steel design project - civil engineering

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Table of Contents

1. Basis of Design .......................... ......... .............................. .................. ......... .......... BD-1 Building Construction ...... ............................ .................................... ......... ......... ....... BD-2

2. Load Calculations a. Dead Load ..................... ......... .................. ................................................. LC-1 b. Live Load .................. ... .............................. ...... ..................... ..................... LC-3 c. Load Combination ............ .................................................................. .......... LC-4 d. Seismic Dead Load .......... ............................................................................. LC-5 e. Wind Analysis ...................................................... ... ..................................... LC-6 f. Seismic Analysis ......................................................................................... LC-16 g. Wind Vs. Seismic Comparison ................................................. ................ ...... LC-21

3. Gravity Framing Design a. 2nd Floor

i. Header. ........ ................................................................................... GD-1 ii. Trimmer Stud .......................................................................... ....... GD-3

b. Floor i. Joist. .......................... .................. ...... ................................ ...... ...... GD-5

ii. Beam ........................... ................................... ............................. GD-8 c. 1st Floor

i. Header. ............................. .................. ......................................... GD-12 ii. Stud Trimmer ............... ................................... ............................... GD-16

iii. King Stud ......... ......................... ......................... ............................ GD-18 d. Foundation

i. Continuous Footing ......... ................................. ......... ...................... GD-23 ii. Footing Spread ............ .................................................... .............. GD-24

4. Lateral Design a. North-South Direction

i. Roof Diaphragm ............ ........................................ ............................. LS-1 ii. Roof Level Shear walls ....................................................................... LS-3 iii. Roof Level Chords ......... ..................................................................... LS-9 iv. Floor Diaphragm .......................................... .................... ......... ....... LS-14 v. Floor Level Shearwall .................. .................................................... LS-16 vi. Floor Level Chords ... .................... ........................ ............................ LS-22

b. East-West Direction i. Roof Diaphragm ............. , ............................................... ......... ......... LS-27 ii. Roof Level Shear walls ........................................... ........................... LS-29 iii. Roof Level Chords ......... ........................... ........................................ LS-32 iv. Floor Diaphragm ............ ............... ................................................... LS-40 v. Floor Level Shear wall. .......................... ............... ...... ...................... LS-42 vi. Floor Level Chords .......... ............. ..................... ... ...................... ...... LS-50

5. Miscellaneous a. Beam and columns in Garage .............. ............................................. ............... M-1 b. Stairs ............ ........... : ... ...... ... ........................ ............... ............................ M-21 c. Stringers ............................................ ......... .................. ............................. M-22

6. Cost Estimate a. Materials/Labor ......... ............... ...... ............................................................. CE-1

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Building Code:

Location:

Occupancy category:

Basic Wind Speed

Surface roughness Categories:

Exposure Categories:

Site Classification:

Latitude and Longitude:

0.2s Spectral Response Acceleration, Ss:

1.0s Spectral Response Acceleration, 81:

Seismic Design Coefficients and Factors

Snow Load

Soil Report

Allowable Soil Bearing Pressure

BD-1

Basis of design

2006 IBC (IBC), ASCE 07-05(ASCE) and AF&PA NOS- 05(NDS).

Mission Ranch Ill in the City of Riverside, California.

II, All building and other structures except those listed in Occupancy Categories I, Ill, and IV (IBC Table 1604.5).

85mph (ASCE, Figure 6-1)

B, Urban and suburban areas, wooden areas, or other terrain with numerous spaced obstructions having the size of single-family closely dwelling or larger (ASCE, 6.5.6.2).

Exposure B, shall apply where the ground surface roughness conditions, and defined by Surface Roughness B. (ASCE, 6.5.6.3)

D, Where the soil properties are not known in sufficient detail to determine the site class D, or geotechnical data determines. (ASCE, 20.1)

33.87545, -117.328535 (Google Maps)

1.5g (USGS Seismic Hazard Program)

0.531g (USGS Seismic Hazard Program)

R= 6.5 Cd = 4 0=3

None

1500 psf

n I I

[-]

Dead Load

Roof (Conventional Framing)

Roof (Truss Framing)

LC-1

Load Calculations

Item Load(psf)

Tile Roofing 10

3-ply Waterproofing 1.5

15/32" Rated Sheathing 1.4

1/2" Gypsum Board 2.5

misc. 0.8

Total 19

�-----i2x4@ 24 o.c Web Chord 1.2

2x4 @ 24 o.c Bot. Chord 0.6

R-11 Insulation 1.5

1/2" Gypsum Board 2.5

misc. 1.8

Total 23

[]

Exterior Wall Detail

Interior Wall Detail

Floor Detail

�

LC-2

7 /8" Stucco 10

Building Paper 0.5

15/32" Rated Sheathing 1.4

Insulation (expanded} 1.1

2x6 @o.c 1.4

1/2" Gypsum Board 0.5

misc. 0.6

Total 17.5

1/2" Gypsum Board 2.5

2x4@ 16 o.c 2.9

1/2" Gypsum Board 2.5

Total 6.5

------- Hardwood Floor 4

19/32" Rated Sheathing 1.8

2x12 @ 16" o.c 2.9

1/2" Gypsum Board 0.5

misc. 0.8

Total 13

LC-3

n Live Load

n Roof Live Load, L,= 20psf, Uniform, (ASCE, Table 4-1)

n Reduction in Roof Live Loads

L,= LoR1R2 where 12 �LR �20 (ASCE, 4.9)

[l 1 for At �200ft2

0 R1 = 1.2-0.001At for 200ft

2 �t ,,600 ft2

0.6 for At �600 ft

D 1 for F �

R2= 1.2-0.05F for 4 �F ,,12

D 0.6 for F�12

F= number of inchers of rise per foot

D 2

nd Floor Live Load, L0= 40psf, Uniform (ASCE, Table 4-1)

Ground Floor, Lo= 40psf, Uniform (ASCE, Table 4-1)

D Reduction in Live Loads.

