Structural Analysis by Hand - VBCOA · PDF fileStructural Analysis by Hand 2 Presenter ......

1

VBCOA – Region 5

May 15, 2014

Structural Analysis by Hand

2

Presenter

Brian Foley, P.E.Fairfax County

Deputy Building Official

[email protected]

703‐324‐1842

3

Logistics

Exits Restrooms

Cell Phones No Smoking

2

4

Questions?

SURE– Go ahead and ask your question!

This class is interactive!

There are no stupid questions!

5

Attendees

■ Inspector

■ Plan reviewer

■ Contractor

■ Engineer

■ Architect

■ Designer

■ Suppliers

■ Attorney

6

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

3

7

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

8

Math 101

9

Operation Sequence

Please Excuse My Dear Aunt Sally

Parenthesis

Exponent

Multiplication

Division

Addition

Subtraction

4

10

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

11

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

12

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

Hei

ght

5

13

You Try It, Find V

Values

2’ w = 3’

h = 6”

14

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

15

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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16

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

17

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

18

Joist/Beam Analysis

7

19

Sample First Floor Framing Plan

2x8 joists

16"

16"

(2)

1¾x9

½ M

icro

lam

Hem-Fir #2

4'-0" 10'-0"

12'-0

"

20

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

21

Step 2: Determine uniform live load (psf)

Live Load: weight produced byuse and occupancy

People

Furniture

Vehicles

Units: pound per squarefoot (psf)

IBC Table 1607.1

IRC Table R301.5

8

22


For floors: IRC Table R301.5

40

23


For roofs, greatest of live load or snow load

Snow load: IRC Section R301.2.3

Northern Virginia counties 20 – 30 psf

24


Roof live load: IRC Table R301.6

9

25

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 of the beam

26


16"

16"

'33.112

16

"162

16

2

16

feettoConvert

TW

EXAMPLE: JOIST

8" 8"

Tributary width

27


EXAMPLE: BEAM

4'-0" 10'-0"

12'-0

"

Tributary width

10

28


EXAMPLE: BEAM

2' 5'

'72

10

2

4TW

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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 plfw = 66.67 plf

BEAM: wLL = (40)(7) = 280.0 plfwDL = (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

11

31

Free Body Diagram

Joist

66.67 plf

10’-0”

Beam

350 plf

12’-0”

w

lR R

32

Example: Free Body Diagram

On plans provided, first floor joist adjacent fireplace hearth extension:

15’-4”

33

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”

12

34

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

36


“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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37


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

38

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

EXAMPLE:

2

6

1bdS

b

d

32 in6.52)5.9)(75.1(6

1)2(:

SBEAM

Section Modulus:

32 in1.13)25.7)(5.1(6

1: SJOIST

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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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40

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

l

w

8

2wlM where: l = span length, ft

w = total linear load, plfM = moment, lbs-ft

41

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

where: l = span length, ftw = total linear load, plfM = moment, lbs-ft

JOIST: w = 66.67 plfl = 10'

BEAM: w = 350 plfl = 12'

ftlbs4.8338

)10)(67.66( 2

M

EXAMPLE:

8

2wlM

ftlbs300,68

)12)(350( 2

M

42

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, psiS

Mfb

12

15

43

Step 5D: Calculate fb (psi)

where: S = section modulus, in3

M = moment, lbs-ft

fb = actual bending stress, psiS

Mfb

12

JOIST: M = 833.4 lbs-ftS = 13.1 in3

BEAM: M = 6,300 lbs-ftS = 52.6 in3

psi4.7631.13

)4.833)(12(bf

EXAMPLE:

psi3.437,16.52

)300,6)(12(bf

44

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 experiencesa slicing action invertical plane

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46

Step 6A: Determine F’v (psi)

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

specified structural member

Units for stress:


lbs/in2

psi

47


Allowable shear stress

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

F’v =Fv CM CD, where:


CD = Load duration (normal or snow)

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Use tables based on:

Species


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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49

Step 6B: Calculate Area, A (in2)

EXAMPLE:

b

d

2in9.10)25.7)(5.1(:JOIST A

2in2.33)5.9)(75.1)(2(:BEAM A

A=bd

Area:

50

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, ftw = total linear load, plfd = member depth, in

122

dlwV

JOIST: w = 66.67 plfl = 10'd = 7.25"

BEAM: w = 350 plfl = 12'd = 9.5"

lbs0.29312

25.7

2

1067.66

V

EXAMPLE:

lbs9.822,112

5.9

2

12350

V

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52

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, psiA = area, in2

V = shear, lbsA

Vfv 2

3

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where: A = area, in2

V = shear, lbs

fv = actual shear stress, psiA

Vfv 2

3

JOIST: V = 293.0 lbsA = 10.9 in2

BEAM: V = 1,822.9 lbsA = 33.2 in2

psi4.82)2.33)(2(

)9.822,1)(3(vf

EXAMPLE:

psi4.40)9.10)(2(

)0.293)(3(vf

Step 6D: Calculate fv (psi)

54

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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55

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:

