Find reaction at C for a frame shown. using Consistent Deformations Method and Least Work Method A 6 m R=3m 10 kN Sin201 Sin 20 de = [2
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- -11 A rubber cube R of a side L = 3 in. and cross- sectional area A = 9 in2 is compressed inside a steel cube S by a force F = 5 lb that applies uniformly distributed pressure to the rubber. Assume E 0.3ksi and,, = 0.45. (a) Calculate the lateral pressure between the rubber and steel (disregard friction between the rubber and the steel, and assume that the steel block is rigid when compared to the rubber). (b) Calculate the change in volume of the rubber.Repeat Problem 11.3-9. Use two C 150 × 12.2 steel shapes and assume that E = 205 GPa and L = 6 m.The inclined ladder AB supports a house painter (85 kg) at C and the weight iq = 40 K/m} of the ladder itself. Each ladder rail (t5= 4 mm) is supported by a shoe (ts= 5 mm) that is attached to the ladder rail by a bolt of diameter d = 8 mm
- A 10-ft rigid bar AB is supported with a vertical translational spring at A and a pin at B. The bar is subjected to a linearly varying distributed load with maximum intensity q0. Calculate the vertical deform at ion of the spring if the spring constant is 4 kips/in.Solve the preceding problem if the collar has mass M = 80 kg, the height h = 0.5 m, the length L = 3.0 m, the cross-sectional area A = 350mm2. and the modulus of elasticity E = 170 GPa..17 A mountain-bike rider going uphill applies torque T = Fd(F = l5lb, d = 4 in.) to the end of the handlebars ABCD by pulling on the handlebar extenders DE. Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L, = 2 in.andL3 = 8.5 in, and with outer diameters and thicknesses d01 = 1.25 in. 101 = 0.125 in. and d03 = O.87in.,i03 = 0.ll5in, respectively as shown. Segment BC’ of length L, = 1.2 in. however. is tapered, and outer diameter and thickness vary linearly between dimensions at B and C. Consider torsion effects only. Assume G = 4000 ksi is constant. Derive an integral expression for the angle of twist of half of the handlebar tube when it is subjected to torque T = Fd acting at the end. Evaluate ‘b1-, for the given numerical1ues.
- Compare the angle of twist 1 for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist 2 calculated from the exact theory of torsion for circular bars, Express the ratio 12terms of the non-dimensional ratio ß = r/t. Calculate the ratio of angles of twist for ß = 5, 10, and 20. What conclusion about the accuracy of the approximate theory do you draw from these results?A block weighing W = 5.0 N drops inside a cylinder from a height h = 200 mm onto a spring having stiffness k = 90 N/m (see figure), (a) Determine the maximum shortening of the spring due to the impact and (b) determine the impact factor.An aluminum bar has length L = 6 ft and diameter d = 1.375 in. The stress-strain curse for the aluminum is shown in Fig. 1.34. The initial straight, line part of the curve has a slope (modulus of elasticity) of 10.6 × 106 psi. The bar is loaded by tensile forces P = 44.6 k and then unloaded. (a) That is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional limit? hint: Use the concepts illustrated in Figs. l.39b and 1.40.
- A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by =18,0001+30000.03(=ksi) in which is nondimensional and has units of kips per square inch (ksi). (a) Construct a stress-strain diagram for the material. (bj Determine the elongation, of the wire due to the Forces P. (c) IF the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?Repeat Problem 2.3-18, but assume that the bar is made of copper alloy. Calculate the displacements SBand Scif P = 50 kips, L = 5 ft = 3/5 in., b1= 2.75 in., b2= 3 in., and E = 16,000 ksi.A sign of weight W is supported at its base by four bolls anchored in a concrete footing. Wind pressure P acts normal to the surface of the sign; the resultant of the uniform wind pressure is force fat the center of pressure (C.P). The wind force is assumed to create equal shear forces F/4 in the y direction at each boll (see figure parts a and c). The overturning effect of the wind force also causes an uplift force R at bolts A and C and a downward force (— R) al bolts B and D (see figure part b). The resulting effects of the wind and the associated ultimate stresses for each stress condition are normal stress in each boll (h — 60 ksi); shear through the base plate (th = 17 ksi); horizontal shear and bearing on each bolt ( tfur = 25 ksi and cr^ = 75 ksi): and bearing on the bottom washer at B (or D) (abor = 50 ksi).