU.R = 1.2 [Pr,DL (1-1 + a) + Pf (0.5 L) + Pw (L, + 0.51-w)] + 0.5 Pr. LL(L1 + a) _
M�,o = 1.6 [F(h + D) + M] = 115 ft -kips
Pu = 1.2 (Pf.DL + Pf + Pw) + 0.5 Pr. LL = 17 kips
eu=0.5L-(Mua-Mu,o)/Pu= 7.26 It
P„ 1 + 6e.,
L ) , fore„ < L
BL 6 = 1.49 ksf
2 P„ L
3B(0.5L-e„)• for e„> 6
[Satisfactory]
ft -kips (resisting moment without live load)
ft -kips (resisting moment with live load)
< 4/3qa
[Satisfactory)
178 ft -kips
L
au,rlc c
is
•
0
sw -*3
Location
A4
Mu,max
d (in)
(OregD
PprovD
Vu,max
OV,,= 2 0 b d (f� )o.s
Top Longitudinal
-39 ft -k
20.69
(cont'd)
BENDING MOMENT & SHEAR AT EACH FOOTING SECTION
9 kips
42 kips
Bottom Longitudinal
5 ft -k
20.69
0.0018
Section 0
1/10 L
2/10 L
3110 L
4110 L
5110 L
6/10 L
7/10 L
8/10 L
9/10 L
L
Xu (ft) 0
2.20
4.40
6.60
8.80
11.00
13.20
15.40
17.60
19.80
22.00
Pu,N,(klf) 0.0
6.8
3.7
0.7
-2.4
-5.4
0.0
0.0
0.0
0.0
0.0
Mu,w(ft-k) 0
0
-17
-52
-91
-118
-124
-127
-130
-133
-136
Vu,w (kips) 0
-1
-13
-18
-16
-7
-1
-1
-1
-1
-1
Put(ksf) 0.3
0.3
0.3
0.3
0.3
0.3
0.3
0.3
0.3
0.3
0.3
Mu.f (ft -k) 0
-2
-7
-15
-27
-42
-61
-83
-108
-136
-168
Vu.t(kips) 0
-2
-3
-5
-6
-8
-9
-11
-12
-14
-15
qu (ksf) -1.5
-1.2
-0.9
-0.6
-0.3
0.0
0.0
0.0
0.0
0.0
0.0
Mu,q (ft -k) 0
7
25
52
85
121
158
195
231
268
305
Vu.G (kips) 0
6
11
14
16
17
17
17
17
17
17
:. Mu (ft
-15
-7
-2
0
L Vu (kips)
0
3
-6
-9
-6
2
6
5
3
2
0
10
0
-10 -
-20
-30
-40
® M
-50
10
0
-41111111
WWI-
OV
-10
0.85 1-
j
1-
"
0.383bd2j
Where
P =
` /
P m;� =
0.0018
f
0.85/3, f c Eu
_
Pnaaa•
=
0.0129
[Satisfactory]
fV
eu+Et
Location
A4
Mu,max
d (in)
(OregD
PprovD
Vu,max
OV,,= 2 0 b d (f� )o.s
Top Longitudinal
-39 ft -k
20.69
0.0009
0.0025
9 kips
42 kips
Bottom Longitudinal
5 ft -k
20.69
0.0018
0.0025
9 kips
42 kips
Bottom Transverse
0 ft -k-/ ft
20.19
0.0000
0.0000
0 kips / ft
21 kips / ft
-k)
0
5
1
-15
-33
-39
-27
-15
-7
-2
0
L Vu (kips)
0
3
-6
-9
-6
2
6
5
3
2
0
Location
A4
Mu,max
d (in)
(OregD
PprovD
Vu,max
OV,,= 2 0 b d (f� )o.s
Top Longitudinal
-39 ft -k
20.69
0.0009
0.0025
9 kips
42 kips
Bottom Longitudinal
5 ft -k
20.69
0.0018
0.0025
9 kips
42 kips
Bottom Transverse
0 ft -k-/ ft
20.19
0.0000
0.0000
0 kips / ft
21 kips / ft
Iw PROJECT: SHEAR WALL 5
CLIENT
. JOB NO DATE
Footi n g i ei ignof:Shear, a- 0,0asedolti;A iI31_$-0 5
IPUT DATA
'ALL LENGTH
LW =
16
ft
ALL HEIGHT
h =
40
ft
ALL THICKNESS
t =
•6
in
DOTING LENGTH
L =
50 :
ft
Lr =
17
ft
)OTING WIDTH
B =
1
.ft
)OTING THICKNESS
T =
`' 12,
in
)OTING EMBEDMENT DEPTH
D =
1 `
ft
.LOWABLE SOIL PRESSURE
q8 =
:. 1:5 .
, ksf
:AD LOAD AT TOP WALL
Pr.DL =
; : "0:1t
" : kips
/E LOAD AT TOP WALL
Pr.LL =
0.1
kips
IP LOAD LOCATION
a =
2;25.:.:.;
ft
4LL SELF WEIGHT
PW =
0.15
kips
TERAL LOAD TYPE (0=wind,1=seismic)
t):.
. wind
NO LOADS AT WALL TOP
F =
'6'.
kips
M =
. 0
ft -kips
)NCRETE STRENGTH
fc, =
2.5..:
ksi
BAR YIELD STRESS
fy =
gp
ksi
P BARS, LONGITUDINAL
3
#
TTOM BARS, LONGITUDINAL
`,g.
#
TTOM BARS, TRANSVERSE
#
3--
@
PAGE:
DESIGN BY:
REVIEW BY:
P,.
L
THE FOOTING DESIGN IS ADEQUATE.
