Question

please answer question 2a and 2b with clear steps and remarks

(a) For the pin-jointed truss under design loads shown in Figure 2(a), member BC, AC and AD using the Method of Sections. Indicate whether the members are in tension (T) or compression (C) calculate the forces in 20 kN 70 kN 45° 45° 45° 45 80 kN 45° 45 4m 4m 4m L 4m 4m Figure 2(a)

Answer 1

We cut a plane through required forces and use equilibrium equation to calculate forces.

Please find the solution in attached images and ask any further query in comment section.

(b)

a) Let R_E & R_F be reactions at E & F. Moment about E = 0 20 times 4 + 80 times 8 + 70 times 16 = R_F times 20 rightwardsdoublearrow R_F = 92 kN Cut a section through BC, CA & DA Moment about C = 0 F_DA times 4 + 70 times 4 = 92 times 8 F_DA = 114 kN (Tensile) Vertical equilibrium F_CA cos 45 degree + 70 = 92 F_CA = 22 squareroot 2 kN (Tensile) Horizontal Equilibrium F_BC + F_CA sin theta + F_DA = 0 F_BC = -(114 + 22 squareroot 2 sin 45) F_BC = -136 kN (Compressive) \r\n (b) (i) Yielding starts at P = 1400 N Area = 50 mm^2 sigma y = load/A = 14000/50 = 280 N/mm^2 (ii) Failure occurs at P = 19600 N A = 50 mm^2 sigma_ultimate = 19600/50 = 392 N/mm^2 (iii) Ductility = % change in length before failure = Elongation/Total length times 100

(iv) Work hardening is strengthening of metal by plastic deformation. Strengthening occurs by movement of dislocation and producing them .

Modulus: It does not change with work hardening because work hardening mainly affects the macro structure of metal and not grain properties.

Yield strength: It increases as it dislocation density is increased by work hardening.

Ultimate strength: Extent of strain hardening increase with work hardening giving a rise to ultimate strength.

Ductility: Since the material gets hardened, capacity to deform decreases and hence ductility decreases.