-1.5 m 1'm -1.5 m. 0.75 m Probs. 8–51/52

Q: There is a P = 188 KN load applied in the mid-span of the below simply supported beam with the given…

A:

Q: Three forces are applied to lever ABD. (a) Draw Mohr's Circle for Point H, (b) determine the max…

A: free body diagram at H Draw Bending stress diagram Shear stress diagram

Q: H.W Replace the loading equivalent resultant force and specify its location on the beam, measured…

A: Consider the given beam. The resultant force will be equal to the algebraic sum of all the forces.

Q: F9-5. The beam is subjected to the load at its end Determine the maximum principal stress at point…

A: Given : Determine the maximum principle stress at point B ? Concept -  Principle stresses are taken…

Q: Determine the force in member CG for the truss using the method of joints. 133 kN 67 kN B F G -4 at…

A: Given Data  Find the force in member CG (FCG) using method of joint.

Q: Determine the resultant force at pins A, B, and C on the three-member frame. P=129 N q=200 N/m

A: Load P = 129 NLoad intensity q = 200 N/m

Q: Situation 1: From the truss shown, each member has a cross-sectional area of A = 400 mm 2, E = 200…

A: Area of each member, A = 400mm2Young's modulus, E = 200GPaThe truss is loaded with a downwards point…

Q: 4. Select member FC and CB in the truss to carry the inclined force P. Set the allowable tensile…

A:

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.