5 /�0 f 22 �to w'C 04-k 5-ee
f)n 44 44
.12 in..j,: < == Not Required
ANALYSIS
CHECK OVERTURNING FACTOR (IBC 06 1605.2.1, 1801.2.1, & ASCE 7-05 12.13.4)
•
F= MR/ MO= 2.83 > 1.610.9 for wind
Where Pf = 7.25 tips (footing self weight)
Mo = F (h + D) + M = 66 ft -kips (overturning moment)
MR = (P1,00 (1-1 + a) + Pt (0.5 L) + PW (L, + 0.51-W) = 187
•
SOIL CAPACITY (ALLOWABLE STRESS DESIGN)
Ps = 5 kips (soil weight in footing size)
P = (Pr.DL + Pr.L0 + PW + (Pf - Ps) = 2.60 kips (total vertical net toad)
MR = (Pr,OI + Pr. LL) (LI + a) + Pf (0.5 L) + PW (Lr + 0.51-W) = 189
e = 0.5 L - (MR - Mo) / P = -22.25 ft (eccentricity from middle of footing)
P 1+6e
Ll, for e < L
9AuX = 6
2P for e>L = -0.09 ksf
—
3B(O.5L—e)' 6
Where e = -22.25 ft, < (L / 6)
FOOTING CAPACITY (STRENGTH DESIGN)
[Satisfactory]
ft -kips (resisting moment without live load)
ft -kips (resisting moment with live load)
< 4/3g8
[Satisfactory]
Mu.R = 1.2 [Pr,DL (L+ + a) + Pt (0.5 L) + PW (Lr + 0.5LW)] + 0.5 Pr, LL(Li + a) = 225 ft -kips
Mu.o = 1.6 [F(h + D) + M] = 106 ft -kips
Pu = 1.2 (Pr,DL + Pt + PW) + 0.5 Pr, LL = 9 kips
eu = 0.51-- (Mu,R - Mru,o) / Pu = 11.78 It u ,w
Pu l 1+6eu1 P f ri
111 L) for eu L
OK
9 u,.awx = BL 6 = 0.46 ksf
2pu for eu > L 0
3B(O.5L —e,J ' 6 1
qu,ma x
.f— Xu ��
•
0
tt
Mu,�
d (in)
PregD
PpraaD
Vu,max
OV, = 2 P b d (f�)os
Top Longitudinal
>K1 *57
8.69
0.0063
0.0089
7 kips
9 kips
Bottom Longitudinal
31 ft -k
8.69
0.0087
0.0089
(conrd)
BENDING MOMENT & SHEAR AT EACH FOOTING SECTION
Bo
Section 0
1110 L
2/10 L
3/10 L
4/10 L
5110 L
6/10 L
7/10 L
8/10 L
9/10 L
L
Xu (ft) 0
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
50.00
Pu,w (k� 0.0
0.0
0.0
0.0
1.6
0.0
-1.5
0.0
0.0
0.0
0.0
M,,(ft-k)0
0
0
0
-10
54
-99
-110
-112
-114
-115
Vu,w (SPS) 0
0
0
0
-6-10
-60
0
0
0
Pu.t (kms 0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
Mu,r (ft -k) 0
-2
-9
-20
-35
54
-78
-107
-139
-176
-218
Vu,t (kips) 0
-1
-2
-0
-3
-0
-5
-6
-7
-8
-9
qu (ksf) -0.5
-0.4
-0.3
-0.3
-0.2
-0.2
-0.1
-0.1
0.0
0.0
0.0
Mu,q (ft -k) 0
5
21
45
76
113
154
197
242
288
333
Vu,q (wPs) 0
2
4
6
7
8
9
9
9
9
9
L Mu (ft -k) 0
3
12
25
31
4
-23
-19
-9
40
20
-
0
v,-
a
-2p
OM
-40
5
0
-5 -
oV
-10
0.85 1_F
j
"' -r
`
O.3836dz j'
Where
P =
-
P min -
0.0018
j r
0.85 f� .%� C Eu
PU4.X =
Et,+E/
=
0.0129
(Satisfactory)
.fy
61
Location
Mu,�
d (in)
PregD
PpraaD
Vu,max
OV, = 2 P b d (f�)os
Top Longitudinal
-23 ft -k
8.69
0.0063
0.0089
7 kips
9 kips
Bottom Longitudinal
31 ft -k
8.69
0.0087
0.0089
7 kips
9 kips
Bo
-2
0
£. Vu (kips)
0
1
2
3
3
-7
3
2
2
1
0
Location
Mu,�
d (in)
PregD
PpraaD
Vu,max
OV, = 2 P b d (f�)os
Top Longitudinal
-23 ft -k
8.69
0.0063
0.0089
7 kips
9 kips
Bottom Longitudinal
31 ft -k
8.69
0.0087
0.0089
7 kips
9 kips
Bo
0M1(A)
SHEAR WALL DESIGN
• C� SW #6 LEFT OF KITCHEN Ipc1}'0
TOTAL LOAD = 216. 00 x 28.00/2 = 3 024. 00 LBS. -t-- i 0 2 7 Q b y v 51
SHEAR WALL LENGTH = 6.50 FT.
SHEAR WALL=4o5f.00 / 6.50 =6:13.Q0 PLF
USE SHEAR WALL TYPE li{ WITH 5/8" x 12" A.B. AT 12" O.0 (3 SCS ll)
MAX. DRAG =1956,.00 LBS. USE (16) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT =5065.00 LBS USE SIMPSON FIDUSHOLDOWN EACH END
SW #7 RIGHT OF KITCHEN AND OUTDOOR .PATIO
TOTAL LOAD = 216.00 x 28.00/2 = 3024.00 LBS.
SHEAR WALL LENGTH = 5.00 + 6.00 = 11.00 FT.
