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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
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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
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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
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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
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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
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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
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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
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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
15
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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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d o n e
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