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Home Work #3 Due Date: 11 Mar, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side.

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Presentation on theme: "Home Work #3 Due Date: 11 Mar, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side."— Presentation transcript:

1 Home Work #3 Due Date: 11 Mar, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side A4 papers only.

2 HW 3-Problem #1 A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound weight at its lower end (see figure). If the diameter of the circular rod is ¼ inch, calculate the maximum normal stress σ max in the rod, taking into account the weight of the rod itself. Weight density of steel: 490lb/ft 3 The maximum force Fmax in the rod occurs at the upper end

3 HW 3-Problem #2 A loading crane consisting of a steel girder ABC supported by a cable BD is subjected to a load P. The cable has an effective cross-sectional area A=0.471 in 2. The dimensions of the crane are H=9 ft, L 1 =12 ft, and L 2 =4 ft. a.If the load P=9000 lb, what is the average tensile stress in the cable? b.If the cable stretches by 0.382 in, what is the average strain?

4 Ray Rax TByTBy TBxTBx P AB C (a) (b)

5 HW 3-Problem #3 The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 0.505 in and a gage length of 2 in. At fracture, the elongation between the gage marks was 0.12 in and the minimum diameter was 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2 in, and percent reduction in area.

6 HW 3-Problem #3

7 Data for the conventional stress-strain curve

8 Ultimate stress 112832.9lb/in 2 Yield stress 69000lb/in 2 Proportinal limit 65000 lb/in 2 0.1%

9 Percentage of reduction in area = % As shown in the plot, Proportinal limit≈ 65000 lb/in 2 Young’s Modulus = Yield stress of 0.1% offset ≈ 69000 lb/in 2 (Any two point in the proportinal line is OK) Ultimate stress ≈ 112832.9 lb/in 2 Percentage of elongation =0.12/2 = 6%

10 HW 3-Problem #4 A bar of monel metal as in the figure (length L=8 in, diameter d=0.25 in) is loaded axially by a tensile force P=1500 lb. Determine the increase in length of the bar and the percent decrease in this cross-sectional area. Monel metal material property: Modulus of elasticity E=25000 ksi Poisson’s Ratio ν=0.32

11 Percentage of reduction in area

12 HW 3-Problem #5 The connection shown in the figure consists of five steel plates, each 3/16 in. thick, jointed by a single ¼-in. diameter bolt. The total load transferred between the plates is 1200 lb, distributed among the plates as shown. a.Calculate the largest shear stress in the bolt, disregarding friction between the plates. b.Calculate the largest bearing stress acting against the bolt.

13 A B C D 360 480 360 600 A 360 600 B 360 A 240 B C 360 480 600 A 240 D 600 A B C 360 480 360 ① ② ③ ④

14 As shown in the figure above, the shear force of A, D= 360 lb The shear force of B,C= 240lb The largest shear force is: The largest bearing force is:

15 HW 3-Problem #6 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test. The pad has dimensions a=150 mm and b=250mm, and the elastomer has thickness t=50 mm. When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene.

16 Shear strain :

17 A steel bar AD has a cross-sectional area of 0.4 in 2 and is loaded by forces P 1 =2700 lb, P 2 =1800 lb, and P 3 =1300 lb. The lengths of the segments of the bar are a=60 in, b=24 in, and c=36 in. a.Assuming that the modulus of elasticity E=30*10 6 psi, calculate the change in length δ of the bar. Does the bar elongate or shorten? b.By what amount P should the load P 3 be increased so that the bar does not change in length when the three loads are applied? HW 3-Problem #7

18 CD BC BA F3 P3F2P2-P3F1P1+P2-P3 P3 has to be increased by 2610-1310=1310 lb

19 A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has diameter d 1 =100 mm, and segment BC has diameter d 2 =60 mm. Both segments have length L/2=0.6 m. A longitudinal hole of diameter d is drilled through segment AB for one-half of its length (distance L/4=0.3 m). The bar is made of plastic having modulus of elasticity E=4.0 Gpa. Compressive loads P=110 kN act at the ends of the bar. If the shortening of the bar is limited to 8.0mm, what is the maximum allowable diameter d max of the hole? HW 3-Problem #8

20 A’


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