Structural Analysis by Hand VBCOA Region 5 May 15, 2014 Presenter - - PDF document

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VBCOA – Region 5 May 15, 2014

Structural Analysis by Hand

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Presenter

Brian Foley, P.E.

Fairfax County Deputy Building Official brian.foley@fairfaxcounty.gov 703‐324‐1842

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Logistics Exits Restrooms Cell Phones No Smoking

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Questions?

SURE– Go ahead and ask your question! This class is interactive! There are no stupid questions!

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Attendees

■ Inspector ■ Plan reviewer ■ Contractor ■ Engineer ■ Architect ■ Designer ■ Suppliers ■ Attorney

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Objectives

 Determine loading requirements for joists and

beams

 Apply loads using free body diagrams  Calculate moment and deflection  Analyze compliance for flexure, shear and

deflection

 Analyze simple steel beam  Analyze spread footings

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Assumptions

 No structural theory  Knowledge of arithmetic

and basic algebra

 Simple loading

 Uniform loads  Point loads at midspan

 Simple spans  Wood, LVL, steel  Residential construction

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Math 101

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Operation Sequence Please Excuse My Dear Aunt Sally

 Parenthesis  Exponent  Multiplication  Division  Addition  Subtraction

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P.E.M.D.A.S.

52 + (3 – 1) / 2 – 4 x 3

Step 1: Parenthesis 52 + 2 / 2 – 4 x 3 Step 2: Exponents 25 + 2 / 2 – 4 x 3 Step 3: Multiply 25 + 2 / 2 – 12 Step 4: Divide 25 + 1 – 12 Step 5: Add 26 – 12 Step 6: Subtract 14

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Mathematical Formulas

 Volume of a box  Length x Width x Height  Variables

 L = Length  W = Width  H = Height  V= Volume

 V = L x W x H  V = LWH

Height

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Mathematical Formulas

 V = LWH  Values

 L = 5’  W = 2’  H = 6’

 V = 5 x 2 x 6  V = (5)x(2)x(6)  V = (5)(2)(6)  V = 60 feet3

Height

5

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You Try It, Find V



•  Values

 2’  w = 3’  h = 6”

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You Try It, Find V



•  Values: 2’, w 3’, h 6”

 Convert h value from inches to

feet: 6” = 0.5’

 Insert values into formula: 2

3 0.5 1 ft3

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A Word About Units

 Always include units with every

calculation

 Ensure all calculations are

completed in the same units (inches, feet)

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Vertical Load Path

Vertical load path transfers gravity load (snow): ■ to roof sheathing ■ to rafters ■ to top plate ■ to studs ■ to bottom plate ■ to foundation & footings ■ to ground

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Design Methodologies

Load & Resistance Factor Design (LRFD)



Applied loads adjusted up



Resistance capacity of structural member adjusted down



Compare values: capacity > loads Allowable Stress Design (ASD)



Actual stress calculated using applied loads



Structural member’s allowable stresses calculated



Compare values: allowable > actual

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Joist/Beam Analysis

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Sample First Floor Framing Plan

2x8 joists 16" 16" (2) 1¾x9½ Microlam Hem-Fir #2 4'-0" 10'-0" 12'-0"

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Step 1: Determine uniform dead load

 Units for uniform dead load

 pounds per square foot  lbs/ft2  psf

 Dead Load: weight of

structure

 Assume 10 PSF for floors  Assume 15 PSF for roofs

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Step 2: Determine uniform live load (psf)

 Live Load: weight produced by

use and occupancy

 People  Furniture  Vehicles

 Units: pound per square

foot (psf)

 IBC Table 1607.1  IRC Table R301.5

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Step 2: Determine uniform live load (psf)

 For floors: IRC Table R301.5

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Step 2: Determine uniform live load (psf)

 For roofs, greatest of live load or snow load  Snow load: IRC Section R301.2.3

 Northern Virginia counties 20 – 30 psf

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Step 2: Determine uniform live load (psf)

 Roof live load: IRC Table R301.6

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Step 3: Calculate tributary width (feet)

 Load influence distance from each side of a

framing member

 Joists: half the distance to next adjacent joists on

each side

 Beams: half of the joists’ span that bear on each side

• f the beam

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Step 3: Calculate tributary width (feet)

16" 16"

' 33 . 1 12 16 " 16 2 16 2 16      feet to Convert TW

EXAMPLE: JOIST

8" 8"