"33.0360

)10)(12(max

"40.0360

)12)(12(max

360

12max

l

20

58

Step 7C: Calculate I (in4)

EXAMPLE:

3

12

1bdI

b

d

43 in250)5.9)(75.1(12

1)2(:

IBEAM

Moment of Inertia:

43 in6.47)25.7)(5.1(12

1: IJOIST

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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:


lbs/in2

psi

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61


Modulus of elasticity

“Raw” value based on wood species: E

Adjusted modulus of elasticity: E’

E’ =E CM, where:


62


Use tables based on:

Species


EXAMPLE:

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

Beam: Microlam, dry: E’ = 1.9 x 106 psi

63

Step 7E: Calculate actual deflection (in)

where:

∆act = actual deflection

l = span length, ftwLL = linear live load, plfE = modulus of elasticity, psiI = moment of inertia, in4

EI

lwLLact

45.22

22

64

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

JOIST: wLL = 53.3 plfl = 10'E = 1.3x106 psiI = 47.6 in4

BEAM: wLL = 280 plfl = 12'E = 1.9x106 psiI = 250 in4

"194.0)6.47)(10x3.1(

)10)(3.53)(5.22(6

4

act

EXAMPLE:

"275.0)250)(10x9.1(

)12)(280)(5.22(6

4

act

65

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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67

Shortcuts…“Dirty numbers”

Span length: no need to be exact

More than one live load on a member?

Use value of live load member mostly sees, or

use highest value

Round values up

Drop decimals

68

Examples – Go to the Plans

69

Example: Garage Rafters

Go to Sheet S4 on the provided plans

Do rafters comply?

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70


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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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

ftlbs059,18

)11)(70( 2

M

psi9671.13

)059,1)(12(bf

72


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 lbs34312

25.7

2

1170

V

25

73


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

psi2.47)9.10)(2(

)343)(3(vf

"733.0180

)11)(12(max

74


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!

"25.0)6.47)(10x1.1(

)11)(40)(5.22(6

4

act

RAFTER PASSES!

75

Example: Second Floor Header

Go to Sheet S3 on the provided plans

Does header comply?

26

76


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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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

ftlbs000,78

)10)(560( 2

M

psi8868.94

)000,7)(12(bf

78


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 lbs275,212

25.11

2

10560

V

27

79


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

psi67)7.50)(2(

)275,2)(3(vf

"333.0360

)10)(12(max

80


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!

"16.0)534)(10x1.1(

)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?

28

82

JOISTS1. 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 psfLL = 40 psf

F’b = 1,070 psiS = 13.1 in3

ftlbs250,18

)10)(100( 2

M

wDL= 2 x 10 = 20 plfwLL = 2 x 40 = 80 plf

w = 100 plf

l = 10’

TW = 2’

83

JOISTS

5D.Actual bending stress:

5E.Compare:

psifb 6.141,1

1.13

250,112

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

JOIST FAILS!

84

BEAM1. Uniform dead load:2. Uniform live load3. Tributary width:

4. Linear load:

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

DL = 10 psfLL = 40 psf

wDL= 5 x 10 = 50 plfwLL = 5 x 40 = 200 plf

w = 250 plfF’b = 1,070 psi

S = (3)(13.1) = 39.3 in3

ftlbs335,38

)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:

psifb 016,1

3.39

335,312

fb = 1,016 psi < F’b = 1,070 psi OK

F’v = 135 psiA = (3)(10.9) = 32.7 in2

lbs 140,112

25.7

2

33.10250

V

86

BEAM

6D.Actual shear stress:

6E.Compare:

7A.Allowable deflection:

7B.Moment of inertia:7C.Modulus of elasticity:

psifv 5.52

7.322

140,13

fv = 52.5 psi < F’v = 135 psi OK

I = (3)(47.6) = 142.8 in4

"344.0

360

33.1012max

E’ = 1.1x106 psi

87

BEAM

7D.Actual deflection:

7E.Compare:

"326.0)8.142)(x101.1(

)33.10)(200)(5.22(6

4

max

Δact = 0.326" < Δmax = 0.344” OK

BEAM PASSES!

30

88

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)

'132

75.12

2

25.13TW

93

Step 5: Calculate Actual Linear Load (plf)

First floor:

Second floor:

Total:w = 1,170 plf

klf 650)1040)(13(1 w

plf 520)1030)(13(2 w

32

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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: CompareActual 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 plfw2= 14 x (30 + 10) = 560 plf

1,260 plf

Actual = 12 x 1,260 = 15.1 kips < 20 kips OK!1,000

'142

33.15

2

75.12TW

99

99

Steel Beams

34

100

Spread Footings

Soil has maximum capacity to withstand pressure

Footing distributes point load to soil based on 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

wlR where: l = span length, ft

w = total linear load, plfR = 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 605,72

)13)(170,1(1 R

lbs 945,92

)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

PQact

lbs 097,1)4)(4(

550,17actQ

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 648,132

)12)(170,1(

2

)33.11)(170,1(P

psf 184,2)5.2)(5.2(

648,13actQ

The End

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