Question 1 Three forces (not to the scale) are acting at point A as in Figure 1. Calculate the resultant force (magnitude and direction) on point A. Clearly show all the force components. (20 marks) 20 N 1000 30 N 100 A Figure 1 10 N Question 2 The curved rod ABC has a radius r and is fixed to the wall at C. A force P is acting downward at point A as shown in the Figure 2. Determine the reactions at C. Solve the problem with a clear free body diagram. (20 marks) C B Strength udents. Queston 3 A svome a ognaty 300 mm long, has a diameter of 12 mm, and is subjected to a force 23N When the force is increased from 2.5 KN to 9 KN, the specimen elongates 0.225 One the modulus of elasticity for the material if it remains linear elastic. (20 Question 4 Roar bars of copper and aluminium are held by pins at their ends, as shown in the gure 3 below. Thin spaces provide a separation between the bars. The copper bars have www.onal dimensions 12 mm by 50mm, and the aluminium bar has dimensions 25 mm by 30mm Deemine the shear stress in the 11mm diameter pins if the temperature is b40(20 marks) For copper E-124 GP and a 20x10 C for aluminium, E,= 69 GPa and a=26x10*/C Copper bar- Aluminum bar Copper bar Figure 3 Useful formula FL Axial lead: Normal stress displacement: $= EA **Qun & Q(ATL End of Examination Page 2 of 2 Question 3 A son is orginally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 25 N When the force is increased from 2.5 kN to 9 kN, the specimen elongates 0.225 Dere the modulus of elasticity for the material if it remains linear elastic. (20 marks Question 4 Redangular bars of copper and aluminium are held by pins at their ends, as shown in the gure 3 below. Thin spacers provide a separation between the bars. The copper bars have cross-sectional dimensions 12 mm by 50mm, and the aluminium bar has dimensions 25 mm by 50 mm. Determine the shear stress in the 11mm diameter pins if the temperature is sed by 40C. (20 marks) For copper E-124 GPa and a=20x10C; for aluminium, E- 69 GPa and a,-26x10C.) Copper bar- Aluminum bar Copper bar Figure 3 Useful formula Axial load: Normal stress - displacement 5 - E FL EA a=a(37) &= Q(ATL