RTU Mechanics of Solids Solutions
📄 RTU Mechanics of Solids Solved Paper 2016
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Rajasthan Technical University (RTU) Solutions

B.Tech. III-Semester (Dec 2016) | Mechanics of Solids (3E1631)

UNIT - I
Q1 (a). Derive an expression between modulus of elasticity and modulus of rigidity. (6 Marks)
Consider a square element ABCD subjected to pure shear stress Ï„.

Derivation Steps:
  • Shear strain = Ï„ / G
  • Linear strain along diagonal AC = Shear strain / 2 = Ï„ / (2G)
  • Linear strain along diagonal AC in terms of E and Poisson's ratio (1/m or μ) = (Ï„ / E) * (1 + μ)
  • Equating both strains: Ï„ / (2G) = (Ï„ / E) * (1 + μ)
Result: E = 2G(1 + μ) Where: E = Young's Modulus G = Shear Modulus / Modulus of Rigidity μ = Poisson's Ratio
Q1 (b). A steel rod and two copper rods together support a load of 370 kN. Find stresses in the rods. (10 Marks)
Given Data: Total Load (P) = 370 kN = 370 x 10^3 N Steel Area (A_s) = 2500 mm² Copper Area (Total A_c for 2 rods) = 2 x 1600 = 3200 mm² E_s = 2 x 10^5 N/mm² | E_c = 1 x 10^5 N/mm² Length of Steel (L_s) = 15 cm + 10 cm = 25 cm = 250 mm Length of Copper (L_c) = 15 cm = 150 mm Step 1: Equilibrium Equation P_s + P_c = 370 x 10^3 N σ_s * A_s + σ_c * A_c = 370000 2500*σ_s + 3200*σ_c = 370000 ----(Equation 1) Step 2: Compatibility Equation (Deflection of Steel = Deflection of Copper) δL_s = δL_c (σ_s * L_s) / E_s = (σ_c * L_c) / E_c (σ_s * 250) / (2 x 10^5) = (σ_c * 150) / (1 x 10^5) σ_s * 1.25 x 10^-3 = σ_c * 1.5 x 10^-3 σ_s = 1.2 * σ_c ----(Equation 2) Step 3: Solve Equations Substitute Eq 2 into Eq 1: 2500*(1.2 * σ_c) + 3200*σ_c = 370000 3000*σ_c + 3200*σ_c = 370000 6200*σ_c = 370000 σ_c = 59.68 N/mm² Now find σ_s: σ_s = 1.2 * 59.68 = 71.61 N/mm² Final Answers: Stress in Steel Rod (σ_s) = 71.61 N/mm² (Compressive) Stress in Copper Rods (σ_c) = 59.68 N/mm² (Compressive)
Q1 (OR) (a). Prove that total expansion of a uniform tapering rod subjected to axial load P is dL = (4PL) / (Ï€ E D1 D2). (8 Marks)
Consider a section of diameter 'dx' at distance 'x' from larger end D1 tapering to D2 over length L.