Tributary width

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Step 3: Calculate tributary width (feet)

EXAMPLE: BEAM

4'-0" 10'-0" 12'-0" Tributary width

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Step 3: Calculate tributary width (feet)

EXAMPLE: BEAM

2' 5'

' 7 2 10 2 4    TW

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Step 4: Calculate linear load (plf)

 w = uniform load x TW  Units: pounds per linear foot = lbs/ft = plf  EXAMPLE:

JOIST: wLL = (40)(1.33) = 53.33 plf wDL = (10)(1.33) = 13.33 plf w = 66.67 plf BEAM: wLL = (40)(7) = 280.0 plf wDL = (10)(7) = 70.0 plf w = 350.0 plf

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Free Body Diagram

 A two‐dimensional graphic symbolization of a

structural member which models bearing locations and loading elements

 Linear load, w:  Point load, P:  Bearing locations (reaction), R:

100 plf 500 lbs

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Free Body Diagram

Joist

66.67 plf 10’-0”

Beam

350 plf 12’-0” w l R R

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Example: Free Body Diagram

 On plans provided, first floor joist adjacent

fireplace hearth extension:

15’-4”

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You Try It

 Draw the free body diagram of the sunroom

beam

 Show load and its value

Total uniform load = 50 psf Tributary width = 5’ Total linear load = (50)(5) = 250 plf

 Show span length

250 plf 10’-4”

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Step 5: Bending Analysis

 Flexure, bending, moment, torque  Highest at midspan for uniform load

Pulling stress or tension on bottom face of member

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Step 5A: Determine F’b (psi)

 Allowable bending stress, F’b  The maximum bending stress permissible for a

specified structural member

 Units for stress:

 pounds per square inch  lbs/in2  psi

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Step 5A: Determine F’b (psi)

 “Raw” value based on wood species: Fb  Adjusted allowable bending stress,

F’b = Fb CM CF Cr CD, where:

 CM = Service condition (wet or dry)  CD = Load duration (normal or snow)  Cr = Repetitive use (joists, 3+ply beams)  CF = Member size (2x?)

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Step 5A: Determine F’b (psi)

 Use tables in chart based on:

 Species  Service condition (wet or dry)  Load duration (normal or snow)  Single (2‐ply beam) or repetitive (joists)  Member size

 EXAMPLE:

 Joists: Repetitive, dry, 2x8, normal load duration,

Hem‐Fir#2: F’b = 1,173 psi

 Beam: Single, dry, (2)1¾x9½, normal load duration,

Microlam: F’b = 2,684 psi

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Step 5B: Determine b (in), d (in), S (in3)

EXAMPLE:

2

6 1 bd S 

b d 3 2

in 6 . 52 ) 5 . 9 )( 75 . 1 ( 6 1 ) 2 ( :         S BEAM Section Modulus:

3 2

in 1 . 13 ) 25 . 7 )( 5 . 1 ( 6 1 :   S JOIST

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EXAMPLE: JOIST:

S=13.1 in3

BEAM:

S=(2)(26.3) =52.6 in3

Step 5B: Determine S (in3)

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Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)

l w

8

2

wl M 

where: l = span length, ft w = total linear load, plf M = moment, lbs-ft

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Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)

where: l = span length, ft w = total linear load, plf M = moment, lbs-ft

JOIST: w = 66.67 plf l = 10' BEAM: w = 350 plf l = 12' ft lbs 4 . 833 8 ) 10 )( 67 . 66 (

2

   M EXAMPLE:

8

2

wl M 

ft lbs 300 , 6 8 ) 12 )( 350 (

2

   M

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Step 5D: Calculate fb (psi)

 Actual bending stress, fb  The bending stress a specified structural member

is experiencing under maximum applied load

where: S = section modulus, in3 M = moment, lbs-ft fb = actual bending stress, psi

S M fb 12 

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Step 5D: Calculate fb (psi)

where: S = section modulus, in3 M = moment, lbs-ft fb = actual bending stress, psi

S M fb 12 

JOIST: M = 833.4 lbs-ft S = 13.1 in3 BEAM: M = 6,300 lbs-ft S = 52.6 in3 psi 4 . 763 1 . 13 ) 4 . 833 )( 12 (  

b

f

EXAMPLE:

psi 3 . 437 , 1 6 . 52 ) 300 , 6 )( 12 (  

b

f

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Step 5E: Compare F’b with fb

If fb ≤ F’b, then member is good for bending EXAMPLE: JOIST: fb = 763.4 psi < F’b = 1,173 psi OK! BEAM: fb = 1,437.3 psi < F’b = 2,684 psi OK!

