(a) What is the supporting force system at A for the cantilever beam shown in the figure ? Neglect the weight of the beam. (10…

(a) Supporting force system at A

A fixed support gives Aₓ, A_y and a fixing moment M_A. Load resultants: the triangular load is ½ × 1.5 × 400 = 300 N, acting 1.0 m from A (at ⅔ of 1.5 m); the 300 N point load acts 2.0 m from A; the 500 N force at B (x = 3.0 m) has components 500 cos 30° = 433.0 N (→) and 500 sin 30° = 250 N (↓).

ΣFₓ = 0: Aₓ = 433.0 N, directed to the left.

ΣF_y = 0: A_y = 300 + 300 + 250 = 850 N, upward.

ΣM_A = 0: M_A = 300(1.0) + 300(2.0) + 250(3.0) = 1650 N·m, anticlockwise. The 433 N component acts along the beam axis and has no moment about A.

Supporting force system: a force of √(433.0² + 850²) = 954 N (433 N ← and 850 N ↑, at 63.0° above the horizontal) together with a fixing couple of 1650 N·m anticlockwise.

(b) Velocity and angle

With Δt = 3.67 − 3.65 = 0.02 s and the average acceleration a = 4î + 6ĵ m/s²,

v = u + aΔt = (6.12 + 0.08)î + (3.24 + 0.12)ĵ = 6.20î + 3.36ĵ m/s.

Its magnitude is √(6.20² + 3.36²) = 7.052 m/s, at 28.45° to the x-axis.

Taking a as the acceleration at t = 3.67 s, a·v = 4(6.20) + 6(3.36) = 44.96 and |a| = √52 = 7.211 m/s². So

cos θ = 44.96/(7.211 × 7.052) = 0.884, giving θ = 27.9°.

Check: a is at 56.31° and v at 28.45° to the x-axis, a difference of 27.86°. Since θ < 90°, the particle is speeding up.

(c) Carbon equivalent (CE)

CE expresses the combined hardening effect of carbon and the alloying elements as one equivalent carbon content. The IIW formula is

CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 (wt %).

Effect on hardening: alloying elements delay the transformation of austenite to ferrite and pearlite, shifting the TTT/CCT nose to the right. A higher CE therefore lowers the critical cooling rate and increases hardenability, so martensite forms to a greater depth and with a milder quench (oil or air instead of water).

A high CE also gives harder, more brittle martensite with larger transformation stresses. Severe quenching then risks cracking and distortion, so prompt tempering is needed. In welding, a high CE (above about 0.40) makes the heat-affected zone harden and crack, so preheating is specified.

(d) Speed of arm G

Since all gears have the same module and the centre distances C–D and E–F are equal, T_C + T_D = T_F + T_E, so T_F = 35 + 65 − 32 = 68.

Take B (and C) as +28 rpm and A (and F) as −60 rpm; let the arm turn at x rpm. With the arm fixed and C turned by y revolutions, D–E turns by −(35/65)y and F by +(35/65)(32/68)y = +(56/221)y. Adding x to every member:

x + y = 28 and x + (56/221)y = −60.

Subtracting gives (165/221)y = 88, so y = 117.87 and x = 28 − 117.87 = −89.87.

Arm G rotates at 89.9 rpm, opposite to shaft B and in the same sense as shaft A.

(e) Elongation of the hanging wire

Here A = (π/4)(8²) = 50.27 mm², L = 50 000 mm, E = 2.1 × 10⁵ N/mm².

Due to the 2000 N load: δ₁ = PL/AE = (2000 × 50 000)/(50.27 × 2.1 × 10⁵) = 9.474 mm.

Due to self-weight, the stress at each section grows with the length hanging below it. With γ = ρg = 8000 × 9.81 = 7.848 × 10⁻⁵ N/mm³,

δ₂ = γL²/2E = (7.848 × 10⁻⁵ × 50 000²)/(2 × 2.1 × 10⁵) = 0.467 mm.

Total elongation = 9.474 + 0.467 ≈ 9.94 mm.