Derivation Steps:
  • Diameter at distance x: D_x = D1 - [(D1 - D2)/L] * x
  • Area at section x: A_x = (Ï€ / 4) * (D_x)²
  • Elongation of small element dx: d(δL) = (P * dx) / (A_x * E)
  • Integrating from 0 to L: δL = ∫ [4P dx] / [Ï€ E (D_x)²]
Result: dL = (4 * P * L) / (Ï€ * E * D1 * D2)
Q1 (OR) (b). A brass bar subjected to axial forces. Find total elongation / stresses. (8 Marks)
Given Data: Area (A) = 900 mm² L_AB = 0.6 m = 600 mm, L_BC = 0.8 m = 800 mm, L_CD = 1.0 m = 1000 mm Forces: A = 40 kN (Left), B = 70 kN (Right), C = 20 kN (Left), D = 10 kN (Right) Step 1: Free Body Diagram of Each Segment - Portion AB: Force = 40 kN (Tension) - Portion BC: Force = 40 - 70 = -30 kN = 30 kN (Compression) - Portion CD: Force = 10 kN (Tension) Step 2: Stresses in Each Section σ_AB = P_AB / A = 40,000 / 900 = 44.44 N/mm² (Tensile) σ_BC = P_BC / A = 30,000 / 900 = 33.33 N/mm² (Compressive) σ_CD = P_CD / A = 10,000 / 900 = 11.11 N/mm² (Tensile) Step 3: Total Change in Length (Assuming E = 1 x 10^5 N/mm²) δL = (1 / (A * E)) * [P_AB * L_AB - P_BC * L_BC + P_CD * L_CD] δL = (1 / (900 * 10^5)) * [(40000 * 600) - (30000 * 800) + (10000 * 1000)] δL = (1 / 9 x 10^7) * [24 x 10^6 - 24 x 10^6 + 10 x 10^6] δL = 10^7 / (9 x 10^7) = 0.111 mm Final Answer: Total Elongation (δL) = 0.111 mm
UNIT - II
Q2 (a). Draw SFD and BMD for the given loaded beam. (12 Marks)
Beam Configuration: Simply supported beam AB = 8 m total length. UDL = 200 N/m over left 4 m. Point load = 3000 N at 4 m from A. Step 1: Support Reactions Total UDL Load = 200 * 4 = 800 N at 2 m from A. Taking moments about A: R_B * 8 = (800 * 2) + (3000 * 4) R_B * 8 = 1600 + 12000 = 13600 R_B = 1700 N R_A = Total Load - R_B = (800 + 3000) - 1700 = 2100 N Step 2: Shear Force Calculations SF at A = +2100 N SF just left of C (x=4m) = 2100 - (200 * 4) = +1300 N SF just right of C (x=4m) = 1300 - 3000 = -1700 N SF at B = -1700 N Step 3: Bending Moment Calculations BM at A = 0 BM at C (x=4m) = R_A * 4 - (200 * 4 * 2) = (2100 * 4) - 1600 = 6800 N-m BM at B = 0 Diagram Characteristics: SFD: Linear drop from 2100 N to 1300 N over A-C, sudden drop of 3000 N at C to -1700 N, constant to B. BMD: Parabolic curve from 0 to 6800 N-m (A to C), straight line drop from 6800 N-m to 0 (C to B). Max BM = 6800 N-m at x = 4m.
Q2 (b). What do you mean by thrust diagram? (4 Marks)
A Thrust Diagram (or Axial Force Diagram) is a graphical representation showing the variation of axial forces (tension or compression) along the length of a structural member or beam.
UNIT - III
Q3. Stresses on an oblique plane at 45° with minor tensile axis. (16 Marks)
Given Data: Major Tensile Stress (σ_x) = 80 N/mm² Minor Tensile Stress (σ_y) = 40 N/mm² Shear Stress (Ï„_xy) = 60 N/mm² Angle with axis of minor stress (θ) = 45° Step 1: Normal Stress (σ_n) Formula σ_n = [(σ_x + σ_y)/2] + [(σ_x - σ_y)/2] * cos(2θ) + Ï„_xy * sin(2θ) cos(90°) = 0, sin(90°) = 1 σ_n = [(80 + 40)/2] + [(80 - 40)/2] * 0 + 60 * 1 σ_n = 60 + 0 + 60 = 120 N/mm² Step 2: Tangential / Shear Stress (Ï„_t) Formula Ï„_t = [(σ_x - σ_y)/2] * sin(2θ) - Ï„_xy * cos(2θ) Ï„_t = [(80 - 40)/2] * 1 - 60 * 0 Ï„_t = 20 N/mm² Step 3: Resultant Stress (σ_r) σ_r = √(σ_n² + Ï„_t²) σ_r = √(120² + 20²) = √(14400 + 400) = √14800 = 121.65 N/mm² Final Answers: Normal Stress (σ_n) = 120 N/mm² Shear Stress (Ï„_t) = 20 N/mm² Resultant Stress (σ_r) = 121.65 N/mm²
UNIT - IV
Q4 (a). Derive the torsion equation: T/J = τ/R = (G*θ)/L. (8 Marks)
Consider a circular shaft fixed at one end and subjected to twisting moment T at the other end.