L= L0( 0.25 + �) (ASCE, Equ. 4-1) LJ

KLLAT

KLL (ASCE, Table 4-2)

At= Tributary area in ft 2

�

. I LJ

Load Combination- Basic Combinations (IBC, Sec.1605.3.1)

Allowable Stress Design

1. D + F

2. D + H + F + L + T

3. D + H + F + (L, or S or R)

4. D + H + F + 0.75(L + T) + 0.75(L, or Sor R)

5. D + H + F + (W or 0.7E)

6. D + H + F + 0.75(W or 0.7E) +0.75L + 0.75(L, or Sor R)

7. 0.6D +W + H

8. 0.6D + 0.7E + H

In this buildi ng Design

Fluid Pressure, F=0

Earth Pressure, H=0

Rain Load, R=0

Snow Load, S=0

Self Straini ng Force, T=0

Simplified Load Combinations

1. D

2. D + L

3. D + L,

4. D + 0.75L + 0.75L,

5. D + (W or 0.7E)

6. D + 0.75(W or 0.7E) +0.75L + 0.75L,

7. 0.6D+W

8. 0.6D + 0.7E

LC-4

[]

[_]

LC-5

Seismic Dead Load

Item Area(sqft) psf Weight(lbs) Floor Diaphragm

Item Weight(lbs)

(1/2) 2nd Floor 15,556.05

walls

Floor 23,985

(1/2) 1st Floor walls 11,611.35

Total weight of Floor for 51152.4*

Diaphraam Desian

Roof Diaphragm

Item Weight(lbs)

Roof 43;171

(1/2) 2nd Floor 15,556.05

walls

Total weight of Roof for 58,727.05*

Diaohraam Desian

*These Values will be used in our seismic design to find the force and acceleration at each floor level

**This value will be used to calculate the base shear in our Seismic Analysis.

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Wind Loads Analysis

General Design Conditions

• Building Type: Enclosed Building

• Location: Riverside, CA

• Wind Velocity: 85 MPH

• Roof Angle: 0 = tan- 1

(4/12) = 18.43 °

LC-6

Building is a simplified diaphragm, low-rise, non-flexible, regular-shape building. The building is enclosed and conforms to wind-borne debris provisions. The building is exempt from torsional load cases since it will be designed with flexible diaphragms. It has an approximately symmetrical cross-section in each direction with a hip roof with identical slopes on all sides. The building is eligible for performing a wind analysis using the simplified procedure.

Figure 1. Basic Wind Speed. (/BC 2006 Figure 1609)

• Angle of Roof: 18.4 ° < 45°

• Hip Roof: 7°<0=18.4 °<27 °

• Height of Structure: 23. 75 ft • Height of Roof: 6.05 ft • Eave Height: 17.7 ft • Roof Mean Height: 20.2 ft < 60 ft • Roof Mean Height: 20.2 ft< 45 ft (least horizontal dimension)

Design Criteria

• Occupancy Category II:

• Importance Factor, I = 1.0:

Non-hurricane prone regions, non-hazardous (/BC 2006 Section 1604.5)

(ASCE 07-05 Table 6-1)

• Surface Roughness B: Urban and suburban areas. Wooded areas or other terrain with closely spaced obstructions the size of a single family dwelling or larger. (/BC 2006 Section 1609.4.3)

LC-7

• Exposure Category B: Where surface roughness conditions of Category B prevails. (/BC 2006 Section 6.5.6.3)

• Adjustment Factor, A = 1.0: Obtained from ASCE 07-05 Figure 6-2 with mean roof height equal to approximately 24 feet. (ASCE 07-005 Figure 6-2)

• Topographic Factor, K21 = 1.0: No escarpment or ridges. (ASCE 07-05 Section 6.5.7.2)

Summary - Building Information

85 R-3 1.0

Enclosed 0.0 8.1

20.2 45.0 49.7 4.0 18.4 4.5 9.0

Method I - Simplified Procedure Requirements

For the design of main wind-force resisting systems (MWFRSs), the building meets all of the required conditions from ASCE 07-05 section 6.4.1.1. For the design of components and cladding the building meets all of the required conditions from ASCE 07-05 section 6.4.1.2.

Main Wind Force Resisting System (MWFRS) Wind Pressures

Simplified design wind pressures, Ps for the MWFRS of a low-rise simple diaphragm building represent the net pressures to be applied on the horizontal and vertical projections of the building surfaces as shown in the following figures.