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Step 6: Shear Analysis

 Shear is similar to a cutting stress  Highest at ends = reaction  Wood shear analysis uses shear value at a

distance from the end equal to member’s depth

Member experiences a slicing action in vertical plane

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Step 6A: Determine F’v (psi)

 Allowable shear stress, F’v  The maximum shear stress permissible for a

specified structural member

 Units for stress:

 pounds per square inch  lbs/in2  psi

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Step 6A: Determine F’v (psi)

 Allowable shear stress  “Raw” stress based on wood species: Fv  Adjusted allowable shear stress: F’v ,

F’v =Fv CM CD, where:

 CM = Service condition (wet or dry)  CD = Load duration (normal or snow)

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Step 6A: Determine F’v (psi)

 Use tables based on:

 Species  Service condition (wet or dry)  Load duration (normal or snow)

EXAMPLE:

Joist: Dry, Hem‐Fir#2: F’v = 150 psi Beam: Dry, Microlam: F’v = 285 psi

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Step 6B: Calculate Area, A (in2)

EXAMPLE:

b d 2

in 9 . 10 ) 25 . 7 )( 5 . 1 ( : JOIST   A

2

in 2 . 33 ) 5 . 9 )( 75 . 1 )( 2 ( : BEAM   A A=bd

Area:

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Step 6B: Calculate Area, A (in2)

EXAMPLE: JOIST:

A=10.9 in2

BEAM:

S=(2)(16.6) =33.2 in2

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Step 6C: Calculate Shear, V (lbs)

where: l = span length, ft w = total linear load, plf d = member depth, in

        12 2 d l w V

JOIST: w = 66.67 plf l = 10' d = 7.25" BEAM: w = 350 plf l = 12' d = 9.5" lbs . 293 12 25 . 7 2 10 67 . 66          V

EXAMPLE:

lbs 9 . 822 , 1 12 5 . 9 2 12 350          V

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Step 6D: Calculate fv (psi)

 Actual shear stress, fv  The shear stress a specified structural member is

experiencing under maximum applied load

where: fv = actual shear stress, psi A = area, in2 V = shear, lbs

A V fv 2 3 

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where: A = area, in2 V = shear, lbs fv = actual shear stress, psi

A V fv 2 3 

JOIST: V = 293.0 lbs A = 10.9 in2 BEAM: V = 1,822.9 lbs A = 33.2 in2 psi 4 . 82 ) 2 . 33 )( 2 ( ) 9 . 822 , 1 )( 3 (  

v

f

EXAMPLE:

psi 4 . 40 ) 9 . 10 )( 2 ( ) . 293 )( 3 (  

v

f

Step 6D: Calculate fv (psi)

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Step 6E: Compare F’v with fv

If fv ≤ F’v, then member is good for shear EXAMPLE: JOIST: fv = 40.4 psi < F’v = 150 psi OK! BEAM: fv = 82.4 psi < F’v = 285 psi OK!

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Step 7: Deflection Analysis

 "Sag" a member experiences  Analysis is based on live load only  Highest at midspan

Deflection, ∆

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Step 7A: Determine allowable deflection (in)

 Use Table R301.7 where l = span length of member, in.

(span length must be converted from feet to inches)

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Step 7B: Calculate allowable deflection (in)

EXAMPLE: Floor, JOIST: BEAM:

" 33 . 360 ) 10 )( 12 (

max

   " 40 . 360 ) 12 )( 12 (

max

   360 12

max

l  

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Step 7C: Calculate I (in4)

EXAMPLE:

3

12 1 bd I 

b d 4 3

in 250 ) 5 . 9 )( 75 . 1 ( 12 1 ) 2 ( :         I BEAM

Moment of Inertia:

4 3

in 6 . 47 ) 25 . 7 )( 5 . 1 ( 12 1 :   I JOIST

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Step 7C: Calculate I (in4)

EXAMPLE: JOIST: I= 47.6 in4 BEAM: I=(2)(125) =250 in4

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Step 7D: Determine E (psi)

 Modulus of elasticity, E  The mathematical description of a structural

member’s elastic characteristics

 Units for modulus of elasticity:

 pounds per square inch  lbs/in2  psi

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Step 7D: Determine E (psi)