SHEAR WALL = 3024.00 / 11.00 = 275.00 PLF
USE SHEAR WALL TYPE 11 WITH 5/8" x 12" A.B. AT 32" O.0
MAX. DRAG = 1442.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 2333.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #8 REAR ELEVATION
TOTAL LOAD = 464.00 x 24.00/2 = 5568.00 LBS.
SHEAR WALL LENGTH = 21.00 + 9.00 + 19.00 + 8.00 + 18.00 = 75.00 FT
SHEAR.WALL = 5568.00 / 75.00 = 75.00 PLF
USE SHEAR WALL TYPE 8 WITH 5/8" x 12" A.B. AT 48" O.0
MAX. DRAG = 558.00 LBS. USE (6) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 318.00 LBS HOLDOWNS NOT REQUIRED
SW #9 AT HALLWAY
TOTAL LOAD = 464.00 x 50.00/2 = 11600.00 LBS.
SHEAR WALL LENGTH = 8.00 + 8.00 + 8.00 = 24.00 FT.
SHEAR WALL = 11600.00 / 24.00 = 484.00 PLF
USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0
MAX. DRAG = 1768.00 LBS. USE (12) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 5015.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #10 REAR OF PATIO AND KITCHEN
TOTAL .LOAD = 464.00 x 43.00/2 x 80/110 = 7255.00 LBS.
SHEAR WALL LENGTH = 5.00 + 8.00 = 13.00 FT.
SHEAR WALL = 7255.00 / 13.00 = 558.00 PLF
USE SHEAR WALL TYPE 13 WITH 5/8" x 12" A.B. AT 8" O.0
MAX. DRAG = 3626.00 LBS. USE (26) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 5136.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
6)71-44 C;' QV &
v
c13� S,O�lbs
4N *- hnofo)0 �F
N -6`T
®-- 4
6
SHEAR WALL DESIGN
•
SW #11 FRONT OF KITCHEN
TOTAL LOAD = 276.00 x 40.00/2 = 5520.00 LBS.
SHEAR WALL LENGTH = 14.00 FT.
SHEAR WALL = 5520.00 / 14.00 = 395.00 PLF
USE SHEAR WALL TYPE 12 WITH 5/8" x 12" A.B. AT 12" O.C.
MAX. DRAG = 2623.00 LBS. USE (20) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 3244.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #12 FONT OF OUTDOOR PATIO
TOTAL LOAD = 276.00 x 20.00/2 = 2760.00 LBS.
SHEAR WALL LENGTH = 12.00 FT.
SHEAR WALL = 2760.00 / 12.00 = 230.00 PLF
USE SHEAR WALL TYPE 10 WITH 5/8" x 12" A.B. AT 48" 0. C.
MAX. DRAG = 1252.00 LBS. USE (10) 16D'S PER TOP PLATE SPLICE
MAX. UPLIFT = 1958.00 LBS USE SIMPSON HDU5 HOLDOWN EACH END
SW #1-_; STEEL COL. AT ENTRY AND PATIO
TOTAL LOAD = (25.00 x 33.00 + 15.00 x 4.00) x 0.145 x 16.00/2 = 102 7. 00
CANTILEVER STEEL COLUMN R= 1.5
LOAD = 1027.00 x 6.5/1.5 = 4900.00 LBS
• SEE STEEL COLUMN DESIN
11
•
•
I- Le.c�. (q/2
Reza PROJECT: S.W. #13Steel Column kt Patio PAGE: .
As har Our CLIENT . !Page Residence _ DESIGN BY: R.A.
JOB NO 11028 DATE : 1.9/28/2011 REVIEW BY: 'R.A.
Cantilever Column & Footina Desian Based oKAISC iso_ Act sla'and iac Isns'z I
INPUT DATA & DESIGN SUMMARY
COLUMN SECTION (Tube, Pipe, or WF)
'HSS84X5/8 . Tube
COLUMN YIELD STRESS
F y = ;
36.
!, ksi
CANTILEVER HEIGHT
H =
10
ft
COLUMN TOP LATERAL LOAD
F =
4.9
kips, ASD
(Strong Axis Bending only)
=
0.62 < 1.0 [Satisfactory]
COLUMN TOP GRAVITY LOAD
P =
5
kips, ASD
DIAMETER OF POLE FOOTING
b =
3
ft
ALLOW SOIL PRESSURE
Qa =
1.5
ksf
LATERAL SOIL CAPACITY
Pp =
VAS
ksf/ft
RESTRAINED @ GRADE ?(1=yes,0=no)
Mry = 0
1
Yes
Use 3 ft dia x 6.14 ft deep footing restrained
@ ground
level
THE DESIGN IS ADEQUATE.
Pc = Pn / f2c =
420 / 1.67 =
251.48 kips, (AISC 360-05 Chapter E)
US,�
ANALYSIS
CHECK COMBINED COMPRESSION AND BENDING
CAPACITY OF COLUMN (AISC 360-05, H1)
Pr+8Mrx+'lL
for
Pr> -0.2
P, 9
Mc, M,y)
P,
=
0.62 < 1.0 [Satisfactory]
Pr +
(A4,
Mrx+ Mr1 for
Pr <0.2
2P�
M,y)
P,
Where Pr = 5.00
kips
Mrx = 49.00
ft -kips
Mry = 0
ft -kips
KL y = 20
ft, weak axis unbraced axial length
Pc = Pn / f2c =
420 / 1.67 =
251.48 kips, (AISC 360-05 Chapter E)
> Pr [Satisfactory]
Mcx = Mn 1.(2b =
134.10 / 1.67 =
80.30 ft -kips, (AISC 360-05 Chapter F)
> Mrx [Satisfactory]
Mcy = Mn 1.0b =
134.10 /1.67 =
80.30 ft -kips, (AISC 360-05 Chapter F)
> Mry [Satisfactory]
DESIGN POLE FOOTING (IBC Sec.1805.7 / UBC Sec.1806.8)
By trials, use pole depth, d = 6.141 ft
Lateral bearing @ bottom, S3 = 2 Pp Min(d , 12') = 1.84 ksf
Lateral bearing @ d/3, S, = 2 Pp Mind/3, 12') = 0.61 ksf
Require Depth is given by 1
A[,+ l+ l+ 4.36h
h J for nonconstrained
d- -
4.25Ph - 6.141 ft [Satisfactory]
for constrained
6S,
Where P= F= 4.90 kips
A= 2.34 P/(b S,) = 4.18
h = Mmax IF= 10.00 ft
CHECK VERTICAL SOIL BEARING CAPACITY (ACI, Sec. 15.2.2)
Cl soil = P I (n b 214) = 0.71 ksf, (net weight of pole footing included.)
