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- The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bending-moment diagram.Space Frame ABC is clamped at A, except it is free to rotate at A about the x and y axes. Cables DC and EC support the frame at C. Force Py= - 50 lb is applied at the mid-span of AS, and a concentrated moment Mx= -20 in-lb acts at joint B. (a) Find reactions at support A. (b) Find cable tension Forces.The inclined beam represents a ladder with the Following applied loads: the weight (W) of the house painter and the distributed weight (u) of the ladder itself. Find support reactions at A and B: then plot axial force (N), shear (V), and moment (M) diagrams. Label all critical N, V, and M values and also the distance to points where any critical ordmates are zero. Plot N, V, and M diagrams normal to the inclined ladder. Repeat part (a) for the case of the ladder suspended from a pin at B and traveling on a roller support perpendicular to the floor at A.
- Situation 8- The beam is supported at A and supported by a cable at B as shown in Figure AP-4.20. 20. Determine the reaction at A, in N. Answer: 513.20 N 21. Determine the tension force in the cable, in N. Answer: 355.56 N 2 m ACable 800 N/m B 30 Flgure AP-4.20A The reaction in the x-direction at joint A in Figure is: O-5.774 KN 5.774 KN O 5.0 kN 5 KN OOKN E 30" 30" F 10 kN D B ALL DIMENSIONS IN METRES (0.5774) (0.5774)Find the reaction at A due to the uniform loading and the applied couple. The force reaction is positive if upward, negative if downward. The moment reaction is positive if counterclockwise, negative if clockwise. 2.5 kN/m 11.1 kN-m A 2.0 m 2.0 m Answers: RA = i kN MA = i kN-m
- Determine the support reactions of the frame shown in the following figure. Also, determine internal forces at C. 3 m A 15 kN/m C P2m +21 -2 m- B -2 m- 60 KNSituation 22 - For the frame shown in AN- 5 m 15, the force F acting upward at C causes a RA horizontal reaction of 100 N at B. 29. Determine the value of the force F in kN. Answer: 120 kN 30. Determine the reaction at A in kN. Answer: 156.2 kN 31. Determine the angle in degrees that the reaction at A makes with the horizontal Figure SP-3.25 axis (positive counter clockwise). Answer: 230.19°The simply supported beam is subjected to the loads as described in the figure and parameter table. Neglecting the thickness of the beam, calculate the horizontal and vertical support reactions at A and B. F A B C -4*- L3 parameter value units L1 L2 L3 ft 4 ft 6 ft theta 65 F 200 lb M 500 lb-ft For the answers, take to the right and up to be the positive directions. The support reaction at A in the z direction is NAz- lb The support reaction at A in the y direction is NAy- lb The support reaction at Bin the z direction is NB= lb The support reaction at Bin the y direction is NBy= lb
- The bar AB has a built-in support at A. An external force F=6kN is applied at B (8,0,0)m, towards to point C (12,4,-2)m. Determine the reactions at A. O O O O O Rx=-4kN Ry=-4kN Rz=2kN Mx=16 kN-m My=16 kN-m Mz=32 kN m 8m Don't Know Rx=-4KN Ry=-4kN Rz=-4kN Mx=0 kN-m My=32 kN-m Mz=32 kN.m Rx=-4kN Ry=-4kN Rz=2kN Mx=0 kN.m My=16 kN-m Mz=32 kN m Rx=-4KN Ry=-4kN Rz=-4KN Mx=0 kN-m My=16 kN m Mz-32 kN.m F=6kN B(8,0,0) m *C(12,4,-2) mThe beam ABCD are resting on a pin support at A and a roller support at E (as shown below). What are the magnitudes of the horizontal and vertical components of the reactions at A and E. A 10 KN B 15 KN C 60 5m 5m+ Horizontal reaction at A = 2.5 kN Vertical reaction at A = 11.64 kN Vertical reaction at E= 27.58 kN O Horizontal reaction at A=5.2 kN Vertical reaction at A= 17.46 kN Vertical reaction at E= 11.48 kN O Horizontal reaction at A = 2.5 KN Vertical reaction at A - 17.46 kN Vertical reaction at E= 22.85 kN O Horizontal reaction at A 7.7 kN Vertical reaction at A= 23.39 kN Vertical reaction at E = 22.85 kN 20 KN 60 D -6 m E Art Microsoft Store > C E na to. Find the internal reactions (forces and moments) at point J. B C 0.5 m 0.5 m D G H 0.6 m J 30 0.2 m E J. 6:25 0.4 m 12 kN 16 KN F