The sign convention signifies positive pressure acting toward the surface and negative pressure acting away from the surface. For slopes other than shown in ASCE 07-05 Figure 6-2 Simplified Design Wind Pressure, interpolation will be conducted.

LC-8

The structure is simplified and a wind load analysis is conducted with the structure being considered as a rectangular shape. This method results in an effective wind area that is larger than it is in actuality and the forces are calculated for each zone using the higher areas. This is considered to be a conservative approach. These results will then be compared to the seismic forces on the structure to determine which horizontal load governs for the design of the MWFRS.

See figure below for an illustration of the different zones.

MWFRS Wind Pressures

End Zone Distance a = .10 x 45ft = 4.5 ft <- governs a = .4 X 20.2ft = 7.08 ft a = .04 X 45 = 1.8 ft 1.8 ft< a= 4.5ft > 3.0 ft 2a = 9 ft

Height of Roof Pressure Zone: Hroof = hstructure - hmean

= 23.75ft- 20.2 ft = 3.55 ft

Height of Wall Pressure Zone: Hwall = hmean

= 20.2 ft

Design Wind Pressures and Forces: Ps = A Kzt I Ps3o = 1.0 Ps3o Ps = (1)(1)(1) Ps30 Ps = Ps30

Fs = Ps Azone

Evaluation of Wind Pressures (ASCE. Figure 6-2, Simplified Design Wind Pressure) Use 0 = 18.4 ° for all horizontal and vertical pressure zones for the design of both the transverse and longitudinal MWFRS. Interpolate between 15° and 20° for basic wind speed of 85 mph.

Horizontal Wind Pressures PA = 15.5 psf Ps = -4.0 psf � use 0 psf Pc = 10.3 psf

Po = -2.4 psf -> use 0 psf

REFERENCE

CORNER

Figure 1- E-W Wind Direction

LC-9

l J

'>, .. ') �C:, / $'

REFERENCE

CORNER

Figure 2- N-S Wind Direction

LC-10

]

AH = 894.6 sf

Forces per Zone: FE = (-13.8 psf) (223.65 sf) = -3086 lb FF = (-9.4 psst) (223.65 sf)= -2102 lb FG = (-9.6 psf) (894.6 sf)= -8588 lb FH = (-7.1 psf) (894.6 sf)= -6352 lb

Total Vertical Force on Roof from N-S Winds: FRNS = FE + FF + FG + FH

= -3086 lb - 2102 lb - 8588 lb - 6352 lb = -2O,128 lb

Summary of MWFRS Wind Pressures and Forces

Wind Direction Wind Zone

East-West Full Heiaht of Structure East-West 2nd Floor East-West 1st Floor North-South Full Heiaht of Structure North -South 2nd Floor North-South 1 st Floor

Components and Cladding Wind Pressures

LC-14

Wind Force ( lb l

11,286 5,109 5,644 10,308 5,160 5,160

Net design wind pressures, Pnet, are applied normal to the surface. Since 0 = 18.4 ° <

25° , Zone 3 is treated as Zone 2. Pnetao are for exposure B, at h = 30 ft, I = 1.0, and Kxt= 1.0. For effective wind areas other than shown in ASCE 07-05 Figure 6-3, interpolation will be conducted.

The structure is simplified for the ease of calculations. It is evaluated as a rectangular shaped structure and the roof is sectioned off as 2 separate hip roofs so that it can be analyzed separately and added together.

Net Design Wind Pressure and Forces:

For roof angle > 7 to 27 degrees P net = ,I Kzt I Pnet30 P net = P net30

F net = P net Azone

[]

Components and Cladding Wind Force on Headers

Effective Wind Area for Header: Tributary Area: AHeader = (6') (6.67') = 40 sq ft

Zone Effective Pnet Fnet Wind Area (psf) (lb)

Pnet (psf)

(sf) Toward Toward Away From Surface Surface Surface

4 20 12.4 496 -13.5 4 50 11.6 464 -12.7 4 40 11.9 476 -13.0

Use 13.0 psf for wind load in the design of headers for 2nd and 1st floors.

Components and Cladding Wind Force on Studs

Effective Wind Area for Stud: Tributary Area: Astud = (14.3' / 2) (16" x 1' / 12") = 9.53 sq ft

LC-15

Fnet (lb)

Away From Surface

-540 -508 -520

Net Design Wind Pressures and Forces for Studs (ASCE 07-05 Fig. 6-3)

Zone Effective Pnet Fnet Pnet Wind Area (psf) (lb) (psf)

(sf) Toward Toward Away From Surface Surface Surface

4 10 13.0 124 -14.1 4 20 12.4 248 -13.5

9.53 13.0 124 -13.0

Use 13.0 psf for wind load in the design of studs for 2nd and 1 st floors.