CHECK STRONG AXIS LATERAL DEFLECTION < Qa [Satisfactory]
FH .3
3El 0.67 in < 2 H/ 240 = 1.00 in [Satisfactory]
�D(
FOUNDATION DESIGN
SOIL BEARING PRESUURE = 1800 PSF PER SOILS REPORT
MAX LOAD = 45.00x25.00/2 + 15.00x 12.00 + 150x 15x 18/144 = 893.00 PLF
FOUNDATION WIDTH REQUIRED = (893.00/1500)x12 = 8"
USE 12" WIDE x 12" DEEP FOOTINGS WITH (I) # 4 TOP AND BOTTOM
FOOTING CAPACITY = (1 500x52xl 2)/(12xl2) = 5000.00 LBS
PADS DESIGN
BM # 11= 8600.00 LBS
USE 2'-6" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY
BM # 23= 7000.00 LBS
USE 2'-6" SQ. x 18" DEEP W/ (3) #4 BARS ECH WAY
BM # 25= 7000.00 LBS
USE 2'-6" SQ. x 12" DEEP W/ (3) #4 BARS ECH WAY
BM # 26 LEFT = 5600.00 LBS
USE 2'-6" SQ. x 12" DEEP W/ (3) #4 BARS ECH WAY
BM # 27 RIGHT + BM # 25 = 11400.00 LBS
USE 3'-0" SQ. x 12" DEEP W/ (3) #4 BARS ECH WAY
.7
wq)zF--(i
0
s
0
0
Reza
PROJECT: Maximum Load For 2 sq'ft Fad Footing (1500ps>] PAGE: j
As har„CLIENT: f i DESIGN BY R:A.;- t
9 M JOB NO.: DATE : ,� - REVI
OUr EW BY RrA.
INPUT DATA
LONGITUDINAL
TRANSVERSE
d
DESIGN SUMMARY
8.50
b
COLUMN WIDTH
c,
= 4 0
in
FOOTING WIDTH
B
= 2,00 ft
COLUMN DEPTH
u
= , 0'
in
FOOTING LENGTH
L
= 2.00 ft
BASE PLATE WIDTH
b;
= 4
in
FOOTING THICKNESS
T
= 12 in
BASE PLATE DEPTH
b,
= s 4
in
LONGITUDINAL REINF.
3 #
4 @ 9 in o.c.
FOOTING CONCRETE STRENGTH
t,'
= 2.5
ksi
TRANSVERSE REINF.
3 #
4 @ 9 in ox
REBAR YIELD STRESS
fy
= 40
ksi
AXIAL DEAD LOAD
PcL
= 2.5
k
P
'
AXIAL LIVE LOAD
PL,
= 2:5
k
LATERAL LOAD (O=WIND. 1=SEISMIC)
= 1
Seismic.SD
SEISMIC AXIAL LOAD
PLAT
= -•0
k, SD
_
117,7777' '
SURCHARGE
qS
= 0;
;ksf
SOIL WEIGHT
WS
= + .:0:1:1
kcf
FOOTING EMBEDMENT DEPTH
Or
= ,2
ft
FOOTING THICKNESS
T
= 12
in
-- - -
- c
ALLOW SOIL PRESSURE
Qa
= • 1.5
ksf
FOOTING WIDTH
B
= 2
ft
_
-74
(/ —•
FOOTING LENGTH
L
= 2
ft
~
BOTTOM REINFORCING
# 4
•-- Ell• .•tee
THE PAD DESIGN IS ADEQUATE.
ANALYSIS
DESIGN LOADS (IBC SEC. 1605.3.2 & ACI 318-05 SEC.9.2.1)
CASE 1: DL + LL P = 5 kips 1.2 DL + 1.6 LL Pu = 7 kips
CASE 2: DL + LL + E / 1.4 P = 5 kips 1.2 DL + 1.0 LL + 1.0 E Pu = 6 kips
CASE 3: 0.9 DL + E / 1.4 P = 2 kips 0.9 DL + 1.0 E Pu = 2 kips
CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2)
P CASE 1 CASE 2 CASE 3
8/ +y, + (0. 15 — u ., )%' — 1.29 kSf, 1.29 ksf, 0.60 ksf
q MAX < k Q a • [Satisfactory]
Where k = 1 for gravity loads. 4/3 for lateral loads.
DESIGN FOR FLEXURE (ACI 318-05 SEC. 15.4.2. 10.2, 10.3.5. 10.5.4, 7.12.2, 12.2, & 12.5)
0.85[3
,f c
11
LONGITUDINAL
TRANSVERSE
d
8.75
8.50
b
24
24
q u.max
1.75
1.75
Mu
1.47
1.47
P
0.000
0.000
Pmin
0.000
0.000
AS
0.07
0.08
RegD
1 # 4
1 # 4
Max. Spacing
18 in o.c.
18 in o.c.
USE
3 # 4 @ 9 in o.c.
3 # 4 @ 9 in o.c.
