Question

Draw the free-body diagram of each member (AB, BC, AD, EDC) and pin (A, B, C, D, E). Then, find all the forces previously reported. Force at C (5 kN) is applied on the pin C 5 kN/m 1 kN/m 1 kN/m 2 kN-m 2 m 1.5 m 1.5 m 5 kN

Reactions of all pins and members must be calculated.

Answer 1

The free body diagram of the beam AB 5 KN/m 1 kN/m 11 kN/m A 1.5m 1.5 m Here, A, A are the horizontal and vertical components of moment of force at point A and BI.B are the horizontal and vertical components of moment of force at point B. The equation about point A is, ΣΜ.-0 3 (5) (1.5 (5) (1.5 B, (3.0 m-1(3) (1.5 1.5 3 3 B 5 kN - The vertical equilibrium equation at point A is, ΣF-0 4+Bx3)5)8)- 0 A 5.5 kN The horizontal reaction at point A is

A B 0 4 B The free body diagram of the beam BC is, B. 2 kN m 2m C, the horizontal and vertical components of moment of force at point C Here, CC are The equation of horizontal reaction at point B is, C B.(2) 2 0 B 1 kN This implies that, A =-1 kN

The horizontal reaction at point C is С, + В, -0 С, %3-В, --(-1 kN) =1 kN The vertical reaction at point C is, В, +С, %3D0 С, %3D-В, (5 kN) - 1 =-5 kN The free body diagram of the beam EDC is, F. AD 22.5 5kN 1.5 E E C 1.5 m 1.5 m The moment of equilibrium equation about point E is, 0 F 2 (1.50 2.5 AD 2 5(3)-5(3)+F (1.5) 0 4T 2.5 F 25 kN AD

The vertical equilibrium equation at point E is, 1 2 -5 0 2.5 E, +C, FAD 2 E, 5-25 +5 2.5 = =-10 kN The horizontal equilibrium equation at point E is,

(1.5 Е -С, - Fp 2.5 AD 1.5 Е, 3С, +F 2.5 1.5 1+25 2.5 16 kN