As shown above, a uniform beam of length d = 5.10 ft and weight 45.6 lb is attac

Question

                    As shown above, a uniform beam of length d = 5.10 ft and weight 45.6 lb is attached to a wall with a pin at point B. A cable attached at point A supports the                    beam. The beam supports a distributed weight w2 = 22.0 lb/ft. IF the support cable can sustain a maximum tension of 300 lb, what is the maximum value of w1?                    Under this maximum weight, what is F_By, the vertical of the support's reaction force at point B?

Learning Goal: To be able to solve for unknown forces and moments in rigid-body problems using the equations of equilibrium. For a rigid body to be in equilibrium, both the sum of the forces and the sum of the moments about an arbitrary point O must be zero: F =0 Mo = 0 When all of the forces lie in the x-y plane, the forces can be resolved into their x and y components, which results in the equations of equilibrium in two dimensions: Fx =0

Explanation / Answer

considering the FBD of horizontal beam, equating the moment about B = 0,

considering the given load as uniform load of W2 and triangular load of W1

and calculating moments separately, T(sin30)xd=[(W2)xd/2]+[weightxd/2]+[1/2xdx(W1-W2)xd/3],

cancelling d on both sides, 300sin30=[22 x5.1/2]+[45.6/2]+[1/6x5.1x(W1-22)],

W1=115.05 lb/ft.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.