(a) A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in Figure 1(a). The crank is…

(a) Let r = 60 mm, l = 240 mm, ω = 10 rad/s. From Figure 1(a), A and B are on the same horizontal line and AO is perpendicular to OB. Let θ be the crank angle measured from AB. The validity condition is that the slider is to the right of A and the connecting rod is not folded, so l > r and the square root below is real.

In right triangle AOB, AB = √(r² + l²) = √(60² + 240²) = 60√17 mm. Hence cosθ = r/AB = 1/√17 and sinθ = l/AB = 4/√17.

For a slider-crank, the slider position from A is x = r cosθ + √(l² − r² sin²θ). Differentiating with respect to time, with dθ/dt = ω, dx/dt = −rω sinθ − r²ω sinθ cosθ / √(l² − r² sin²θ) = −rω sinθ [1 + r cosθ / √(l² − r² sin²θ)].

Now √(l² − r² sin²θ) = √(240² − 60²·16/17) = 960/√17 mm. Therefore the speed magnitude is v_B = rω sinθ [1 + r cosθ / √(l² − r² sin²θ)] = 60·10·(4/√17) [1 + 60(1/√17)/(960/√17)] = 2400/√17 · (17/16) = 150√17 mm/s ≈ 618.5 mm/s.

As a check, the component of the crank-pin velocity along the connecting rod is −600 mm/s; since the slider velocity component along the rod must equal this, v_B(4/√17) = −600, giving the same result.

Speed of slider = 150√17 mm/s ≈ 618.5 mm/s. For the drawn configuration with O above AB and CCW rotation, dx/dt is negative, so the slider moves toward A.

(b) The given two-dimensional principal stresses are +160 MPa and −80 MPa. For a plane stress state, the third principal stress is zero, so the ordered principal stresses are σ₁ = 160 MPa, σ₂ = 0 MPa, σ₃ = −80 MPa. The elastic limit in simple tension is σ_e = 300 MPa. Factor of safety is dimensionless. The maximum principal stress theory is most appropriate for brittle materials, while the maximum shear stress and distortion energy theories are commonly used for ductile materials.

• (i) Maximum principal stress theory: failure is predicted when the largest tensile principal stress reaches σ_e. FOS = σ_e/σ₁ = 300/160 = 15/8. FOS = 1.875.

• (ii) Maximum shear stress theory: the maximum shear stress is half the difference between the largest and smallest principal stresses. τ_max = (σ₁ − σ₃)/2 = (160 − (−80))/2 = 120 MPa. In a simple tension test, the allowable maximum shear stress is σ_e/2 = 150 MPa. FOS = 150/120 = 5/4. FOS = 1.25.

• (iii) Maximum distortion energy theory: the equivalent von Mises stress is σ_vm = √(1/2[(σ₁−σ₂)² + (σ₂−σ₃)² + (σ₃−σ₁)²]) = √(1/2[160² + 80² + 240²]) = √44800 = 80√7 MPa ≈ 211.66 MPa. FOS = σ_e/σ_vm = 300/(80√7) = 15/(4√7) = 15√7/28 ≈ 1.417. FOS = 1.417.

(c) From Figure 1(c), take A as the origin, +x to the right, +y upward, and +z out of the page. The right-handed system is consistent with the described axes. The force positions are in metres and force vectors in newtons.

• 900 N downward at (0, 0, 10): F₁ = (0, −900, 0)
• 200 N downward at (0, −4, 20): F₂ = (0, −200, 0)
• 300 N downward at (5, −8, 20): F₃ = (0, −300, 0)
• At the free end (10, −8, 20): F₄ = (−600, 0, 0), F₅ = (400, 0, 0), F₆ = (0, −500, 0)

The resultant force is R = ΣF = (−200, −1900, 0) N. Its magnitude is |R| = √(200² + 1900²) = 100√365 N ≈ 1910.5 N. The direction is 84.0° below the negative x-axis, since arctan(1900/200) = 84.0°.

The moment about A is M_A = Σ r × F, where r × F = (yF_z − zF_y, zF_x − xF_z, xF_y − yF_x). Evaluating each force:

• 900 N: (9000, 0, 0) N·m
• 200 N: (4000, 0, 0) N·m
• 300 N: (6000, 0, 0) N·m
• 600 N: (0, −12000, −4800) N·m
• 400 N: (0, 8000, 3200) N·m
• 500 N: (10000, 0, 0) N·m

Summing, M_A = (29000, −4000, −1600) N·m. Its magnitude is |M_A| = √(29000² + 4000² + 1600²) = 200√21489 N·m ≈ 29318 N·m.

Equivalent system at A: R = (−200 i − 1900 j) N and M_A = (29000 i − 4000 j − 1600 k) N·m. If the question requires only the resultant force, R = 100√365 N ≈ 1910.5 N in the stated direction.

(d)

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(e) Given standard module m = 10 mm, pinion teeth z₁ = 14, gear teeth z₂ = 54, and addendum a = m = 10 mm. Pitch radii: r₁ = mz₁/2 = 70 mm, r₂ = mz₂/2 = 270 mm. Addendum radii: r_a1 = r₁ + a = 80 mm, r_a2 = r₂ + a = 280 mm. Centre distance: C = r₁ + r₂ = 340 mm.

For involute gears, the line of action is tangent to the base circles. The distance from the pitch point to the pinion base-circle tangent point is r₁ sinφ. The gear addendum circle intersects the line of action at a distance √(r_a2² − r_b2²) − r₂ sinφ from the pitch point on the pinion side, where r_b2 = r₂ cosφ. No interference of the gear addendum on the pinion requires this distance not to exceed r₁ sinφ: √(r_a2² − r_b2²) − r₂ sinφ ≤ r₁ sinφ. This simplifies to r_a2² ≤ r_b2² + C² sin²φ.

At φ = 20°: r_a2² = 280² = 78400 mm². RHS = 270² cos²20° + 340² sin²20° ≈ 64372.3 + 13522.6 = 77894.9 mm². Since 78400 > 77894.9, interference occurs.

The opposite possible interference, pinion addendum on the gear, is checked by r_a1² ≤ r_b1² + C² sin²φ. At 20°, 6400 mm² ≤ 70² cos²20° + 340² sin²20° ≈ 17849.4 mm², so it is not governing.

To eliminate interference, set the governing condition to equality: (r₂ + m)² = r₂² cos²φ + C² sin²φ. Using cos²φ = 1 − sin²φ, r₂² + 2r₂m + m² = r₂² + (C² − r₂²) sin²φ. Therefore sin²φ_min = (2r₂m + m²)/(C² − r₂²) = (2·270·10 + 10²)/(340² − 270²) = 5500/42700 = 55/427. φ_min = arcsin(√(55/427)) ≈ 21.03°.

Interference occurs at 20°; the pressure angle must be increased by at least 1.03° to 21.03° to eliminate interference. A smaller pressure angle would increase interference.