 Modulus of elasticity  “Raw” value based on wood species: E  Adjusted modulus of elasticity: E’

E’ =E CM, where:

 CM = Service condition (wet or dry)

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Step 7D: Determine E (psi)

 Use tables based on:

 Species  Service condition (wet or dry)

EXAMPLE:

Joist: Hem‐Fir#2, dry: E’ = 1.3 x 106 psi Beam: Microlam, dry: E’ = 1.9 x 106 psi

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Step 7E: Calculate actual deflection (in)

where: ∆act = actual deflection l = span length, ft wLL = linear live load, plf E = modulus of elasticity, psi I = moment of inertia, in4

EI l wLL

act 4

5 . 22  

22

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Step 7E: Calculate actual deflection, Δact (in)

JOIST: wLL = 53.3 plf l = 10' E = 1.3x106 psi I = 47.6 in4 BEAM: wLL = 280 plf l = 12' E = 1.9x106 psi I = 250 in4 " 194 . ) 6 . 47 )( 10 x 3 . 1 ( ) 10 )( 3 . 53 )( 5 . 22 (

6 4

  act

EXAMPLE:

" 275 . ) 250 )( 10 x 9 . 1 ( ) 12 )( 280 )( 5 . 22 (

6 4

  act

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Step 7F: Compare ∆max to ∆act

If Δact ≤ Δmax, then member is good for deflection. EXAMPLE: Joist: Δact = 0.194" < Δmax = 0.333" OK! Beam: Δact = 0.275" < Δmax = 0.400" OK! JOIST PASSES! BEAM PASSES!

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When good beams go bad…

 When one component fails,

the structural member fails

 RULE OF THUMB: the

bending stress is the gauge

 In a uniformly loaded

member, shear will pass if flexure passes

 Deflection is critical when

flexure barely passes

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Shortcuts…“Dirty numbers”

 Span length: no need to

be exact

 More than one live load

• n a member?

 Use value of live load

member mostly sees, or

 use highest value

 Round values up  Drop decimals

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Examples – Go to the Plans

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Example: Garage Rafters

 Go to Sheet S4 on the

provided plans

 Do rafters comply?

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Example: Garage Rafters

 Step 1: Determine uniform dead load, DL = 15 psf  Step 2: Determine uniform live load, LL = 20 psf  Step 3: Determine tributary area, TW = 2’  Step 4: Calculate linear load

 wDL = (2)(15) = 30 plf 

wLL = (2)(20) = 40 plf



w = 70 plf

 Step 5A: Determine F’b

 Repetitive, dry, snow load duration, SPF So. #2  F’b = 1,230 psi

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Example: Garage Rafters

 Step 5B: Determine S

 For 2x8, from chart, S = 13.1 in3

 Step 5C: Determine l, calculate M

 l = 11’  w = 70 plf 

 Step 5D: Calculate

ft lbs 059 , 1 8 ) 11 )( 70 (

2

   M psi 967 1 . 13 ) 059 , 1 )( 12 (  

b

f

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Example: Garage Rafters

 Step 5E: Compare:



fb = 967 psi < F’b = 1,230 psi

 Flexure analysis OK!

 Step 6A: Determine F’v

 Dry, SPF So. #2  F’v = 135 psi

 Step 6B: Determine A (in2)

 For 2x8, from chart, A = 10.9 in2

 Step 6C: Calculate

lbs 343 12 25 . 7 2 11 70          V

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Example: Garage Rafters

 Step 6D: Calculate  Step 6E: Compare:



fv = 47.2 psi < F’v = 135 psi

 Shear analysis OK!

 Step 7A: Determine ∆max

 Roof, slope > 3:12, no ceiling finish  l/180

 Step 7B: Calculate

psi 2 . 47 ) 9 . 10 )( 2 ( ) 343 )( 3 (  

v

f " 733 . 180 ) 11 )( 12 (

max

  

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Example: Garage Rafters

 Step 7C: Determine I

 For 2x8, from chart, I = 47.6 in4

 Step 7D: Determine E’



Dry, SPF So. #2

 E’ = 1.1 x 106 psi

 Step 7E: Calculate  Step 7F: Compare:



∆act = 0.25” < ∆max = 0.733”

 Deflection analysis OK!