Fnet (lb)

Away From Surface

-134 -129 -24

11 11

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LJ I I

Seismic Design

Ss= 1.5g (Basis of Design)

S1= 0.531g (Basis of Design)

Site Classification

Stiff Soil (Site Class D) (/BC, Table 1613.5.2)

Site Coefficient

Fa= 1.0 ( /BC, Table 1613.5.3(1))

Fv= 1.5 (/BC, Table 1613.5.3(2))

Maximum Considered Earthguake (MCE)

SMs= F.s.= (1 )(1.5g)= 1.5g (/BC, Eq. 16-37)

SM1 = FvS1 = (1.5)(0.5319)= 0.797g (/BC, Eq. 16-38)

Design Spectral Acceleration Parameters

2 2 Sos= 3 S Ms = 3 (1.Sg) = lg (IBC, Eq. 16-39 )

2 2 Sv1= 3 SM1 = 3 (0.797g) = 0.531g

Design Response Spectrum

T 0 = 0.2 G::) = 0.2 (0·�31) = 0.11s

T _ s01 _ o.531 _ 0 531 s-----. s Sns 1

h =8s,

(/BC, Eq. 16-40)

(ASCE, 11.4.5)

(ASCE, 11 . .4.5)

(ASCE, Figure 22-16)

LC-16

�I

Approximate Fundamental Period

Ta = Cth;; , (ASCE, Eq. 12.8-7)

C t = 0.02, x= 0.75, All other structural systems (ASCE, Table 12.8-2)

Ta = (0.02)(20°·75) = 0.19 sec

k =1, for structures having a period of 0.5 or less. (ASCE, sec.12.8.3)

Spectrum F.e.zp=,:.e Ac:c:elention, _g Design Response Spectrum

1.2

Sm= o.s 0.53-1 - - - -

0,4

0.2

' '

I I

I I ... -· 1

....... . --,-

' I

I I

· I I

-�-------J _______ _ I I

I I I .

. . .. ,. • 1--·

I I

0,2 0.4 r 0.6

To=u T,= 0.5-31

0.8

Importance Factor/ Design Category

Occupancy Category 11

1.0 1.2

Importance Factor; 1.0 (/BC, Table 1604.5)

Seismic Design Category (SOC)

Sos= 1g, So1 =0.6g, Occupancy Category II

SOC= D, (/BC, Table 1613.5.5)

1.4 1.6 1.8 2.0 2.2

LC-17

2.4 2.6 2.8 Perio..i,s

n Design Coefficients and Factors

A. Bearing Wall System

LC-18

n 13. Light Framed walls sheathed with wood structural panels rated for shear resistance or steel sheets

R= 6.5

Q=4 (ASCE, Table 12.2-1)

h s;65' For SDC=D, E or F

Design Base Shear. V

V = C5 W, (ASCE, Eq. 12.8-1)

Seismic Response Coefficient. Cs

Cs = (;) = c•:s) = 0.154, (ASCE, Eq. 12.8-2)

Cs must not be greater than

C5 = s(ni) = (1.·') = 0.81 (ASCE, q. 12.8-3) T J .19 l

Cs must not be less than 0. 01 (ASCE, Eq. 12. 8-5)

W= 121490.8 lbs (Seismic Dead Load, LC3)

V = (0.154)(121,490.8 lb)= 18,709.58 lbs

Both boundaries check off

Vertical Distribution of Seismic Forces

Fx = Cvx V (ASCE, Eq. 12.8.11)

Cvx = �y�;;;hf (ASCE, Eq. 12.8.12)

Horizontal Distribution of Forces

V,, = Lf=xFi (ASCE, Eq. 12.8.13)

Diaphragm Design Forces

D F, - �f=x Fi (ASCE E 12 10 1} px - �Y=xW< , q. . .

]

LC-19

Seismic Load Distribution

Level Wxh/ Wxh! Fpx = CvxV N-S Direction E-W Direction hx h/ Wx (lbs) c., Fpx/Wx

X (ft-lb) tf=1 Wthf (lbs) Wox (lbs) Fox (lbs) W-x llbsl F-x(lbs)

R 20' 20' 58,727.05 1,174,541 0.718 13,433.48 0.20 52,504.63 10,862.28 49,393.42 10,218.63

2 9' 9' 51,152.4 460,371.6 0.282 5,276.10 0.154 40,285.44 6,203.96 34,851.96 5,367.20

I=109,879.45 I=1,634,912.6 I=1 I=18,709.58

F orces or Iao ra f o· h am A I . naIvsIs Direction/ Item Total Seismic Load Force, Total Seismic Load for

Fnx (lbs) entire building (lb)

N-S Roof Level 13,433.48 19,637.44

Floor Level 6,203.96

E-W Roof Level 13,433.48 18,800.68

Floor Level 5,367.20

Lateral Force Svstem

Seismic Direction Seismic Force on Each Level Total Seismic Load for entire building (lb)

N-S Roof Level 10,862.28 17,066.24

Floor Level 6,203.96

E-W Roof Level 10,218.63 15,585.83

Floor Level 5,367.20

LC-20

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GD-5

Design Floor Joist (Max Span 14'-3")

**14'-3 is the maximum length of the floor joists that will typically support the second floor right above the first Floor, we will later analyze the floor joists that will support the second floor right above the garage.

