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Score: 1. 0/100 0 Points % Award: 0 out of 15.00 points Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12-mm and the inner diameter of each washer is 16-mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers, knowing that the average normal stress in the bolts is 32 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa. The smallest allowable outer diameter d of the washers is 17 mm. References Numeric Response Difficulty: Medium Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12-mm and the inner diameter of each washer is 16-mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers, knowing that the average normal stress in the bolts is 32 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa. The smallest allowable outer diameter d of the washers is Explanation: Bolt: A Bolt πd = 2 π(0.012 = 4 Tensile force in bolt: σ = P = = m) 2 4 = 1.13097 × 10 −4 m P A σA (32 × 10 6 Pa) (1.13097 × 10 −4 2 m ) 3 = 3.6191 × 10 N Bearing area for washer: Aw = and Aw = π 4 2 2 (d o − d ) i P σ BRG Therefore, equating the two expressions for Aw gives π 4 2 (d o 2 do = 2 − di ) 4P πσ BRG P = σ BRG 2 + di 3 4(3.6191 × 10 2 do 2 do = π(8.5 × 10 = 7.9812 6 × 10 do = 28.251 do = 28.3 mm N) + (0.016 Pa) −4 × 10 −3 m m 2 m) 2 2 28.3 ± 2% mm. 2. Award: 0 out of 20.00 points A load P is applied to a steel rod, whose flange diameter is 40 mm, supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 177 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. The largest load P that can be applied to the rod is kN. 6720 References Numeric Response Difficulty: Medium A load P is applied to a steel rod, whose flange diameter is 40 mm, supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 177 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod. The largest load P that can be applied to the rod is 66.7 ± 2% kN. Explanation: For steel: A1 = τ1 = P = πdt = π (0.012 m) (0.010 m) = 376.99 × 10 −6 m 2 P A1 A 1 τ1 = 376.99 ( × 10 −6 2 m ) (177 × 10 6 Pa) = 66.727 × 10 3 N For aluminum: A2 = τ2 = P = π dt = π (0.040 m) (0.008 m) = 1.00531 × 10 −3 m 2 P A2 A 2 τ2 = (1.00531 × 10 −3 2 70 m )( Limiting value of P is the smaller value, so P = 66.7 kN × 10 6 Pa) = 70.372 × 10 3 N A 6-mm-diameter pin is used at connection C of the pedal shown. It is given that P = 750 N. References Section Break 3. Difficulty: Medium Award: 0 out of 6.00 points Determine the average shearing stress in the pin. The average shearing stress in the pin is MPa. References Numeric Response Difficulty: Medium Determine the average shearing stress in the pin. The average shearing stress in the pin is 34.4836 ± 2% MPa. Explanation: Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces. Using geometry, − −−−−−−−−−−−−−−−−−− 2 C E = √ (300 mm) 2 + (125 mm) = 325 mm Using the free-body diagram of BCD, +↑ΣFy = 0 : 1 τpin = AP 325 mm 1 C 2 125 mm = C 2 π 4 d 2 = C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N 2C πd 2 = 2(1950 N) π(6×10 −3 2 m) = 34.4836 × 10 6 Pa = 34.4836 MPa 4. Award: 0 out of 6.00 points Determine the nominal bearing stress in the pedal at C. The nominal bearing stress in the pedal at C is MPa. References Numeric Response Difficulty: Medium Determine the nominal bearing stress in the pedal at C. The nominal bearing stress in the pedal at C is 36.1111 ± 2% MPa. Explanation: Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces. Using geometry, − −−−−−−−−−−−−−−−−−− 2 C E = √ (300 mm) 2 + (125 mm) = 325 mm Using the free-body diagram of BCD, +↑ΣFy = 0 : σb = C Ab = C dt 125 mm 325 mm C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N (1950 N) = (6×10 −3 m)(9×10 −3 = 