05 Force Stress

7/29/2019 05 Force Stress

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Force vs stress Copyright Prof Schierle 2011 1

Cause:exter

nalforceP

Effect:internalst

ressf

Uplift Force generates fabric Stress


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Type of Force

1 Axial (tension / compression)

2 Shear 3 Bending

4 Torsion

5 Images

6 Symbol (+ -)

A Tension (elongates +)

B Compression (shortens -)

C Shear (clockwise couple +)D Bending (concave + convex -)

E Torsion (right-hand rule +)

Tension elongatesCompression shortens

Shear tends to slide

Bending (+) top compression

bottom tensionTorsion twists (right-hand +)


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Force vs. Stress

Force P (external action)

Stress f = P/A (internal reaction)

Force P (absolute)

US units pound (lb, #)

kip (k) = 1000 pounds

SI units N (Newton)

kN (kilo Newton = 1000 Newton)

Stress f = P/A (relative allows to compare material)

US units psi (pound / square inch)

ksi (kip / square inch)

SI units Pa (Pascal = N/m2)

kPa (kilo Pascal = 1000 Pascal)

Note:

SI units (System International = metric units)

Stress f = P/A (force / area, helps to compare)

f F fa = axial-; fb = bending-; fv = shear-stress

f = actual stress

F = allowable stress


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Shear

Shear stress tends to slide

1 Single shear (1 shear plane)

2 End-block shear

3 Double shear (2 bolted shear planes)4 Double shear (2 glued shear planes)

5 Double shear (twin beam / column)

6 Shear wall

A Shear plane(s)

B Shear crack (diagonal)

P Load


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Bending

1 Simple beam

2 Deformed beam under load

3 Bending stress

Bending moment

M = w L2/ 8

where

L = span

w = uniform load

Bending stress

f = M / S

where

S = Section Modulus

Note:

Derivations will be introduced later

Gravity load causes:

Concave bending (+)

Top shortens in compression

Neutral axis has zero stress

Bottom elongates in tension


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Torsion

1 Door handle

P = Forcee = lever arm

M = P e (torsion moment)

2 Building subject to torsion, caused by

Seismic force & eccentric resistance.

Shear wall at B but one side open;

(tuck-under parking)

Assume: P = 12 k

e = 10

Torsion moment:

M = P e = 12k x 10 M = 120 k



Compression / tension

1 Wood column Fa = 1200 psi

P = 800 #

A = 2x2 A = 4 in2

f = P/A = 800/4 f = 200 psi

200 < 1200, ok

2 Steel rod ( ) Fa = 30 ksi

P = 5 k

A=r2 = 3.14 x 0.252 A = 0.2 in2

f = P/A = 5 / 0.2 f = 25 ksi

25 < 30, ok

3 Heel

Allowable cross-fiber stress F = 400 psi

Impact load P = 200 #

A = 0.2 x 0.2 A = 0.04 in2

f = P/A = 200 / 0.04 f = 5,000 psi

5,000 > 400

Not ok

Heel would sink into wood due to overstress



Compression

Wood column / concrete footing

Assume:

Allowable soil pressure Fs = 2000 psfColumn 6 x 6 (nominal, 5.5x5.5 actual)

Allowable stress Fa =1000 psi

3x3x1 concrete footing @ 150 pcf

Load P = 15,000 #Column analysis

A = 5.5 x 5.5 A = 30 in2

f = P/A = 15000/30 f = 500 psi

500 < 1000, ok

Footing analysisDL = 150 pcf x 3x 3 x 1 DL = 1350 #

Load on soil

Ps = P+ DL = 15,000 + 1,350 Ps = 16,350 #

Soil pressuref = Ps/A = 16350 / (3x3) f =1817 psf

1817 < 2000, ok

Note:

Ignore light column DL but not footing DL



Compression

Concrete slab, CMU wall

8 slab, L = 20, DL=100 psf, LL=40 psf

8 nominal CMU wall, 8 high, 80 psf

2 x 1 concrete footing @ 150 pcf

Allowable soil pressure Fs = 1500 psf

Allowable CMU stress Fa = 80 psi

Analyze 1 ft wide strip

Slab weight on wallw = (100+40) x 20 / 2 w = 1400 plf

CMU wall (8 nominal = 7 5/8 = 7.625)

w = 80 psf x 8 w = 640 plf

Wall area (per foot)A = 12 x 7.625(per linear foot) A = 92 in2

Wall stress

f = P/A = (1400+640) / 92 f = 21 psi



Tension

1 Cable ( strand)

Assume:

Allowable cable stressFa = Fy/3 = 210 ksi/3 Fa = 70 ksi

Load P = 8 k

Metallic area (70% metallic)

Am = .7r2= .7 0.252 Am = 0.14 in2

Stress

f = P/A = 8 / 0.14 f = 57 ksi



Shear1 Single shear

Assume:

Load P=3k2x4 woodbarswith bolt

Allowablebolt shear stress Fv =20ksi

Shear area (bolt crosssection)

A= r2

= (0.5/2)2

A=0.2 in2

Shear stress

fv = P / A = 3/ 0.2 f v = 15 ksi

15 < 20, ok

2 Check end shear block (A)Assume: P=3000#

Length of shear blockA e=6

Allowable wood shear stress Fv = 85 psi

Endblock shear areaA=2x2 x6 A=24 in2

Shear stress

fv = P/A = 3000 # / 24 f v=125 psi

125 > 85, NOT OK

Required block length

e = 6 x 125/85 = 8.8 use e = 9



Shear

3 Bolted double shear

Assume:Load P = 22 k

2 5/8 bolts, allowable stress Fv = 20 ksi

Shear area

A = 4 r2 = 4 (0.625/2)2 A = 1.2 in2

Shear stressfv = P / A = 22 / 1.2 f v = 18 ksi

18 < 20, ok

4 Glued double shearAssume:

Load P = 6000 #

Wood bars, allowable stress Fv = 95 psi

Shear areaA = 2 x 4 x 8 A = 64 in2

Shear stress

fv = P / A = 6000 / 64 f v =94 psi

94 < 95, ok



Shear

5 Twin beam double shear

Assume: P=R=12k

2bolts, Fv =20ksi

Shear area

A=4 r2 =4 (0.5/2)2 A=0.79 in2

Shear stressfv =P/A=12/ 0.79 f v =15ksi

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Shear walls:1 Plywood on wood studs2 Plywood on metal studs3 Reinforced concrete wall4 Reinforced CMU wall

Twin bolts at double shear


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05 Force Stress