Floor Dead Load= 13.0 psf Floor Live Load= 40.0 psf Partition Dead Load= 10.0 psf

16"' Tributary width (T\N) = (1)

16" w = (FDL + FLL + PDL)(TW) = (13psf + 40psf + lOpsf) (12,.) = 84p!f

Vmax = wl = (84plf)(14.25') _ 599 #2 2

Mm ax

= wlz = (84plt)(14.2S)' (12) = 25 586 lb - in 8 8

Try 1.5" X 9.25" Section Properties Fb= 1300 psi Fv= 400 psi E= 1.3x106 psi b= 1.5 in d= 9.25 in A= 13.88 in2 S= 21.39 in3 I= 98.93 in4 Co, CM, Ct, CL, C;, C,, CF, Cru= 1

F'b= Fb(CoCMCtCLCFC1uC;C,) = 1300 psi( 1)= 1300 psi

F'v= Fv( CoC MC1C;) = 400psi (1) = 400psi

f' - 3V - C3lC599#) - 64 8 '< F' - 400 . v - 2bd - (2)(13.8Bin2) - · PSL

v- PSI

f' = � = <5l<25'586 lb-in) 1,196 psi < F'b= 1300 psi OK <-b bd2 (1.5")(9.25")2

4 4 3 ( ') • _ 5wl _ (5)(84plt)(14.25) (12 ) _ O 606 . I _ 14.25 (12) _ O 7125 , OK '-'TL ____ -��--��- Ln <-- ��- Ill <-384E/ (384)(1.3x106)(98.93in4) · 240 (240) · --

*USE 1.5" X 9.25" 1.3E Timberstand LSL Floor

W/ moisture content less than 16%@ 16" o.c for up to a max span of 14'-3"

Design Floor Joist (Max Span 15'-6") **15' -6" is the Maximum Length of the floor joists that are above the garage.

FOL= 13.0 psf FLL= 40.0 psf PDL= 10.0 psf

16" Tributary width (JW) = (12,,)

16" w = (FDL + FLL + PDL)(TW) = (13psf + 40psf + l0psf)

(12,,) = 84plf

Vmax = wL = (84pLf)(15.5? = 651 # 2 2

M _ wL 2

(84pLf)(15.5)\12) max-

8 -

Try 1.5" X 9.25" Section Properties Fb= 1300 psi Fv= 400 psi E= 1.3x106 psi b= 1.5 in d= 9.25 in A= 13.88 in2 S= 21.39 in3 I= 98.93 in4 Co, CM, C1, CL, C;, C,, CF, C1u= 1

30,272 lb - in

F'b= Fb(CoCMCtCLCFCruC;C,) = 1300 psi( 1)= 1300 psi

] F'v = Fv( CoCMC1C1) = 400psi (1) = 400psi

]

f, = � = C3lC651#) = 70 4 s· < F' = 400 i OK+-,, 2bd (2)(13.88in2) . p L V pS

f' - 6M - (6)(30,272 Lb-in) - 1415 . < F' - 1300 . OK <-b - bd2 - (1.5")(9.25")2 - , PSL b- psi __

!J. - 5wi• - (5)(84pLf)(15.s)•(123) = 0 784 in < _1_ = (14.25')(12) 0.7125 in NG +-TL - 384fil - (384)(1.3x106)(98.93in4) · 240 (240)

GD-6

Try 1.5" X 12" Section Properties b= 1.5 in d= 11.25 in A= 16.88 in2 S=31.64 in3 I= 178.0 in4

f' = � = C3lC55l#) = 57 85 . < F' = 400 .v 2bd (2)(16.ssin2) · psi v PSI

f, = � = C5lC30•2721b-in) = 956 74 · < F' = 1300 . OK <-b bd2 (1.5")(11.25")2 · psi b psi --

!!,. _ 5wl 4

_ (5)(s4p!f)(1s.s)4(123) = 0_389 in < _1_ = (15.s')c12) = 0_775 in OK,_ TL - 384EI - (384)(1.3X106)(17Bin4) 240 (240)

*USE 1.5" X 12" 1.3E Timberstand LSL Floor

W/ moisture content less than 16%@ 16" o.c for up to a max span of 15'-6"

Location of Floor Joist

;ri� ..... ®�

vs"� ;o �,, g, °lL f!

°lL •, h�

p�

(II) ll.00 kJOISTBQ \!1' OJl .

§ fR - ;o � c::11::::= :s�

'"ii ciJ ¼'x9}( 1.3E !2 II TIM ERSTRAND LSL

F OORJOJSTS =! @16"0.C. " ,,

,� .. 0 � � ;o "JL

��

Wx12" 1.3E � � ii kl

TIM BER STRAND - � ��ISTS@12'

0 ..

·� (Ji 1

GD-7

]

Floor Beam Design

Design Floor Beam - 1:

Maximum Span of 9'-0"

Floor Dead Load = 13.0 psf Floor Live Load = 40.0 psf Partition Dead Load = 10.0 psf

( 15ft 6ft) w= (13.0psf +40.0psf +10.0psf) 2

+ 2

= 662lbl ft

P = 3,025 lb

P=3,025 lb

w=662Ib/ft

k---4''---J'------ ----

Trial: 8" x 12":

Vmax = 5,566 lb M max = 16,967 lb-ft

f' -- l.5(VMAX) (1.5)(5,566/b) F' O OK102.3 p

si < , = 18 psi� ' A (7 .25in )(l 1.25in)

f' = M MAX = (l 6•9671h - ft)(lZ)(6) = 1331 si < F' = 1500 si � OK

b S (7 .25in )(l l.25in )2 ' p b p

I I /',. TL= -<-�OK814 240

Provide 8" x 12" Douglas-fir Select Structural Floor Beam with C M <16% for up to a maximum span of 9'-0"

GD-8

17 I I . I

Design Floor Beam - 2:

Maximum Span of 11 '-0"

Floor Dead Load = 13.0 psf Floor Live Load = 40.0 psf Partition Dead Load = 10.0 psf

(15.fi 8in. ) w=(l3.0psf+40.0psf+l0.0psf) -+-. =515/blji

2 12zn.