Pmax
0.019
0.019
Check ppm, , pmax
[Satisfactory]
[Satisfactory]
11
w
ul
[7
CHECK FLEXURE SHEAR (ACI 318-05 SEC. 9.3.2.3, 15.5.2. 11.1.3.1, & 11.3)
06'lt = 20hdF.
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2, 11.12.1.2. 11.12.6. & 13.5.3.2)
54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3 )
Its = ratio of long side to short side of concentrated load = 1.00
bo = C1 * C2 + 131 + b2 + 4d = 42.5 in
AP = by d = . 366.6 in'
y tf = MIN(2 , 4 / (3c. 40 d / bp) = 2.0
Vu = l'u.max(I --(h ci+c/l(h: ca +c/l1 5.63 kips < ¢ V n [Satisfactory)
a
(oont'd)
LONGITUDINAL
TRANSVERSE
Vu
0.66
0.73
P
0.75
0.75
0Vn
15.8
15.3
Check V„ < Wn
[Satisfactory]
[Satisfactory]
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2, 11.12.1.2. 11.12.6. & 13.5.3.2)
54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3 )
Its = ratio of long side to short side of concentrated load = 1.00
bo = C1 * C2 + 131 + b2 + 4d = 42.5 in
AP = by d = . 366.6 in'
y tf = MIN(2 , 4 / (3c. 40 d / bp) = 2.0
Vu = l'u.max(I --(h ci+c/l(h: ca +c/l1 5.63 kips < ¢ V n [Satisfactory)
a
(oont'd)
•
Reza
TRANSVERSE
d
8.75
8.50
PROJECT:
Max.`Load For 2'5 sq: ft Pa'd Fooling (1500psf)`
PAGE
q u.max
„ CLIENT:
As har our
DESIGN BY . ,R.A.;
,_
Mu
N JOB NO.:
L t DATE: }
REVIEW BY: R.A.
0.000
Ndif 'Desi n`6asei'`on'ACi318-05(
.
0.001
0.001
A,
0.17
0.18
RegD
INPUT DATA
1 # 4
DESIGN SUMMARY
18 in ox,
18 in o.c.
COLUMN WIDTH c, =
0
in FOOTING WIDTH
B = 2.50 it
0.019
COLUMN DEPTH C2 =
;; •0
in FOOTING LENGTH
L = 250 It
BASE PLATE WIDTH b, =
' '4
in FOOTING THICKNESS
T = 12 in
BASE PLATE DEPTH b, =
4 '
in LONGITUDINAL REINF, 3
4 4 @ 12 in o.c
FOOTING CONCRETE STRENGTH fc =
2:5
ksi TRANSVERSE REINF. 3
9 4 @ 12 in o.c
REBAR YIELD STRESS fy =
• =40
ksi
3
AXIAL DEAD LOAD P, =
4.5,
k
AXIAL LIVE LOAD Pig =
,
,4.5
k
LATERAL LOAD (O=WIND, 1=SEISMIC) =
1
I Seismic,SD
SEISMIC AXIAL LOAD PL,,, =
0
k, SD
SURCHARGE qs
to
ksi
SOIL WEIGHT W, =
0,11
kcf -
FOOTING EMBEDMENT DEPTH Or =
p
ft
4 -
FOOTING THICKNESS T =
12 `
in
c'
ALLOW SOIL PRESSURE Oa =
<1.5
`ksf
FOOTING WIDTH B =
FOOTING LENGTH L =
25
ft
rr
'
2:5
tt
•,y<
BOTTOM REINFORCING #
'4
_l
THE PAD DESIGN IS ADEQUATE.
ANALYSIS
DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1)
CASE 1: OL + LL P =
CASE 2: DL + LL + E / 1.4 P =
9
9
kips 1.2 DL + 1.6 LL
kips
Pu = 13 kips
1.2 DL + 1.0 LL + 1.0 E
Pu = 10 kips
CASE 3: 0.9 DL + E / 1 4 P =
4
kips 0.9 OL + 1.0 E
Pu = 4 kips
CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC.15.2.2)
P
CASE 1
CASE 2 CASE 3
1 5 — �r,.,.) % _
1.48
ksf, 1.48 ksf. 0.69
ksf
q mAx < k Q a . (Satisfactory]
where k = 1 for gravity loads, 4/3 for lateral loads.
DESIGN FOR FLEXURE (ACI 318-05 SEC.15.4.2, 10.2,
10.3.5.
10.5.4, 7.12.2, 12.2, & 12.5)
-
1
•
LONGITUDINAL
TRANSVERSE
d
8.75
8.50
b
30
30
q u.max
2.02
2.02
Mu
3.43
3.43
P
0.000
0.001
Pmin
0.001
0.001
A,
0.17
0.18
RegD
1 # 4
1 # 4
Max. Spacing
18 in ox,
18 in o.c.
USE
3 # 4 @ 12 in o.c.
3 # 4 @ 12 in o.c.
Pmax
0.019
0.019
Check Pprod , Pmax
[Satisfactory]
[Satisfactory]
El
•
CK FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3)
Or 20bdFf.
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2)
Ov,=(2+ v)o fc..4P = 54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3 )
0, = ratio of long side to short side of concentrated load = 1.00
bo = c1+G2+b1+b2+4d = 42.5 in
Ap = b0 d = 366.6 in'
Y = MIN(2 . 4 / (1c, 40 d / bo) = 2.0
u = Pu. maxL I H/ ( 2 d )! 2 + d I I = 11.02 kips < 0 V „ [Satisfactory]
(conrd)
LONGITUDINAL
TRANSVERSE
Vu
2.21
2.31
4)
0.75
0.75
Wn
19.7
19.1
Check V„ < Wn
[Satisfactory]
[Satisfactory]
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2)
Ov,=(2+ v)o fc..4P = 54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3 )
0, = ratio of long side to short side of concentrated load = 1.00
bo = c1+G2+b1+b2+4d = 42.5 in
Ap = b0 d = 366.6 in'
Y = MIN(2 . 4 / (1c, 40 d / bo) = 2.0
u = Pu. maxL I H/ ( 2 d )! 2 + d I I = 11.02 kips < 0 V „ [Satisfactory]
(conrd)
(6
•
•
Reza
PROJECT: ,_
#P.
