(a) The coefficient of friction for all contacting surfaces in Figure 3(a) is 0·2. Does the 25 kg force move the block A up, hold…

(a) Let θ = 15°, μ = 0.2. Work in kg force: W_A = 100 kgf, W_B = 50 kgf, applied force F = 25 kgf leftwards. Let N be the normal reaction between A and B, H the wall reaction on A, and R the floor reaction on B.

For impending motion of A up and B left: interface friction on A acts down the incline, wall friction on A acts down, and floor friction on B acts right. For A:

• x: H = N sinθ + μN cosθ = N(sinθ + μ cosθ)
• y: N cosθ = W_A + μN sinθ + μH Substituting H: N[(1 − μ²)cosθ − 2μ sinθ] = W_A N = 100 / [0.96 × 0.9659 − 0.4 × 0.2588] = 121.4 kgf H = 121.4(0.2588 + 0.2 × 0.9659) = 54.9 kgf For B: R = W_B + N(cosθ − μ sinθ) = 50 + 121.4(0.9659 − 0.2 × 0.2588) = 161.0 kgf Horizontal equilibrium: F_up = H + μR = 54.9 + 0.2 × 161.0 = 87.1 kgf Since 25 kgf < 87.1 kgf, the force cannot move A up.

For impending motion of A down and B right: interface friction on A acts up the incline, wall friction on A acts up, and floor friction on B acts left. For A:

• x: H = N sinθ − μN cosθ = N(sinθ − μ cosθ)
• y: N cosθ + μN sinθ + μH = W_A Substituting H: N[(1 − μ²)cosθ + 2μ sinθ] = W_A N = 100 / [0.96 × 0.9659 + 0.4 × 0.2588] = 97.0 kgf H = 97.0(0.2588 − 0.2 × 0.9659) = 6.37 kgf For B: R = W_B + N(cosθ + μ sinθ) = 50 + 97.0(0.9659 + 0.2 × 0.2588) = 148.7 kgf For impending right motion of B, the required external left force would be: F_right = H − μR = 6.37 − 0.2 × 148.7 = −23.4 kgf The negative sign means no left force is needed to prevent this motion; a rightward force of 23.4 kgf would be required to cause it.

Final answer: the 25 kg force neither moves A up nor is too small. It holds the system in equilibrium.

(b) Take x horizontal right and y vertical up. From the stress element: σ_x = −80 N/mm² (compression), σ_y = +20 N/mm² (tension), τ_xy = −30 N/mm². Clockwise rotation through 15° means θ = −15°, so 2θ = −30°. cos 2θ = 0.8660, sin 2θ = −0.5.

Use the plane-stress transformation equations: σ_x′ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 cos 2θ + τ_xy sin 2θ = −30 + (−50)(0.8660) + (−30)(−0.5) = −30 − 43.30 + 15 = −58.30 N/mm²

σ_y′ = (σ_x + σ_y)/2 − (σ_x − σ_y)/2 cos 2θ − τ_xy sin 2θ = −30 − (−50)(0.8660) − (−30)(−0.5) = −30 + 43.30 − 15 = −1.70 N/mm²

τ_x′y′ = −(σ_x − σ_y)/2 sin 2θ + τ_xy cos 2θ = −(−50)(−0.5) + (−30)(0.8660) = −25 − 25.98 = −50.98 N/mm²

Thus the rotated element has normal stresses σ_x′ = 58.30 N/mm² compressive, σ_y′ = 1.70 N/mm² compressive, and shear stress magnitude 50.98 N/mm², with sign corresponding to the negative τ_x′y′ direction. Check: σ_x′ + σ_y′ = −60 N/mm², equal to the original invariant σ_x + σ_y.

(c) Pearlite has a lamellar eutectoid microstructure of alternating ferrite and cementite (Fe₃C) layers, formed by diffusion-controlled transformation of austenite at about 550–727°C. Finer interlamellar spacing, obtained at lower transformation temperature, gives higher strength and hardness. It has moderate strength and hardness, good ductility and toughness, and good machinability; typical UTS is about 600–900 MPa and hardness about 15–30 HRC.

Bainite is a non-lamellar aggregate of ferrite and fine carbides formed isothermally at about 250–550°C. Upper bainite (about 400–550°C) has carbides between ferrite laths, while lower bainite (about 250–400°C) has fine carbides inside ferrite laths. Bainite gives a better combination of strength and toughness than pearlite at the same hardness; lower bainite is stronger and tougher than upper bainite. Hardness is typically about 40–55 HRC.

Martensite forms by diffusionless shear transformation of austenite on rapid quenching below the Ms temperature. It is a body-centred tetragonal supersaturated solid solution of carbon in α-iron. Low-carbon martensite is lath-like; high-carbon martensite is plate-like or twinned. It has the highest hardness and strength but low ductility and toughness, is brittle, and develops high residual stresses and cracking tendency. Hardness can reach about 65 HRC for high carbon. Tempering is required to improve toughness.