According to Beam - 3:

P = 424 lb

P=424 lb

,,,_1, __ __,,,,5•·-----,l'J----h6'---7

,064 lb

489 lb

3,025 lb

8,884.1 lb-ft

M (lb-ft) IL-----------_.::,,.

JJL--' ____..,5. 8 73�· ______,,J

GD-9

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Trial: 6" x 1 0":

Vmax = 3,064 lb Mmax = 8,884.1 lb-ft

f , = I.5(VMAX) = (I.5)(3,0641b) = 88 si < F' = 180 si � OK v A (52.25in 2 ) p v p

f'. = M MAX = (8,884· llb � 1) (l2) = l,289

p si < F'• = 1500(1.0) = l500psi � OKS (121.2zn )

CL = 1.0 ( d / b :,;2)

ti = (5)(515/b/ .ft)(ll.ft)4 (12)

+ (424lb)(ll.ft)(l2) = 0_23in. + O.OO l = 0_23 lin. TL 384(1.9xl06 )(393in 4) 48(1.9xl06 )(393in 4)

. l l

tin= 0.23lzn. = -<-� OK 517 240

GD-10

Provide 6" x 1 0" Douglas-fir Select Structural Floor Beam with CM <16% for up to a maximum span of 11 '-0"

[]

First and Second Floor King Stud

Species and Commercial Grade

Douglas Fir-Larch Sawn Lumber Stud

Loads on Stud (1 st

Floor)

Roof Dead Load: D, = 23.0 psf = 24.3 psf (20'-7")

19'-6"

Floor Dead Load: D1 = 13.0 psf

Exterior Wall Dead Load: Dw = 17.5 psf

Floor Live Load: L1 = 40 psf

Design Loads on Stud (1 st

Floor)

Roof Dead Load: D, = 24.3 psf (35' / 2) = 425 plf

Floor Dead Load: D1 = 13.0 psf (14.3' / 2) = 93.0 plf

Exterior Wall Dead Load: Dw = 17.5 psf (9.1') = 159 plf

Floor Live Load: L1 = 40.0 psf (14.3' / 2) = 286 plf

Roof Live Load: L, = 20.0 psf (35' / 2) = 350 plf

Horizontal Wind Load: W = 13.0 psf (16" x 1'/12") = 17.3 plf (Applied at east corner)

Considerations for Live Load Reduction: ASCE 07-05 Table 4-2

KLL = 4 (Exterior columns without cantilever slabs}

Tributary Area: Ar = (14.3' I 2) (16" x 1' / 12") = 9.53 sq ft

KLLAr = 4 (9.53 sq ft)= 38.1 sq ft KLLAr = 38.1 sq ft < 400 sq ft (No reduction is necessary)

GD-18

Considerations for Roof Live Load Reduction: ASCE 07-05 Section 4.9

R1 = 1.0 F = 4/12 = 1/3<4 R2 = 1.0

L r = Lo R1 R2 = 20 psf

(No reduction is necessary)

Design Load

D = 425 + 93.0 + 159 = 677 plf L = 286 plf Lr= 350 plf W = 17.3 plf

D + L = 677 + 286 = 963 plf D + Lr = 677 + 350 = 1027 plf D + 0.75L + 0.75Lr = 677 + 0.75 (286) + 0.75 (350) = 1154 plf � governs D + W = 677 + 17 .3 = 694 plf D + 0.75W + 0.75L + 0.75Lr = 677 + 0.75 (17.3) + 0.75 (286) = 904.5 plf

Governing Design Load for Compression:

P = 1154 plf = 1154 ( 16" X 1' / 12")

= 1539Ib

Design Values for Douglas-Fir Larch (2" - 4" thick)

Bending: Fb = 700 psi Compression: Fe = 850 psi Modulus of Elasticity: Emin = 510,000 psi

Adjustment Factors

Load Duration Factor: Co = 1.0 Wet Service Factor: Cm = 1.0 Temperature Factor: Ct = 1.0 Beam Stability Factor: CL = See calculations that follow Size Factor: CF = 1.0 Flat Use Factor: C1u = 1.0

GD-19

Incising Factor: Ci = 1.0 Repetitive Member Factor: C, = 1.15 Column Stability Factor: Cp = See calculations that follow

Trial Design

Try 2 x 6 DF-L Studs @ 16" OC

Section Properties: b = 1.5 in d = 5.5 in A= 8.250 in2 Sxx = 7 .563 in3

Axial Compression Check

Design Requirement: fc ::;Fe' fc = P / A F c' = F c ( Co Cm C1 Ct Ci Cp )

Effective Study Length, le: I= 8.1 ft Ke= 1.0 le= Ke I= 1.0 (8.1 ')(12"/1 ') = 97.2 in le Id = 97.2" I 5.5" = 17.7 < 50 (Slenderness ratio is OK)

