Q.6/ Determine the horizontal and vertical components of reactions on the member at the pin "A" and normal reaction at the roller "B" in the figure below. * 3336.2 N 1 m 1 m A 0.6 m B 30 O FB=2416.8 N, Ax=1168.4 N, Ay=1328.88 N O FB=2516.86 N ,Ax=1258.4 N, Ay=1339.8 N O FB=2316.8 N, Ax=1158.44 N, Ay=1529.8N O FB=2316.8 N, Ax=1158.4 N, Ay=1329.8 N Other:
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- A circular steel rod has a diameter of 50 mm. It touches the unyielding supports at point A and B when ?1 = 80 °C. Find the applied force P to the collar at its midsection so that, when ?2 =30 °C, the reaction at B is zero. Use E st = 200 GPa and the coefficient of linear expansion of steel is at 12 x 10^-6 m/m C.The J-shaped member shown in the figure (Figure 1) is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces A_y and A_z at support A required to keep the system in equilibrium. The cylinder has a weight WBWBW_B = 24.0 N, andF = 7.60 N is a vertical force applied to the member at C. The dimensions of the member are w = 0.450 m , l= 2.00 m , and h = 0.600 m .Shown in the following figure is a foldable table loaded by weight “W”. The contact points, B, D, and E are frictionless. 1. Find the safest load W if the allowable tensile force in the cable is 1800 N. a. 600Nb. 180 Nc. 482Nd. 630N2. If W = 1500 N, find the reaction at D. a. 500 Nb. 1000 Nc. 800 Nd. 1200 N3. If W = 1500 N, find the reaction at B.a. 500 Nb. 1000 Nc. 800 Nd. 1200 N
- A steel rod is stretched between two walls. At 20C, the tensile force in the rod is 5000N. If the stress is not to exceed 130MPa at -20C, find the minimum allowable diameter of the rod. Use a=11.17x10^-6C and E=200GPa. -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.- Direction of the assumption of the equilibrium equation must be shown, no direction will be marked wrong. -Show complete solution and explanationThe rigid bar ABC is supported by pin at A and aluminum wire BD, is horizontal before the Load P is applied. Find the vertical displacement of point C caused by the load P=35KN. Where E of aluminum equals 70GPA and the area of the wire is 200mm^2 -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.- Direction of the assumption of the equilibrium equation must be shown, no direction will be marked wrong.25 - In the plane truss system, the loading status of which is evident in the figure, the support reactions and all rod forces will be found. A is the fixed joint and B is the sliding joint. Single upload P = 35 kN, dimensions a = 1 m and bar numbers are given. Accordingly, S5 = ? A) 49.5B) 35C) 99D) 70E) 140
- PRIOR ANSWERS: Reaction Forces: By = 76.809 Cy = 15.94 Shear Forces: V (× = 10.5- ft (i.e., just to the left of support B)) = -36.75 kips V ( x = 10.5+ ft (i.e., just to the right of support B)) = 40.059 kips V (x = 25.5 ft) = -12.441 kips V (x = 26.5- ft (i.e.. just to the left of support C))= = -15.941 kips Bending moment: M (x = 10.5 ft (i.e., at support B)) = -192.9375 kips-ft M (x = 25.5 ft) = 14.1975 kips-ftI need help with this review problem Engineering Statics Take F1 = 40 kN, F2 = 20 kN, and d = 8 m (a) Draw a free body diagram of the entire truss. (b) Find the support reaction at A (Give the x and y components) (c) Consider a cut through members CD, DI and HI. Draw the free body diagram of the left part of cut. (d) Use the FBD of the previous question to find the tensions in members CD, DI and HI.A bar of square cross section is subjected to point loads of P1, P2, P3 and P4 as shown here. Find the magnitude of the force P3 necessary for the equilibrium, if P1= 110 kN, P2 = 230 kN & P4 = 160 kN. Also, determine the stresses in each section & the net change in the length of steel bar. square of size a1=60mm, L1=34 cm, size a2 = 35mm, L2=36 cm, size a3=60 mm & L3 =31 cm Take E = 173 GPa. The force P3 to make it to equilibrium in kN is The Stress in section 1 in N/mm2 is The Compressive Stress in section 2 in N/mm2 is The stress in section 3 in N/mm2 is The total change in length in x10-3 mm is
- A brass link (Eb=105 GPa, αb=20.9 3 10–6/°C) and a steel rod (Es= 200 GPa, αs=11.7 3 10–6/°C) have the dimensions shown at a temperature of 20°C. The steel rod is cooled until it fits freely into the link. The temperature of the whole assembly is then raised to 45°C. Determine (a) the final normal stress in the steel rod, (b) the final length of the steel rodMember AB (helical spring) and Member AC (axial rod) are carrying a weight at point A. Member AB has the following properties: R = 90 mm, n =4 turns, G = 70 GPa, d = 30 mm, and allow = 105 MPa while Member AC has the following properties: L = 2 , A = 500 mm2, E = 200 GPa, and allow = 30 MPa. Determine the maximum safe value of W (in N). Use simplified formula for the helical spring.Free-Body Diagram Involving Internal Reactions