ENGR Statics and Mechanics of Materials 1 (2161) Homework #8 Solution Set

Transcription

1 NGR Statics and Mechanics of Materials 1 (2161) Homework #8 Solution Set 1. One should begin by drawing a free-body diagram for the beam, as shown below. here are horizontal and vertical reaction forces due to the pin support at and a vertical reaction force due to the roller support at. istributed loads in a free-body diagram are replaced with their statically equivalent concentrated loads the distributed load in this problem was treated at two separate distributed loads. 8 kn y 18 kn x x y 3m m y he support reactions are then determined by the equilibrium equations for the freebody diagram. Fx = x = 0 = x = 0 M = 6 y (3)(18) (4.6667)(8) = 0 = y = kn Fy = y + y 18 8 = 0 = y = kn hus, x = 0, y = 10.8 kn, y = 15.2 kn

2 2. here are horizontal and vertical reaction forces and a reaction couple due to the fixed support at lb in M x 1 ft 2 ft y Fx = x = 0 = x = 0 Fy = y 200 = 0 = y = M = M (1500 lb in) + (3)() = 0 = M = 5700 lb in hus, x = 0, y =, M = 5700 lb in 3. Let 2 be the tension in rope 2 and consider the free-body diagrams shown below W W From the free-body diagram on the left, Fy = W = 0 = 2 = 1 2 W hen, from the free-body diagram on the right, Fy = W = 0 = = 1 4 W Since W = (500 kg)(9.81 m/s 2 ) = 4905 N, = 1226 N

3 4. Let F 1 and F 2 be the forces exerted on the cylinder by the ground and curb, respectively, and consider the free-body diagram shown below. Moment equilibrium about the center of the cylinder shows that the line of action of F 2 must pass through the center of the cylinder. Note the angle θ = sin 1 3/6 = lb P θ F 2 F 1 Summing forces in the horizontal x-direction, Fx = P F 2 cos θ = 0 = F 2 = P cos θ hen, summing forces in the vertical y-direction, Fy = F 1 + F 2 sin θ 50 = 0 = F 1 = 50 F 2 sin θ = 50 P tan θ he force exerted by the ground is F 1 = 50 lb when the force P = 0 and decreases as the force P is increased. he cylinder will lift off of the ground when F 1 = 0, which corresponds to a force P = 50 = 86.6 lb tan θ

4 5. onsider first a free-body diagram of the entire frame: y x F y F Since this frame is statically determinate, the support reactions at and F can be determined from the equilibrium equations for this free-body diagram: Fx = x 100 = 0 = x = M = 12F y (24)(100) = 0 = F y = Fy = y + F y = 0 = y = Now, take the frame apart and consider free-body diagrams for each of its members, remembering to obey Newton s third law at each pin joint where the members are attached to each other. he directions of the internal reaction forces are chosen arbitrarily their actual directions will be determined by their signs in the solutions below. y y x x y x y y F x x y x Note that there are a total of nine remaining equilibrium equations (three for each free-body diagram), but only six unknown forces ( x, y, x, y, x, and y ). his

5 means that not all of the equilibrium equations will be required and that there are multiple ways to determine the unknown forces. One way is shown here. First consider equilibrium of member F : M = 6 y + (12)(200) = 0 = y = Fy = y + y = 0 = y = Next consider equilibrium of member : Fy = y y = 0 = y = + M = 6 x 6 y = 0 = x = + Fx = x x = 0 = x = Finally, again consider equilibrium of member F : Fx = x + x = 0 = x = + he results can now be shown using report diagrams: F

6 6. (a) Note that members and F are two-force members and consider the following free-body diagram. From symmetry, the vertical force at is half of the weight. 9 in F F 45 F F F y x x From the free-body diagram for members and F, Fx = F cos 45 + F F cos 45 = 0 = F F = F 50 lb Fy = 100 F sin 45 F F sin 45 = 0 = F = lb hen, from the free-body diagram for member, Fy = F sin 45 + y 50 = 0 = y = 0 M = 12 x (9)(50) (12)(F cos 45 ) (6)(F sin 45 ) = 0 = x = lb Fx = x + x + F cos 45 = 0 = x = lb Using a report diagram to show the forces acting on member, 70.7 lb lb lb 50 lb

7 (b) he shearing stress on the cross section of the 1/4-in.-diameter pin at is τ = lb π(0.125 in 2 ) = 1441 psi 7. (a) Let F and F be the axial loads in bars and and let L = 100 mm and L = 200 mm be their lengths. hey each have elastic modulus = 210 GPa and cross-sectional area = 100 mm 2. hen, their changes in length are δ = F L, δ = F L However, the kinematic constraint on the motion of the system is that δ = 2δ, so ( ) F L = 2F L L = F = 2 F = 4F Now, consider a free-body diagram for member : L 4F F 50 mm 100 mm 100 mm x y P = 10 kn M = (200)(10) (50)F (100)(4F ) = 0 = F = kn hus, the stresses in bars and are = F = kn σ = F = MPa, σ = F = 44.4 MPa (b) he deflection of pin is twice the change in length of bar, (17, N)(0.1 m) δ = 2 ( m 2 )( N/m 2 ) = m