" 25 . ) 6 . 47 )( 10 x 1 . 1 ( ) 11 )( 40 )( 5 . 22 (

6 4

  act

RAFTER PASSES!

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Example: Second Floor Header

 Go to Sheet S3 on the

provided plans

 Does header comply?

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Example: Second Floor Header

 Step 1: Determine uniform dead load, DL = 10 psf  Step 2: Determine uniform live load, LL = 30 psf  Step 3: Determine tributary area, TW = 14’  Step 4: Calculate linear load



wDL = (14)(10) = 140 plf



wLL = (14)(30) = 420 plf



w = 560 plf

 Step 5A: Determine F’b

 Repetitive, dry, normal duration, SPF So. #2  F’b = 891 psi

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Example: Second Floor Header

 Step 5B: Determine S

 For 2x12, from chart, S = (3)(31.6) = 94.8 in3

 Step 5C: Determine l, calculate M

 l = 10’  w = 560 plf 

 Step 5D: Calculate

ft lbs 000 , 7 8 ) 10 )( 560 (

2

   M psi 886 8 . 94 ) 000 , 7 )( 12 (  

b

f

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Example: Second Floor Header

 Step 5E: Compare:



fb = 886 psi < F’b = 891 psi

 Flexure analysis OK!

 Step 6A: Determine F’v

 Dry, SPF So. #2  F’v = 135 psi

 Step 6B: Determine A

 For 2x12, from chart, A = (3)(16.9) = 50.7 in2

 Step 6C: Calculate

lbs 275 , 2 12 25 . 11 2 10 560          V

27

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Example: Second Floor Header

 Step 6D: Calculate  Step 6E: Compare:



fv = 67 psi < F’v = 135 psi

 Shear analysis OK!

 Step 7A: Determine ∆max

 Floor  l/360

 Step 7B: Calculate

psi 67 ) 7 . 50 )( 2 ( ) 275 , 2 )( 3 (  

v

f " 333 . 360 ) 10 )( 12 (

max

  

80

Example: Second Floor Header

 Step 7C: Determine I

 For 2x12, from chart, I = (3)(178) = 534 in4

 Step 7D: Determine E’



Dry, SPF So. #2

 E’ = 1.1x106 psi

 Step 7E: Calculate  Step 7F: Compare:



∆act = 0.16” < ∆max = 0.333”

 Deflection analysis OK!

" 16 . ) 534 )( 10 x 1 . 1 ( ) 10 )( 420 )( 5 . 22 (

6 4

  act

HEADER PASSES!

81

You Try It

 Using the house plans provided, do the sunroom

joists and beam comply?

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JOISTS

• 1. Uniform dead load:
• 2. Uniform live load:
• 3. Tributary width:
• 4. Linear load:
• 5A. Allowable bending stress:
• 5B. Section modulus:
• 5C. Span length, moment:

DL = 10 psf LL = 40 psf F’b = 1,070 psi S = 13.1 in3 ft lbs 250 , 1 8 ) 10 )( 100 (

2

   M wDL= 2 x 10 = 20 plf wLL = 2 x 40 = 80 plf w = 100 plf l = 10’ TW = 2’

83

JOISTS

5D.Actual bending stress: 5E.Compare:

  

psi fb 6 . 141 , 1 1 . 13 250 , 1 12  

fb = 1,141.6 psi > F’b = 1070.0 psi NG

JOIST FAILS!

84

BEAM

1. Uniform dead load: 2. Uniform live load 3. Tributary width: 4. Linear load:

• 5A. Allowable bending stress:
• 5B. Section modulus:
• 5C. Span length, moment:

DL = 10 psf LL = 40 psf wDL= 5 x 10 = 50 plf wLL = 5 x 40 = 200 plf w = 250 plf F’b = 1,070 psi S = (3)(13.1) = 39.3 in3 ft lbs 335 , 3 8 ) 33 . 10 )( 250 (

2

   M l = 10.33’ TW = 5’

29

85

BEAM

5D.Actual bending stress: 5E.Compare: 6A.Allowable shear stress: 6B.Area: 6C.Shear:

  

psi fb 016 , 1 3 . 39 335 , 3 12  

fb = 1,016 psi < F’b = 1,070 psi OK F’v = 135 psi A = (3)(10.9) = 32.7 in2 lbs 140 , 1 12 25 . 7 2 33 . 10 250          V