Column Stability Fa ctor: F' = Fe ( Co Cm Ct C1 Ci)

= 850 psi ( 1.0 X 1.0 X 1.0 X 1.0 X 1.0 ) = 850 psi

F _ 0.822E'nun

CE - (ljd)2 0.822(510,000psi)

= 1338 . ( )2 psi 17.7 FCE / F:

1338/850 = 1.968 C 0.8

(c = 0.8 for sawn lumber)

GD-20

'1 I , I

= 1.609- �(1.609 ) 2 -1.968

= 0.82

Fe' = F• Cp = 850 psi (0.82) = 697 psi

n fe --

P -- 1539lb A 8.250in2

187 psi

Fe'= 697 psi> fe = 187 psi ( Capacity > Demand OK )

Bending Check

Governing Design Load for Bending: W = 17.3 plf

Mmax = wL 2

= (l?.3plf)(9.lft) 2

= 179 lb-ft= 2148 lb-in 8 8

Design Requirement: fb �Fb'

fb = Mmax = 2148lb- i n_ 284 psi

S xx 7.563in 3

Effective Length: I n / d = 9.1'(12"/1){5,,= 19.85 "?.7 le = 1.63 lu + 3d

= 1.63 (9.1 ') + 3 (5.5" X 1 '/12") = 16.2 ft = 194 in

R8 = {z;d = (l 94)(S.5) = 26.67 < 50

'Vb' 1.5 (Slenderness ratio is OK)

Fb. = Fb ( Co Cm Ci Cr C; C,) = 700 psi ( 1.0 X 1.0 X 1.0 X 1.0 X 1.0 X 1.15 ) = 805 psi

Beam Stability Factor:

F _ 120E' min _ 120(510,000 psi)CE - - (Rb )' (26.67)

2 860 psi

GD-21

l+FbE/F/ = 1+860/805 = 1.09 1.9 1.9

= 1.09- (1.09)2 860/805 0.95

= 0.838

Fb ' =F· CL 805 psi (0.838) = 675 psi

Fb' = 675 psi > fb = 284 psi ( Capacity > Demand OK )

Combined Bending and Compression Check

[fc ] 2

fb _ [ 187psi]

284psi Fe' + p •[i-( ¾cE)] - 697psi + 675psi[l-(187psi/1338psi)]

= 0.563 �1.0 (Bending and compression is proportionate OK)

GD-22

Use 2 x 6 Douglas-Fir Larch king stud for entire wall framing structure.

Note: Corner stud design is considered to be conservative. Use same size studs throughout the 1 st and 2nd floor for ease of construction.

Design of Concrete Footing

Loads:

wDL = (13.0ps/Xl8ft)+(16.0ps/XI7 ft)+ (23.Sps/XlS.fi)= 930plf Wu =(40.0psfXI8ft)=720plf Wn = 930 plf + 720 plf = 1,650 plf

Calculation:

q. = 1,SOOpsf(I.B.C.2006) J; = 2,500psi f, = 60ksi

Assume 12in. thick footing (d=8.5 in)

q, =1500-G�)(lso)=1275ps/ Use B =18in

B = 16SOplf =l.3ft=15.5in

1275psf

Longitudinal steel:

A,= (o.001s)(lsX12)= 0.389in'

Check seismic loads:

Use 4- # 4 Bars

GD-23

Overturning Moment: M = (1500 lb) x (10 ft)= 15,000 lb - ft

Equivalent Eccentricity: e = lS,OOO lb - ft/7,600 lb= 1.97 ft

The kern limit for footing: k = lO ft/6 = 1.67 ft

Distance of load from end of footing: D = Sf t - 1.97 ft = 3.03 ft Therefore: x = (3)(3.03 ft)= 9.09 ft

P (2)(7600 lb)

= ��----'-- =1 115 psf (1.5 ft)(9.09 ft)

P < qa --> a. k.

GD-24

• Provide 12 inches thick and 18 inches wide wall footing for all bearing walls with 2- # 4 bars at top and bottom of the footing ( 4-total). No transverse steel is required, because entire base is within 45° of frustum. Where shear wall occur, extend footing 2ft min. on each side of boundary chord.

Design Spread Footing

PDL = l 0,000/b

Pu =5,000/b

J; = 2500psi f, = 60000 psi q

a = l500psf -(per.I.B,C-2006)

( 6") q, =1500- lz'' (150)=1425ps/ B=

10000+5000 1425

= 3.24ft = 42in

Two-way shear:

b 0

= (4X5.5 + 8.5) = 56in = (1.2X10000)+(1.6X5ooo) = 1633 sifq. 12.25 p

v.2 = (12.25 - (1. 17 )2 \1633) = 17769 /b d

17769 2 0· 35· k = ( X =x )

= . zn < . zn _._ o .. 0.75 4-v2500 42

One-way shear:

v. 1

= ( 3.5X2.6X1633) = 14861/b

Use D= 12in

14861) . . . d = ( = ) =4.3hn <8.5zn . . o.k.

0.75 2-v2500 42 use-D =l2in

µ" = (1.52X3.5X1633{1.�2) = 66o31b- ft �

= (12X6603) =29 01 si¢bd2 (o.9X42X8.s)2 · P

p < Pm;n For flexure 20 3.J2500 Use p = larger of -- = 0.0033or--- 0.0027460000 60000

A, = (0.0033 X42X8.5) = l.18in 2 Use 6- #4 bars in both directions.

