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4 in. 7 in. D 75° В 20° 8 in. 400 IE F 1S in

4 in. 7 in. D 75° В 20° 8 in. 400 IE F 1S in

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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- GET THE GIVEN, WHAT IS ASKED, AND THE ANSWER.

- MAKE A PROPER FREE BODY DIAGRAM. (IT DOES HAVE THE GREATER POINTS IN THIS ACTIVITY)

- YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES.

- YOUR FINAL ANSWER MUST BE IN 4 DECIMAL PLACES.

- TELL THE DIRECTION AND WHETHER IT IS TENSILE OR COMPRESSION.

- WRITE YOUR SOLUTION PROPERLY (EASY TO UNDERSTAND).

- PLEASE BASE YOUR SOLUTIONS ON THE SAMPLE PROBLEM I'VE GIVEN.

 
3. The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 5/8in thick and is connected to the vertical rod by a 3/8in diameter bolt. Determine the average shearing stress in the bolt and the bearing stress on member BD due to Pin C. 4 in.- 7 in. D 75° В C 20° E 8 in. 400 lb A F 1.8 in.
Transcribed Image Text:3. The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 5/8in thick and is connected to the vertical rod by a 3/8in diameter bolt. Determine the average shearing stress in the bolt and the bearing stress on member BD due to Pin C. 4 in.- 7 in. D 75° В C 20° E 8 in. 400 lb A F 1.8 in.
Sample Problem 8 1 2 4. Bearing Stress 6 7.5 mm For the joint shown in the figure, calculate (a) the largest bearing stress between the pin and the members; (b) the average shear stress in the pin; and (c) the largest average normal stress in the members. 25 kN 25 kN Sample Problem 8 12 mm 12 mm 7,5 mn + Consi der tmember 2 Normal Stress 12.5 mm A• (12 )(1B ) : 216mm2 A. (12)(25-125):/5omm? 7.5 mm 25 kN 25 kN 25 kN 25 kN 12 mm 12 mm - 18 mm 125 mm 18 mm 7.5 mm F Ozmay 25000 12.5 mm A ISD Tizmm Ozmax 166 646 25 kN 25 kN - 18 mm 25 mm 18 mm 1 4. Bearing Stress 7.5 mm 25 kN 25 kN Sample Problem 8 12 mm 12 mm 1 2 3 4. Bearing Stress 6. 7.5 mm Given: Sample Problem 9 12.5 mm Required: omar, 7, Tmax .Ozmax Solution: Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is -140 MPa , and that the average shearing stress in each of the two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the average bearing stress in the link. A 25 kN 25 kN 18 mm 25 mm 18 mm + Consider Bearing Stress Pb 25 O00 133 3333 MPa %3D Abi 2(125)(5) 125 Pb 25000 %3D : 166. 6667 MP a Apz Tes)(12) 12mm Oomax:166 6667 mpal 1 2 4. Bearing Stress 7,5 mm 1 2 3 4. Bearing Stress 25 kN 25 kN Sample Problem 8 + Consider 12 mm 12 mm Sample Problem 9 7.5 mm Shear Stress P : 125 KN 12.5 mm Given: * Consider FAB 2 b = 50mm 12500 FAB 25 kN 25 kN t = 6mm て A 18 nm 25 mm 18 mm AAB (2s) 7. 101.8592MPa * Consider Norma! Stress on Member I = Z16 mm o = -140MPa T = 80MPA FAB Required: d, oo Solution: ー140 A (12)() (Sanm conm) Fe: -42 000 N, В O A: [(15)(25 - 12s1 5mm? 12'0 + Consider d V: FAB 4200ON 2 S000 187-5 : 133-3333MPO Oimax こ 4200ON A 80 N 7. A d: 25 8544 m 1 2 3 4. Bearing Stress 6. 3 4. Bearing Stress 5 6 7.5 mm 25 kN 25 kN Sample Problem 9 Bearing Stress Sample Problem 8 12 mm mm 7.5 mm + Consi der Normal Stress tmember 2 + Consider Po A 12.5 mm @ A• (12)(1B) : 216mm2 © A• (2)(25-12 5)somm? td 25 kN 25 kN 18 mm 125 mm 18 mm 42000 (6)(25-8544) F Ozmax 25000 B A MPa סך7 ב ISO A Tizmm Ozmax 166 6467al
Transcribed Image Text:Sample Problem 8 1 2 4. Bearing Stress 6 7.5 mm For the joint shown in the figure, calculate (a) the largest bearing stress between the pin and the members; (b) the average shear stress in the pin; and (c) the largest average normal stress in the members. 25 kN 25 kN Sample Problem 8 12 mm 12 mm 7,5 mn + Consi der tmember 2 Normal Stress 12.5 mm A• (12 )(1B ) : 216mm2 A. (12)(25-125):/5omm? 7.5 mm 25 kN 25 kN 25 kN 25 kN 12 mm 12 mm - 18 mm 125 mm 18 mm 7.5 mm F Ozmay 25000 12.5 mm A ISD Tizmm Ozmax 166 646 25 kN 25 kN - 18 mm 25 mm 18 mm 1 4. Bearing Stress 7.5 mm 25 kN 25 kN Sample Problem 8 12 mm 12 mm 1 2 3 4. Bearing Stress 6. 7.5 mm Given: Sample Problem 9 12.5 mm Required: omar, 7, Tmax .Ozmax Solution: Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is -140 MPa , and that the average shearing stress in each of the two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the average bearing stress in the link. A 25 kN 25 kN 18 mm 25 mm 18 mm + Consider Bearing Stress Pb 25 O00 133 3333 MPa %3D Abi 2(125)(5) 125 Pb 25000 %3D : 166. 6667 MP a Apz Tes)(12) 12mm Oomax:166 6667 mpal 1 2 4. Bearing Stress 7,5 mm 1 2 3 4. Bearing Stress 25 kN 25 kN Sample Problem 8 + Consider 12 mm 12 mm Sample Problem 9 7.5 mm Shear Stress P : 125 KN 12.5 mm Given: * Consider FAB 2 b = 50mm 12500 FAB 25 kN 25 kN t = 6mm て A 18 nm 25 mm 18 mm AAB (2s) 7. 101.8592MPa * Consider Norma! Stress on Member I = Z16 mm o = -140MPa T = 80MPA FAB Required: d, oo Solution: ー140 A (12)() (Sanm conm) Fe: -42 000 N, В O A: [(15)(25 - 12s1 5mm? 12'0 + Consider d V: FAB 4200ON 2 S000 187-5 : 133-3333MPO Oimax こ 4200ON A 80 N 7. A d: 25 8544 m 1 2 3 4. Bearing Stress 6. 3 4. Bearing Stress 5 6 7.5 mm 25 kN 25 kN Sample Problem 9 Bearing Stress Sample Problem 8 12 mm mm 7.5 mm + Consi der Normal Stress tmember 2 + Consider Po A 12.5 mm @ A• (12)(1B) : 216mm2 © A• (2)(25-12 5)somm? td 25 kN 25 kN 18 mm 125 mm 18 mm 42000 (6)(25-8544) F Ozmax 25000 B A MPa סך7 ב ISO A Tizmm Ozmax 166 6467al
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