An aluminum ring has an outside diameter of 12.0 mm and an inside diameter of 10 mm. If the ring is short and is made of 2014-T6, compute the 3-13. force required to produce ultimate compressive failure in the ring. Assume that s, is the same in both tension and compression.

Need help on all questions. Please include all units, steps to the problem and information such as its direction or if it is in compression or tension. Thx.

number 9

Repeat the question but with the load at joint 1 acting horizontally? the answers are: (i) F1-2 100 kN (compression) F1-3 100 kN (tension) F2-3 50 kN (tension)(ii) R2=87 kN vertically upwards, R3 = 130kN at -41o (below horizontal) please show all workings.

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6.10 - This is a statics problem that deals with ‘Zero-Force Members’.

5 Please help. Provide Free Body Diagram, labeled with appropriate quantities to better understand. Also solve. Thank you.

The bars AB, AC, and AD are pinned together as shown in the Horizontal movement of the pin at A is prevented by the rigid horizontal strut AE. Calculate the axial force in the strut caused by the 10-kip load. For each steel bar, A = 0.3 in.2 and E = 29 x 106 psi. For the aluminum bar, A = 0.6 in.2 and E = 10 x 106 psi.