1 of 7. Law of Sines: Stress = E = G. Deformation due to Temperature: Δ
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1 NME: ES30 STRENGTH OF MTERILS FINL EXM: FRIDY, MY 1 TH 4PM TO 7PM Closed book. Calculator and writing supplies allowed. Protractor and compass allowed. 180 Minute Time Limit GIVEN FORMULE: Law of Cosines: C b a Law of Sines: c Stress = E = G Deformation due to Temperature: Δ ending M/EI = 1/ = -y/ =My/I M=EI M/EI = dv /d I = (/4)r 4 for a solid circular cross-section Torsion = Tc/J = TL/GJ Power P = Twhere T is torque and is angular frequency (rad/s) 1hp = 550 ft-lb/s 1Watt = 1 N-m/s J = (/)c 4 for a solid, circular shape of radius c. eam Shear Stress = VQ/It q = VQ/I q = F/s I = (/4)r 4 for a solid, circular shape of radius r Pressure Vessels hoop = Pr/t long = Pr/(t) Stress Transformation Equations σ ' cos θ τ τ ' ' sin θ τ y cos θ y Principal Stresses 1, y sin 1, Occur on Principal Planes at ngle p tan Maimum Shear Stress is: Strain Transformation Equations ' cos sin ' y' y sin cos ( )sin cos (cos sin ) y y y y θ 1 of 7
2 GIVEN FORMULE (continued): uckling EXM QUESTIONS (10 QUESTIONS: 100 POINTS): 1. (1 point). Given: () 5-hp motors. Motor operates at 600 rpm. Motor operates at 3600 rpm. Which motor would require a larger shaft diameter, assuming the shear strengths of the shaft are the same? Circle the correct nswer: 1. Motor. Motor 3. oth motors would require the same shaft size 4. Cannot be determined Shaft Shaft Motor 5 hp at 600 rpm Motor 5 hp at 3600 rpm. (1 point) Determine the internal shear stress on the cross section at Point J in the drum lifter. Point J is located at the centroid of the bar, which is a 1 ¼ steel plate Given: The gripping action on the top of the drum has horizontal and vertical force components, only. of 7
3 1 in. 3. (4 points). For a ductile metal, which state of stress is most severe (bringing the material closest to failure)? Circle the correct answer:. State of Stress. State of Stress C. State of Stress C D. Gary usey C of 7
4 4. (13 points) Compute the horizontal end-reaction forces, and and indicate their directions Given: Fied at and 0 kip and 40 kip forces at C and D, as shown Cross-sectional area: = in, Modulus of Elasticity, E = ksi C D 5. (7 points). Determine the maimum normal stress in the pile. concrete pile with a circular crosssection is embedded in the soil. It is known that the frictional resistance of the soil varies linearly with a function f() = 0.0 (units of kips, inches). Hence, the frictional resistance is 10 kips/inch at point, while it is zero at point. Given: The pile is known to be elastic. Point does not move. P is resisted by frictional forces, eclusively (the end of the pile resists no forces). The cross-sectional area of the pile = 500 in and E = 3000 ksi P Ground 500 inches Frictional resistance, f() = 0.0 (kips/inch Concrete Pile Final nswer: 4 of 7
5 6. (6 points) Determine the radius of curvature of the wooden (E=100ksi) beam, between points and C due to the two 5-kip point loads. The beam has a cross-section of 6 1, as shown. 5 kips 5 kips C D 1 4 in. 4 in. 4 in. 6 CROSS-SECTION 7. (30 points) The machine component is subjected to a load of 5000 lb. Determine the principal stresses at point H and draw the principal element at its proper orientation, labeling each stress and the angle lb CROSS-SECTION of 7
6 8. (9 points) ¼ steel plate is shown with a 3/8 diameter bolt in a 3/8 diameter hole, with a force P applied to the bolt. The center of the hole is located ½ from the edge of the plate, as shown. Determine the force P that caused this failure. Given: Ultimate tensile strength of the bolt is 100 ksi Ultimate shear strength of the bolt is 60 ksi Ultimate tensile strength of the steel plate is 50 ksi Ultimate shear strength of the steel plate is 30 ksi P Top View 3/8 diam bolt Side View P 3D View The Failure: The end tears out 9. (8 points) Determine the maimum deflection of the beam below. Given: Fied at. Free at. E=1000 ksi, I = 10 in 4, L=100 inches, M = 1 kip-inch. 1 kip-in = M of 7
7 10. (1 points) Given: wooden (E=1500 ksi) cross-section with a ½ 10 steel plate (E=30000 ksi) that is securely glued to the bottom of the wood. is a pinned support. is a roller support. Determine the maimum shear stress in the glue. 0.5 kip/ft 10 ft. 4 ft Wood ½ 10 Steel eam Cross Section 7 of 7
1 of 12. Law of Sines: Stress = E = G. Deformation due to Temperature: Δ
NAME: ES30 STRENGTH OF MATERIALS FINAL EXAM: FRIDAY, MAY 1 TH 4PM TO 7PM Closed book. Calculator and writing supplies allowed. Protractor and compass allowed. 180 Minute Time Limit GIVEN FORMULAE: Law
1 of 12. Given: Law of Cosines: C. Law of Sines: Stress = E = G
ES230 STRENGTH OF MATERIALS FINAL EXAM: WEDNESDAY, MAY 15 TH, 4PM TO 7PM, AEC200 Closed book. Calculator and writing supplies allowed. Protractor and compass required. 180 Minute Time Limit You must have
NAME: Given Formulae: Law of Cosines: Law of Sines:
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