PAGE
CLIENT: DESIGN BY
. -R.A.AS har OVr
JOB NO.: DATE : ` . ' . REVIEW BY: R.A.
INPUT DATA
LONGITUDINAL
TRANSVERSE
d
DESIGN SUMMARY
COLUMN WIDTH
c, =
0
in
FOOTING WIDTH
COLUMN DEPTH
c7 =
0
in
FOOTING LENGTH
BASE PLATE WIDTH
b, =
4
in
FOOTING THICKNESS
BASE PLATE DEPTH
b, =
,4
. in
LONGITUDINAL REINF
FOOTING CONCRETE STRENGTH
fc' =
w'2,5
ksi
TRANSVERSE REINF
REBAR YIELD STRESS
fy =
40
ksi
0.019
AXIAL DEAD LOAD
P, =
6:5
k
AXIAL LIVE LOAD
PL: =
6.5
k
LATERAL LOAD (O=WIND, 1=SEISMIC)
=
1
Seismic,SD
SEISMIC AXIAL LOAD
PLAT =
0
k. SO
SURCHARGE
qs =
:0
ksf
SOIL WEIGHT
ws =
0.11
kcf
FOOTING EMBEDMENT DEPTH
Dr =
2
ft
FOOTING THICKNESS
T =
12
in
ALLOW SOIL PRESSURE
Qa =
1.5
ksf
FOOTING WIDTH
B =
3
ft
FOOTING LENGTH
L =
3
ft
BOTTOM REINFORCING
#
4
THE PAD DESIGN IS ADEQUATE.
B = 3.00 It
L = 3.00 h
T = 12 in
3 # 4 @ 15 in o.c.
3 # 4 @ 15 in ox
n
r
ANALYSIS
DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1)
CASE 1: DL + LL P = 13 kips 1.2 DL + 1.6 LL Pu =
CASE 2: DL + LL + E / 1.4 P = 13 kips 1.2 DL + LO LL + 1.0 E Pu =
CASE 3: 0.9 OL + E 11.4 P = 6 kips 0.9 DL + 1.0 E Pu =
CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2)
L. CASE 1 CASE 2 CASE 3
ry,r,,x=—+q.,+(0.15-Tr,, )% = 1.48 ksf, 1.48 ksf. 0.69 ksf
6L.
q MAx < k Q a , [Satisfactory]
where k = 1 for gravity loads, 4/3 for lateral loads.
DESIGN FOR FLEXURE (ACI 318-05 SEC.1 5A.2,10.2, 10.3.5, 10.5.4, 7.12.2. 12.2, & 12.5)
r
0.851', I- 1-7/u 0.85 i
1 0.383h,/= - r err / /. 4
f,r r'rr Lr Pura l d 3 �1
18 kips
14 kips
6 kips
0
LONGITUDINAL
TRANSVERSE
d
8.75
8.50
b
36
36
Cl u,max
2.02
2,02
Mu
6.09
6.09
P
0.001
0.001
Pmm
0.001
0.001
As
0.31
0.32
RegD
2 # 4
2 # 4
Max. Spacing
18 in o.c.
18 in o.c.
USE
3 # 4 @ 15 in o.c.
3 # 4 @ 15 in o.c.
Pmax
0.019
0.019
Check Pprcw < Pmax
[Satisfactory]
[Satisfactory]
18 kips
14 kips
6 kips
0
w
•
•
FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3)
of ' 11= 20hdF
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2, 11.12.6, & 13.5.3.2)
OI'n=('—+ v)o A = 54.98 kips
where 0 = 0.75 (ACI 318-05, Section 9.3.2.3)
01, = ratio of long side t0 Short side of Concentrated load = 1.00
bo = Cl * C2 * bl + b2 + 4d = 42.5 in
Ap = bo d = 366.6 in7
Y = MIN(2 , 4 / fig. 40 d / b 2.0
o) =
I _P II_ I hi+,'i+(/�(b LL d)ll= 16.61 kis <
u— u, mai BL( 2 l 2 /J P m n [Satisfactory]
0
(cont'd)
LONGITUDINAL
TRANSVERSE
Vu
4.17
4.30
4
0.75
0.75
Wn
23.6
23.0
Check V„ < Wn
[Satisfactory)
[Satisfactory]
PUNCHING SHEAR (ACI 318-05 SEC. 15.5.2. 11.12.1.2, 11.12.6, & 13.5.3.2)
OI'n=('—+ v)o A = 54.98 kips
where 0 = 0.75 (ACI 318-05, Section 9.3.2.3)
01, = ratio of long side t0 Short side of Concentrated load = 1.00
bo = Cl * C2 * bl + b2 + 4d = 42.5 in
Ap = bo d = 366.6 in7
Y = MIN(2 , 4 / fig. 40 d / b 2.0
o) =
I _P II_ I hi+,'i+(/�(b LL d)ll= 16.61 kis <
u— u, mai BL( 2 l 2 /J P m n [Satisfactory]
0
(cont'd)
•
•
/� /Reza
PROJECT: Max:Yload For 3 5 sq.fk Pad FooLng'(15OOpsf),,
A�''/ har OUr CLIENT : ii
:7 P JOB NO.: d~�° e` ,,q 1-� _ �. �h.e r
u_ _:z -DATE : �"' `
INPUT DATA
B
= 3.50 ft
COLUMN WIDTH
C, =o
• in
COLUMN DEPTH
Ce =
1 4, in
BASE PLATE WIDTH
b, =
4 -'in
BASE PLATE DEPTH
bz =
4 7 in
FOOTING CONCRETE STRENGTH
fc, _
"2,5: ksi
REBAR YIELD STRESS
fy =
40' ksi
AXIAL DEAD LOAD
Poi =
8.5 1 k
AXIAL LIVE LOAD
PLL =
8.5 k
LATERAL LOAD (O=WIND, 1=SEISMIC)
_
-11. Seismic.SD
SEISMIC AXIAL LOAD
PLAT =
0 k. SD
SURCHARGE
qs =
0 ksf
SOIL WEIGHT'rs
="k0;11
kCf
FOOTING EMBEDMENT DEPTH
D) =
r `''2 -ft
FOOTING THICKNESS
T =
f •', ,17-, in
ALLOW SOIL PRESSURE
Qa =
�1.5 ksf
FOOTING WIDTH
B =
3.5- ft
FOOTING LENGTH
L =
Y'3.5` . ft
BOTTOM REINFORCING
#
4
THE PAD DESIGN IS ADEQUATE.
