Type of Force 1 Axial (tension / compression) Shear. 3 Bending 4 Torsion 5 Images 6 Symbol (+ )


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1 Cause: external force P Force vs. Stress Effect: internal stress f 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 1 Type of Force 1 Axial (tension / compression) 2 3 Bending 4 Torsion 5 Images 6 Symbol (+ ) A Tension (elongates +) B Compression (shortens ) C (clockwise couple +) D Bending (concave + convex ) E Torsion (righthand rule +) Tension elongates Compression shortens tends to slide Bending (+) top compression bottom tension Torsion twists (righthand +) 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 2 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) US units psi (pound / square inch) ksi (kip / square inch) SI units Pa (Pascal = N/m 2 ) kpa (kilo Pascal = 1000 Pascal) SI units (System International = metric units) Stress f = P/A (force / area, helps to compare) f F f = actual stress F = allowable stress 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 3 1
2 tends to slide 1 Single shear (1 shear plane) 2 Endblock shear 3 Double shear (2 bolted shear planes) 4 Double shear (2 glued shear planes) 5 Double shear (twin beam / column) 6 wall A B P plane(s) crack (diagonal) 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 4 Bending 1 Simple beam 2 Deformed beam under load 3 Bending stress Bending moment M = w L 2 / 8 where L = span w = uniform load Bending stress f = M / S where S = Section Modulus Derivations will be introduced later Gravity load causes: Concave bending (+) Top shortens in compression Neutral axis has zero stress Bottom elongates in tension 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 5 Torsion 1 Door handle P = Force e = lever arm M = P e (torsion moment) 2 Building subject to torsion, caused by Seismic force & eccentric resistance. wall at B but one side open; (tuckunder parking) P = 12 k e = 10 Torsion moment: M = P e = 12k x 10 M = 120 k 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 6 2
3 Compression / tension 1 Wood column F a = 1200 psi P = 800 # A = 2 x2 A = 4 in 2 f = P/A = 800/4 f = 200 psi 200 < 1200, ok 2 Steel rod (φ ½ ) F a = 30 ksi P = 5 k A= πr 2 = 3.14 x A = 0.2 in 2 f = P/A = 5 / 0.2 f = 25 ksi 25 < 30, ok 3 Heel Allowable crossfiber stress F = 400 psi Impact load P = 200 # A = 0.2 x 0.2 A = 0.04 in 2 f = P/A = 200 / 0.04 f = 5,000 psi 5,000 > 400 Not ok Heel would sink into wood due to overstress 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 7 Compression Wood column / concrete footing Allowable soil pressure F s = 2000 psf Column 6 x 6 (nominal, 5.5x5.5 actual) Allowable stress F a =1000 psi 3 x3 x1 concrete 150 pcf P = 15,000 # Column analysis A = 5.5 x 5.5 A = 30 in 2 f = P/A = 15000/30 f = 500 psi 500 < 1000, ok Footing analysis DL = 150 pcf x 3 x 3 x 1 DL = 1350 # on soil P s = P+ DL = 15, ,350 P s = 16,350 # Soil pressure f = P s /A = / (3 x3 ) Ignore light column DL but not footing DL f =1817 psf 1817 < 2000, ok 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 8 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 150 pcf Allowable soil pressure F s = 1500 psf Allowable CMU stress F a = 80 psi Analyze 1 ft wide strip Slab weight on wall w = (100+40) x 20 / 2 w = 1400 plf CMU wall (8 nominal = 7 5/8 = ) w = 80 psf x 8 w = 640 plf Wall area (per foot) A = 12 x (per linear foot) A = 92 in 2 Wall stress f = P/A = ( ) / 92 f = 21 psi<80, ok Footing w = 150 pcf x 2 x 1 w = 300 plf Soil load per foot P = P = 2340 plf Soil pressure f = P/A = 2340 / 2 f =1170 psf 1170<1500, ok 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 9 3
4 ~ 70% metallic ~ 60% metallic Tension 1 Cable (φ ½ strand) Allowable cable stress F a = F y /3 = 210 ksi/3 F a = 70 ksi P = 8 k Metallic area (70% metallic) A m =.7πr 2 =.7 π A m = 0.14 in 2 Stress f = P/A = 8 / 0.14 f = 57 ksi<70, ok 2 Hong Kong Shanghai Bank Allowable hanger stress F a = 30 ksi per level P = 227 k Level 1 hanger (pipe) A = 12 in 2 Level 1 stress f = 227 / 12 f = 19 ksi Level 6 hanger (pipe) A = 75 in 2 Level 6 stress f = 6 x 227 / 75 f = 18 ksi 18 < 19 < 30, ok 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 10 1 Single shear P = 3 k 2 x4 wood bars with ½ bolt Allowable bolt shear stress F v = 20 ksi (bolt cross section) A = π r 2 = π (0.5/2) 2 A = 0.2 in 2 f v = P / A = 3/ 0.2 f v = 15 ksi 15 < 20, ok 2 Check end shear block (A) Length of shear block A e = 6 Allowable wood shear stress F v = 85 psi End block shear area A = 2 x 2 x 6 A = 24 in 2 f v = 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 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 11 3 Bolted double shear P = 22 k 2 5/8 bolts, allowable stress F v = 20 ksi A = 4 π r 2 = 4 π (0.625/2) 2 A = 1.2 in 2 f v = P / A = 22 / 1.2 f v = 18 ksi 18 < 20, ok 4 Glued double shear P = 6000 # Wood bars, allowable stress F v = 95 psi A = 2 x 4 x 8 A = 64 in 2 f v = P / A = 6000 / 64 f v =94 psi 94 < 95, ok 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 12 4
5 5 Twin beam double shear P = R = 12 k 2 bolts, φ ½ F v = 20 ksi A = 4π r 2 = 4 π (0.5/2) 2 A = 0.79 in 2 f v = P / A = 12 / 0.79 f v = 15 ksi 15 < 20, ok 6 wall P = 20 k 8 CMU wall, t = F v = 30 psi Wall length L = 8 A = x 12 x 8 A =732 in 2 f v = P / A = 20,000 # / 732 f v = 27 psi 27 < 30, ok 05 Force vs. Stress Copyright G G Schierle, press Esc to end, for next, for previous slide 13 5
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