86

BEAM

6D.Actual shear stress: 6E.Compare: 7A.Allowable deflection: 7B.Moment of inertia: 7C.Modulus of elasticity:

     

psi fv 5 . 52 7 . 32 2 140 , 1 3  

fv = 52.5 psi < F’v = 135 psi OK I = (3)(47.6) = 142.8 in4

  

" 344 . 360 33 . 10 12

max

   E’ = 1.1x106 psi

87

BEAM

7D.Actual deflection: 7E.Compare:

" 326 . ) 8 . 142 )( x10 1 . 1 ( ) 33 . 10 )( 200 )( 5 . 22 (

6 4 max

  

Δact = 0.326" < Δmax = 0.344” OK

BEAM PASSES!

30

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Steel Beams

 Wide flange section  A992, high strength  A36, normal strength  Design check for

uniformly loaded beam

 Basement location  Simple spans

89

Step 1: Determine Span Length (ft)

 From plans,

choose longest span of a beam type, W8x18

 For example,

assume 17’

90

90

Step 2: Determine Allowable Load (kips)

31

91

Step 3: Determine Uniform Loads (psf)

 First floor

 Live load = 40 psf  Dead load = 10 psf

 Second floor

 Live load = 30 psf (if sleeping)  Dead load = 10 psf

92

Step 4: Determine Tributary Width (ft)

' 13 2 75 . 12 2 25 . 13    TW

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Step 5: Calculate Actual Linear Load (plf)

 First floor:  Second floor:  Total:

w = 1,170 plf

klf 650 ) 10 40 )( 13 (

1

   w plf 520 ) 10 30 )( 13 (

2

   w

32

94

Step 6: Calculate Actual Total Load (kips)

 Actual Load = (17)(1,170) = 19,890 lbs = 19.9 kips

17’ 1,170 plf

95

Step 7: Compare Actual and Allowable Loads

 Actual = 19.9 kips > Allowable = 14 kips NG!

Try W8x28, Lmax = 23 kips > 19.9 kips, OK! BEAM FAILS!

96

Rules of Thumb

 The deeper the beam the

more it can span

 A steel beam can usually

span 2 feet for every inch of depth

 W8, span≈16’  W10, span≈20’

33

97

You Try It

 Go to plans  Check steel beam in

basement

98

Solution

 W8x18, span length = 12’  Allowable load = 20 kips   Total uniform load:

w1= 14 x (40 + 10) = 700 plf w2= 14 x (30 + 10) = 560 plf 1,260 plf

 Actual = 12 x 1,260 = 15.1 kips < 20 kips OK!

1,000

' 14 2 33 . 15 2 75 . 12    TW

99

99

Steel Beams

34

100

Spread Footings

 Soil has maximum

capacity to withstand pressure

 Footing distributes

point load to soil based

• n maximum capacity

101

Step 1: Determine Presumptive Soil Bearing Pressure, Qmax (psf)

 Determine using Table R401.4.1  Maximum assumable often identified by AHJ  For example, Qmax = 1,500 psf

102

How to Calculate Reactions

l w

2 wl R 

where: l = span length, ft w = total linear load, plf R = reaction, lbs

R R

35

103

Step 3: Calculate Point Load

 Point load = R1 + R2  For example:  P = 17,550 lbs.

1,170 plf 17’ 10’ 13’

lbs 605 , 7 2 ) 13 )( 170 , 1 (

1

  R lbs 945 , 9 2 ) 17 )( 170 , 1 (

2

  R

104

Step 2: Calculate Qact (psf)

 Calculate actual bearing

pressure:

 P = column load, lbs  L = footing length, ft  W = footing width, ft

 For example:

LW P Qact  lbs 097 , 1 ) 4 )( 4 ( 550 , 17  

act

Q

105

Step 4: Compare Qact to Qmax

 Qact= 1,097 psf < Qmax = 1,500 psf, OK!

FOOTING PASSES!

36

106

You Try It

 Go to the plans

provided

 Are the spread

footings supporting the steel columns adequate?

107

Solution

 Soil type = silty sand (SM), Qmax = 2,000 psf   L = W = 2.5'   Qact = 2,416 psf > Qmax = 2,000 psf, NG!

lbs 648 , 13 2 ) 12 )( 170 , 1 ( 2 ) 33 . 11 )( 170 , 1 (    P psf 184 , 2 ) 5 . 2 )( 5 . 2 ( 648 , 13  

act

Q

The End