IANALYSIS
DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1)
CASE 1: DL + LL P = 17 kips
CASE 2: DL + LL + E / 1.4 P = 17 kips
CASE 3: 0.9 DL + E / 1.4 P = 8 kips
CHECK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2)
DESIGN SUMMARY
PAGE
DESIGN BY R A.
REVIEW BY: ,R-A`.�
FOOTING WIDTH
B
= 3.50 ft
FOOTING LENGTH
L
= 3.50 ft
FOOTING THICKNESS
T
= 12 in
LONGITUDINAL REINF
4 #
4 12 in o.c
TRANSVERSE REINF.
4 #
4 Q 12 in o c
P
0.001
0.001
Pmm
0.001
0.001
As
0.48
0.50
RegD
3 # 4
3 # 4
Max. Spacing
18 in O.C.
18 in o.c.
USE
l
4 # 4 @ 12 in o.c.
Pmax
pJ` _
0.019
Check Pproa < Pmax
[Satisfactory]
[Satisfactory]
%/."j
_-
t
D_..
1.2 DL + 1.6 LL Pu = 24 kips
1.2 DL+LOLL+1.OE Pu = 19 kips
0.9 DL + 1.0 E Pu = 8 kips
P CASE 1 CASE 2 CASE 3
K7 (0.15 - i„ ,) % - 1.43 ksf, 1.43 ksf. 0.66 ksf
4 MAx < k O a . [Satisfactory]
where k = 1 for gravity loads, 4/3 for lateral loads.
DESIGN FOR FLEXURE (ACI 318-05 SEC. 15.4.2. 10.2, 10.3.5. 10.5.4, 7.12.2, 12.2, & 12.5)
Ox 'f/tr
/r. I 11 0.3R3hd'./,.) - O.RS�i✓�• er: -
11 f" t ° ll d 3 1
!.2
LONGITUDINAL
TRANSVERSE
d
8.75
8.50
b
42
42
q u.max
1.94
1.94
Mu
9.44
9.44
P
0.001
0.001
Pmm
0.001
0.001
As
0.48
0.50
RegD
3 # 4
3 # 4
Max. Spacing
18 in O.C.
18 in o.c.
USE
4 # 4 @ 12 in o.c.
4 # 4 @ 12 in o.c.
Pmax
0.019
0.019
Check Pproa < Pmax
[Satisfactory]
[Satisfactory]
!.2
.f '
•
s
0
FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3, 15.5.2, 11.1.3.1, & 11.3)
Or, = 2obd f .
PUNCHING SHEAR (ACI 318-05 SEC.15.5.2, 11.12.1.2. 11.12.6, & 13.5.3.2)
Tia=(2+!'�� ./r. !IP = 54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3)
Oc = ratio of long side to short side of concentrated load = 1.00
bo = C1 + C2 + b1 + b2 + 4d = 42.5 in
AID = bo d = 366.6 in -
y = MIN(2. 4/13c. 40d/bo) = 2.0
I ' u = Pu, max L I 81.(-2 * d II 2 * d1J = 22.28 kips < 0 v n [Satisfactory]
9
(cont'd)
LONGITUDINAL
TRANSVERSE
Vu
6.38
6.52
Op
0.75
0.75
0Vn
27.6
26.8
Check Vu < 0Vn
[Satisfactory]
[Satisfactory]
PUNCHING SHEAR (ACI 318-05 SEC.15.5.2, 11.12.1.2. 11.12.6, & 13.5.3.2)
Tia=(2+!'�� ./r. !IP = 54.98 kips
where 0 = 0.75 (ACI 318-05. Section 9.3.2.3)
Oc = ratio of long side to short side of concentrated load = 1.00
bo = C1 + C2 + b1 + b2 + 4d = 42.5 in
AID = bo d = 366.6 in -
y = MIN(2. 4/13c. 40d/bo) = 2.0
I ' u = Pu, max L I 81.(-2 * d II 2 * d1J = 22.28 kips < 0 v n [Satisfactory]
9
(cont'd)
•
CK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2)
P CASE 1 CASE 2 CASE 3
BL t q + (0. 15 - 11,07 = 1.48 ksf. 1.48 ksf, 0.69 ksf
q MAX < k O a . [Satisfactory]
where k = 1 for gravity loads. 4/3 for lateral loads.
IGN FOR FLEXURE (ACI 318-05 SEC. 15.4.2. 10.2, 10.3.5. 10.5.4, 7.12.2. 12.2, & 12.5)
11.85.