Development length: C = 6in = 3in2 c + k12 3 60000 (1.0 Xr.o Xr.o Xr.o) 32 86d. db = 40 X .J2500 X 2.5 = . zameterS I d

As,,q,fred (32 86{1.18) 31 27d' t -x--'--- = . - = . zame ersdb As 1.24 I d

= (31.27X0.625) = 16.2in > l d available = 42in - 5,5in -3in = 15.25in2 2

Use: B = 45in . . 45 5.5 I

d = 16.2zn < ld avazlable = ----3 = 16.75 :. o.k.2 2

GD-25

Provide 45 inches square spread footing 12" deep below grade with 6 - #4 bars in eachway centered below post and 3" min. concrete cover over reinforcing steel.

LS-1

Roof Diaphragm N-5 Design

4'-4" 30'-8 "

5'-0"

Loading:

iii 1'1'1'1'1'1'1'1'1'iW2= 320.4 lb/� � = 252 lb/ft Shear Diagram:

V, V2 V,

I 4.9)3.3 lb 1,805 lb r-------__

r-------.._ I ------------

�Sib �

4,913.3 lb

Moment Diagram:

37,672.9 lb-ft 6,468.Slb-ft �

�

Calculations (in ASD):

Area =(44.5'x 45')-(9.5'x14.33') = 1,866 ji2 Reference Seismic Load DistributionQ

Seismic shear = 13,443.48 lb

13,443.48/b 7 21b/1,866.2 Ji 2 . I Ji2

cu, =(7.2¾2)35fi)=252¾ W2 =( 7.2¾2 )(44.5ji)= 320.4% Diaphragm Section 1:

Shear:

wl ( 252¾2 j(14.33')Vi-2 =2 = 2 Vi-2 = 1,805 lb Moment:

w/2 ( 252¾2 }14.33')2M,_2 = -8-= 8 M1_2 =6,468.5 /b- ft Unit Shear:

V = Ji; = 1,805 lb = 51,6 /b/

l 35ft /Ji

Column C Supporting Garage Beams A and B

Structural Species and Commercial Grade

Sawn Lumber Douglas-fir Larch No. 1

Reference Design Values (NDS Table 4D. DF-L. No. 1 Posts)

Bending About X-X Axis: Fb = 1200 psi Ft = 825 psi Fv = 170 psi F c-1- = 625 psi Fe = 1000 psi E = 1.6 x 106 psi Emin = 580,000 psi

Adjustment Factors (NDS Sec. 4.3.1) Co = 1.0 Load duration factor CM = 1.0 Wet service factor C1u = 1.0 Flat use factor C; = 1.0 Incising factor C, = 1.0 Repetitive member factor Ct = 1.0 Temperature factor C: Sawn Lumber = 0.80 CF and C

p will be calculated later

Column Design Attempts

Loads on Column

Pu = PA+ Ps = ½ Vmax (from Beam A)+½ Vmax (from Beam B) = (12,028 I 2) lb+ (10,215 / 2) lb =11,121Ib

Calculations

Try 3 x 6

Section Properties L = lu = 8.1 ft b = 2.5 in d = 5.5 in A = 13.75 in2

I x = 34.66 in 4

Length Width Depth Area Moment of Inertia

M-15

fl I ' ' !

Size Factor for Compression

CF = (12/d)119 = (12 / 5.5)119 = 1.09 -,.

Column Capacity

Use 1.0

(�) =(K,l ) =((1. 0)(8.lft)�l2in/ ft))= 17_67 < 50 d "'" d y 5.Szn

n Column Stability Factor

E'min = Emin (CM)(Ct)(CT)(Ci) = 580,000 psi (1.0)(1.0)(1.0)(1.0) = 580,000 psi

F _ 0.822E'min

eE -

(l,ld)2

= 0.822(580,000 psi) = 1526 psi(97.2in 15.Sin )2

F\ = Fe (Co)(CM)(Ct)(CF)(Ci) = 1000 psi (1.0)(1.0)(1.0)(1.0)(1.0) = 1000 psi

F,E = 1526psi _ 1 _526 F'c l000psi

1 + F,E I F'c =

1 + 1.526 1.579 2c 2(0.8)

Cp = 1 + F cE I F'c _ (1 + F,E Ip',

J ' F,E I F'c

2c 2c c

= 1.579 - (1.579 )' - l.SZ6 = 0.814 0.8

F'c = Fe (Co)(CM)(Ct)(CF)(Cp)(Ci) = 1000 psi (1.0)(1.0)(1.0)(0.814)(1.0) = 814 psi

M-16

(NOS Eq. 4.3-1)

(NOS Table 4.3.1)

(NOS Section 3.7.1)

(NOS Table 4.3.1)

(NOS EQN. 3.7-1)

(NOS Table 4.3.1)

[]

Compression Check

Pa11ow = F'cA = (814 psi) (13.75in

2 )

= 11,193 lb

Pallow = 11,193 lb > Pu= 11,122 lb OK

Provide 3 x 6 DF-L No. 1 for Column C supporting Beam A and Beam B

M-17