Supplementary problem 1

1. A B A B B A 200 Nm 200 Nm 200 Nm 100 N 100 N 100 N 160 N 160 N 160 N 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m 0.1 m Supplementary problem 1 2. Replace bearings with reactions Ax Az Bz 1. Assign the co-ordinate system 3. Use equilibrium conditions ∑Fx = 0 → Ax = 0 ∑My = 0 → 0.4Bz-0.3×160-0.2×100- 200 = 0 → Bz= 670 N ∑Fz = 0 → Az+Bz = 100+160 = 260 N → Az = -410 N X Y Z

2. B B A A A B B A 50 N 50 N 50 N 1 1 4 4 4 40 N 1 1 3 3 3 3 m 3 m 9 m 9 m 3 m 3 m 9 m 9 m Supplementary problem 2 1 3 X Y Z 4. Resolve forces along X and Y directions 2 Ax Ax Bz Bx Az Az Bxz 30 N

3. B A 40 N 3 m 9 m Supplementary problem 2 5. Use equilibrium conditions Ax Bz • ∑Fx = 0 → Ax-Bx-30 = 0 • ∑My = 0 → 3×40-12Bz = 0 → Bz= 10 N • ∑Fz = 0 → Az+Bz = 40 N → Az = 30 N • Bx = Bz→ Bx= 10 N → Ax = 40 N Bx Az 30 N

4. p p p A B C B C A B C A a a a l/2 l/2 l/2 l/2 l/2 l/2 Supplementary problem 3 1 3 X Y Z 2 Cx Ax Az Cz

5. p C A B a l/2 l/2 Supplementary problem 3 4. Make free body diagrams Bz Bx Cx CMy Ax Bx B Az Bz Cz 5 • ∑Fx = 0 → Ax = Bx = Cx • ∑My = 0 → Bz= -p/2 • → Az= p/2 • → CMy= -pa/2 • ∑Fz = 0 → Bz+Cz = 0 → Cz = p/2

6. 10 kN/m 10 kN/m 10 kN/m 2 m 2 m B B A A 3 m 3 m Supplementary problem 4 15 kN Ax Bz • Equivalent force generated by line load = area under the force distribution curve • Line of action – through centroid of the force distribution curve Az 3×2/3 m • Bz = 6 kN, Az = 9 kN, Ax = 0

7. Problem 3 1. Movement of axes along body axis X X X X C B Y Y Y Y Z Z Z Z D A

8. Problem 3 2. Dissection of body in segment CD x C X X B Y Y Q negative area M Z Z L L M F3 c-x Q F2 D A 3. Use equilibrium conditions for the segment and calculate IFVs

9. 4. Repeat for segments BC and AB 5. Draw the diagrams for Q,L and M • Draw the body axis • Draw the IFV distribution • Show the sign of IFV in the diagram • Show the important IFV magnitudes Q + F1 • F1 only