/ .I I I 0.383hc/'/,) - 0.85�i.%c err / l
P va.r - �• l c rr -r t r P.ur, = P
:b//,%' 0.0018 d
Reza
TRANSVERSE
d
8.75
8.50
b
48
48
PROJECT:
Max.JLoad For 4'O.sg7-f Pad Footin' '(1500psf) .
PAGE
Mu
A
A� har Olir
P
CLIENT :
..%
-
,, t " +:
- •- :.>
DESIGN BY R.A., `
0.002
0.002
JOB NO.:
I_ �.
;r .i
DATE: :._ ..: '
REVIEW BY: R.A.
4 # 4
Pad Footii ;Desi h"Sa d on}'ACi 318-05
18 in o.c.
18 in o.c.
USE
4 # 4 @ 14 in o.c.
4 # 4 @ 14 in o.c.
Pmax
0.019
0.019
Check pproe ` Pmax
INPUT DATA
[Satisfactory]
DESIGN SUMMARY
COLUMN WIDTH
c, =
0:
in
FOOTING WIDTH
B = 4.00 It
COLUMN DEPTH
q2 =
0
in
FOOTING LENGTH
L = 4.00 It
BASE PLATE WIDTH
b, =
4
in
FOOTING THICKNESS
T = 12 in
BASE PLATE DEPTH
b: =
4 -
. in
LONGITUDINAL REINF 4
# 4 @ 14 in o.c
FOOTING CONCRETE STRENGTH
fc' =
2.5
. ksi
TRANSVERSE REINF 4
# 4 @ 14 in o.c
REBAR YIELD STRESS
fy =
4.6
ksi
AXIAL DEAD LOAD
Po, =
-11.5
k
AXIAL LIVE LOAD
PLL =
1.1:5
k
'
LATERAL LOAD (O=WIND• 1=SEISMIC)
=
1.
Seismic,SO
SEISMIC AXIAL LOAD
PLAT =
0,.
k, SD
SURCHARGE
qs =
9
ksf
7
I" to
SOIL WEIGHT
ws =
X0.11.
kcf
FOOTING EMBEDMENT DEPTH
Of =
2
g
FOOTING THICKNESS
T =
12
in
ALLOW SOIL PRESSURE
Oa =
.1:5:
ksf
'-
FOOTING WIDTH
B =
q
It
FOOTING LENGTH
L
BOTTOM REINFORCING
#
4
b.
THE PAD DESIGN IS ADEQUATE.
v
ANALYSIS
DESIGN LOADS (IBC SEC.1605.3.2 & ACI 318-05 SEC.9.2.1)
CASE 1: OL + LL
CASE 2: DL + LL + E / 1.4
P =
P =
23
23
kips
kips
1.2 DL + 1.6 LL
1.2 DL + 1.0 LL + 1.0 E
Pu = 32 kips
Pu = 25 kips
CASE 3: 0.9 DL + E / 1.4
P =
10
kips
0.9 DL + 1.0 E
Pu = 10 kips
•
CK SOIL BEARING CAPACITY (ACI 318-05 SEC. 15.2.2)
P CASE 1 CASE 2 CASE 3
BL t q + (0. 15 - 11,07 = 1.48 ksf. 1.48 ksf, 0.69 ksf
q MAX < k O a . [Satisfactory]
where k = 1 for gravity loads. 4/3 for lateral loads.
IGN FOR FLEXURE (ACI 318-05 SEC. 15.4.2. 10.2, 10.3.5. 10.5.4, 7.12.2. 12.2, & 12.5)
11.85.
/ .I I I 0.383hc/'/,) - 0.85�i.%c err / l
P va.r - �• l c rr -r t r P.ur, = P
:b//,%' 0.0018 d
A
0
LONGITUDINAL
TRANSVERSE
d
8.75
8.50
b
48
48
q u.max
2.01
2.01
Mu
14.79
14.79
P
0.001
0.001
Pmin
0.002
0.002
As
0.76
0.78
RegD
4 # 4
4 # 4
Max. Spacing
18 in o.c.
18 in o.c.
USE
4 # 4 @ 14 in o.c.
4 # 4 @ 14 in o.c.
Pmax
0.019
0.019
Check pproe ` Pmax
(Satisfactory]
[Satisfactory]
A
0
FLEXURE SHEAR (ACI 318-05 SEC.9.3.2.3. 15.5.2. 11.1.3.1. 8 11.3)
01'n = 30hd f .
PUNCHING SHEAR (ACI 318-05 SEC.15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2)
0I'n=��+ 00j,- Ap = 54.98 kips
where 0 = 0.75 (ACI 318-05, Section 9.3.2.3)
[3, = ratio of long side to short side of concentrated load = 1.00
bo = C1 + c2 + bi + b2 + 4d = 42.5 in
Ap = bo d = 366.6 in'
Y = MIN(2 . 4 / (1c , 40 d / bo) = 2.0
I',u = Pu. max I -- +d I( +d)] 30.62 kips < 0 V „ [Satisfactory]
,
(cont'd)
LONGITUDINAL
TRANSVERSE
Vu
9.56
9.73
0
0.75
0.75
ovn
31.5
.30.6
Check Vu < 0Vn
[Satisfactory)
[Satisfactory]
PUNCHING SHEAR (ACI 318-05 SEC.15.5.2. 11.12.1.2. 11.12.6. & 13.5.3.2)
0I'n=��+ 00j,- Ap = 54.98 kips
where 0 = 0.75 (ACI 318-05, Section 9.3.2.3)
[3, = ratio of long side to short side of concentrated load = 1.00
bo = C1 + c2 + bi + b2 + 4d = 42.5 in
Ap = bo d = 366.6 in'
Y = MIN(2 . 4 / (1c , 40 d / bo) = 2.0
I',u = Pu. max I -- +d I( +d)] 30.62 kips < 0 V „ [Satisfactory]
,
(cont'd)