(a) A rod of length 2 m and weight 450 N rests on ground at one end. The other end is supported by a cord through which a force…

(a) Let T = 200 N, W = 450 N and L = 2 m. Take moments about the lower end A. If the cord makes an angle β with the horizontal, the moment of T about A is T L sin(α + β), and the moment of the weight is W (L/2) cos α. For equilibrium:

T L sin(α + β) = W (L/2) cos α.

For the minimum possible α, sin(α + β) must be maximum, i.e. 1, which means the cord is perpendicular to the rod. Hence:

T L ≥ W (L/2) cos α

cos α ≤ 2T/W = 2(200)/450 = 8/9.

Therefore:

α_min = arccos(8/9) ≈ 27.27°.

Check friction at this angle. sin α = √(1 − (8/9)²) = √17/9. Vertical component of cord force = T cos α = 200(8/9) = 177.78 N upward. Normal reaction:

N = 450 − 177.78 = 272.22 N.

Required friction = T sin α = 200(√17/9) = 91.63 N.

Maximum static friction = μN = 0.4(272.22) = 108.89 N.

Since 91.63 N < 108.89 N, slip does not occur. Hence equilibrium is possible.

(b) Given L = 1.5 m, d = 20 mm = 0.02 m, m = 125 kg, E = 200 GPa = 200 × 10⁹ N/m². Neglect shaft mass. For a simply supported beam with central mass, stiffness is:

k = 48EI/L³.

Moment of inertia:

I = πd⁴/64 = π(0.02)⁴/64 = 7.854 × 10⁻⁹ m⁴.

EI = (200 × 10⁹)(7.854 × 10⁻⁹) = 1570.8 N m².

k = 48(1570.8)/(1.5³) = 75398.4/3.375 = 22340.2 N/m.

Natural frequency:

ω_n = √(k/m) = √(22340.2/125) = 13.369 rad/s.

Fundamental frequency:

f_n = ω_n/(2π) = 13.369/(2π) = 2.128 Hz.

f_n ≈ 2.13 Hz.

(c) Nut movement per one-fourth turn:

δ = pitch/4 = 1.8/4 = 0.45 mm.

Bolt area:

A_b = π(12)²/4 = 36π = 113.10 mm².

Copper tube area:

A_c = π(20² − 14²)/4 = 51π = 160.22 mm².

Let F be the tensile force in the bolt and compressive force in the tube. Compatibility gives:

δ = F L/(A_b E_s) + F L/(A_c E_c).

Substitute L = 120 mm, E_s = 200000 N/mm², E_c = 105000 N/mm².

1/(A_b E_s) = 1/(113.10 × 200000) = 4.421 × 10⁻⁸ N⁻¹.

1/(A_c E_c) = 1/(160.22 × 105000) = 5.944 × 10⁻⁸ N⁻¹.

Sum = 1.0365 × 10⁻⁷ N⁻¹.

δ = F(120)(1.0365 × 10⁻⁷) = F(1.2438 × 10⁻⁵ mm/N).

F = 0.45/(1.2438 × 10⁻⁵) = 36179 N.

Bolt stress:

σ_b = F/A_b = 36179/113.10 = 319.9 N/mm² = 320 MPa tension.

Copper stress:

σ_c = F/A_c = 36179/160.22 = 225.8 N/mm² = 226 MPa compression.

σ_b ≈ 320 MPa (tension), σ_c ≈ 226 MPa (compression).

(d) A continuous cooling transformation (CCT) diagram plots temperature against time on a logarithmic scale for austenite cooled continuously. Unlike an isothermal TTT diagram, CCT curves are shifted to lower temperatures and longer times because transformation occurs during continuous cooling. The critical cooling rate is the minimum cooling rate from the austenitizing temperature that suppresses diffusion-controlled transformations such as ferrite, pearlite and bainite, and ensures that austenite transforms largely to martensite. On the CCT diagram, it is the cooling curve just tangent to the start of the pearlite/ferrite transformation region. If cooling is slower, some soft pearlite or bainite forms; if faster, the curve misses the transformation nose and martensite forms at M_s.

Importance in hardening of alloy steel: It is a measure of hardenability. Alloying elements such as Cr, Ni, Mo and Mn shift the CCT curves to the right, lowering the critical cooling rate. This allows through-hardening of thicker sections with milder quenching media like oil or air, reducing quench cracking and distortion. It also guides selection of quenching medium and cooling rate to obtain the desired martensitic structure, hardness and toughness, and helps design heat-treatment cycles that avoid undesirable phases and control residual stresses.

(e) Given N = 250 rpm, stroke = 300 mm, so crank radius r = 150 mm = 0.15 m. m_rec = 60 kg, m_rev = 50 kg at r = 0.15 m. Balanced fraction of reciprocating parts = 2/3.

(i) Total mass to be balanced at crank radius:

m_rev + (2/3)m_rec = 50 + (2/3)(60) = 50 + 40 = 90 kg.

Balance mass B at radius R = 400 mm = 0.4 m:

B R = 90 r

B = 90(0.15)/0.4 = 33.75 kg.

B = 33.75 kg at radius 400 mm.

(ii) Angular speed:

ω = 2πN/60 = 2π(250)/60 = 25π/3 = 26.18 rad/s.

m_rec ω² r = 60(25π/3)²(0.15) = 625π² = 6168.5 N.

For balanced fraction c = 2/3, residual primary force components at crank angle θ = 60° from TDC are:

Along line of stroke:

F_x = (1 − c)m_rec ω² r cos θ = (1/3)(6168.5) cos 60° = 1028.1 N.

Perpendicular to line of stroke:

F_y = c m_rec ω² r sin θ = (2/3)(6168.5) sin 60° = 3561.4 N.

Resultant residual unbalanced force:

F = √(F_x² + F_y²) = 6168.5 √((1/3)²cos²60° + (2/3)²sin²60°)

F = 6168.5 √(13/36) = 6168.5(√13/6) = 3706.8 N.

F ≈ 3.707 kN, acting at tan⁻¹(3561.4/1028.1) ≈ 73.9° to the line of stroke.