36.1111 × 10 m) 6 Pa = 36.1111 MPa 5. Award: 0 out of 6.00 points Determine the nominal bearing stress in each support bracket at C. The nominal bearing stress in each support bracket at C is MPa. References Numeric Response Difficulty: Medium Determine the nominal bearing stress in each support bracket at C. The nominal bearing stress in each support bracket at C is 32.5 ± 2% MPa. Explanation: Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces. Using geometry, − −−−−−−−−−−−−−−−−−− 2 C E = √ (300 mm) 2 + (125 mm) = 325 mm Using the free-body diagram of BCD, +↑ΣFy = 0 : 125 mm 325 mm C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N 1 C σb = 2 Ab = C 2dtb (1950 N) = 2(6×10 −3 m)(5×10 −3 = 32.5000 × 10 m) 6 Pa = 32.5000 MPa 6. Award: 0 out of 16.00 points Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 19-kN load P as shown. Knowing that the ultimate load for cable BD is 100 kN, determine the factor of safety with respect to cable failure. The factor of safety with respect to cable failure is . References Numeric Response Difficulty: Medium Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 19-kN load P as shown. Knowing that the ultimate load for cable BD is 100 kN, determine the factor of safety with respect to cable failure. The factor of safety with respect to cable failure is 2.9024 ± 2% . Explanation: Use member ABC as a free body and note that member BD is a two-force member. + ↺ ΣMC = 0 : (P cos 40°)(1.2 m) + (P sin 40°)(0.6 m) − (FBD cos 30°)(0.6 m) − (FBD sin 30°)(0.4 m) = 0 ⇒ 1.304926P − 0.719615 FBD = 1.813366P = (1.813366)(19 × 10 F.S. = FU F BD = 100×10 3 N 34.4540×10 3 = 2.9024 N 3 N) = 34.4540 × 10 3 N 7. Award: 0 out of 16.00 points Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 150 MPa in either rod, determine the smallest allowable values of the diameters d1 and d2. Take P = 40 kN. The smallest allowable value of the diameter d1 is The smallest allowable value of the diameter d2 is mm. mm. References Numeric Response Difficulty: Easy Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 150 MPa in either rod, determine the smallest allowable values of the diameters d1 and d2. Take P = 40 kN. The smallest allowable value of the diameter d1 is The smallest allowable value of the diameter d2 is 18.4 ± 2% mm. 42.2 ± 2% mm. Explanation: Rod AB: Force: P = 40 × 103 N Stress: σ AB = 150 × 106 Pa Area: A = π4 d12 σ AB A π 4 = 2 d1 2 d1 = d1 = P = A P σ AB = P σ AB (4)(40 × 10 4P = πσ AB π(150 × 10 18.426 × 10 18.4 mm Rod BC: Force: P = 40 d1 3 N) = Pa) 339.53 × 10 −6 m 2 m = Stress: σ BC Area: A = σ BC 2 −3 6 = P A 4P π 4 d 3 N − (2) (125 × 10 − 150 × 10 6 3 N) = 210 – Pa 2 2 = 4P πd 2 2 (4)(-210 × 10 d2 = d2 = 42.22 d2 = 42.2 mm πσ BC × 10 = = π(−150 × 10 × 10 −3 m 6 3 N) Pa) = 1.78 × 10 −3 m 2 × 10 3 N 8. Award: 0 out of 15.00 points Two wooden members are joined by plywood splice plates that are fully glued on the contact surfaces. Knowing that the clearance between the ends of the members is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for the loading shown. Take P = 15 kN. The length L is mm. References Numeric Response Difficulty: Easy Two wooden members are joined by plywood splice plates that are fully glued on the contact surfaces. Knowing that the clearance between the ends of the members is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for the loading shown. Take P = 15 kN. The length L is 138 ± 2% mm. Explanation: τall 2.5 = MPa = 2.75 0.90909 MPa On one face of the upper contact surface, A L − 0.006 = m 2 (0.125 m) Since there are 2 contact surfaces, τall = P 2A 0.90909 × 10 L = 0.138 m L = 138 mm 6 Pa = 15 × 10 (L − 0.006 3 N m)(0.125 m)