Derivation Steps:
  • Shear strain at outer surface = (R * θ) / L
  • From Hooke's Law: Ï„ = G * strain = (G * R * θ) / L => Ï„ / R = (G * θ) / L
  • Torque on elemental ring of radius 'r' and thickness 'dr': dT = Ï„_r * (2Ï€ r dr) * r = [(G * θ)/L] * 2Ï€ r³ dr
  • Total Torque T = ∫ dT = [(G * θ)/L] * J => T / J = (G * θ) / L
Final Torsion Equation: T / J = τ / R = (C * θ) / L
Q4 (b). Maximum shear stress in a solid shaft of dia 20 cm transmitting 187.5 kW at 200 RPM. (8 Marks)
Given Data: Diameter (D) = 20 cm = 200 mm Power (P) = 187.5 kW = 187.5 x 10^3 W Speed (N) = 200 RPM Step 1: Calculate Mean Torque (T) P = (2 * Ï€ * N * T) / 60 187.5 x 10^3 = (2 * Ï€ * 200 * T) / 60 T = (187.5 x 10^3 * 60) / (400 * Ï€) T = 8952.46 N-m = 8.952 x 10^6 N-mm Step 2: Polar Modulus (Z_p) Z_p = (Ï€ / 16) * D³ Z_p = (Ï€ / 16) * (200)³ = 1.57 x 10^6 mm³ Step 3: Max Shear Stress (Ï„) Ï„ = T / Z_p Ï„ = (8.952 x 10^6) / (1.57 x 10^6) Ï„ = 5.70 N/mm² Final Answer: Maximum Shear Stress (Ï„) = 5.70 N/mm² (or MPa)
UNIT - V
Q5. Thin cylinder: Change in dia, length, and volume. (16 Marks)
Given Data: Diameter (d) = 120 cm = 1200 mm Thickness (t) = 1.5 cm = 15 mm Length (L) = 6 m = 6000 mm Pressure (p) = 2.5 N/mm² E = 2 x 10^5 N/mm², μ = 0.3 Step 1: Stresses Circumferential Stress (σ_1) = (p * d) / (2 * t) = (2.5 * 1200) / (2 * 15) = 100 N/mm² Longitudinal Stress (σ_2) = (p * d) / (4 * t) = 100 / 2 = 50 N/mm² Step 2: Change in Diameter (δd) Hoop Strain (e_1) = (1 / E) * [σ_1 - μ * σ_2] e_1 = (1 / 2x10^5) * [100 - (0.3 * 50)] = (1 / 2x10^5) * 85 = 4.25 x 10^-4 δd = e_1 * d = (4.25 x 10^-4) * 1200 = 0.51 mm Step 3: Change in Length (δL) Longitudinal Strain (e_2) = (1 / E) * [σ_2 - μ * σ_1] e_2 = (1 / 2x10^5) * [50 - (0.3 * 100)] = (1 / 2x10^5) * 20 = 1.0 x 10^-4 δL = e_2 * L = (1.0 x 10^-4) * 6000 = 0.60 mm Step 4: Change in Volume (δV) Volumetric Strain (e_v) = 2*e_1 + e_2 = 2*(4.25 x 10^-4) + (1.0 x 10^-4) = 9.5 x 10^-4 Initial Volume (V) = (Ï€/4) * d² * L = (Ï€/4) * (1200)² * 6000 = 6.785 x 10^9 mm³ δV = e_v * V = (9.5 x 10^-4) * (6.785 x 10^9) = 6,445,750 mm³ = 6445.75 cm³ Final Answers: (i) Change in Diameter = 0.51 mm (ii) Change in Length = 0.60 mm (iii) Change in Volume = 6445.75 cm³
Q5 (OR). Beam deflection under point couple. (16 Marks)
Given Data: Span (L) = 6 m, Clockwise Couple (M) = 300 kN-m at x = 4m from left end A E = 2 x 10^5 N/mm² = 2 x 10^8 kN/m² I = 2 x 10^8 mm^4 = 2 x 10^-4 m^4 EI = (2 x 10^8) * (2 x 10^-4) = 40,000 kN-m² Step 1: Reactions R_A + R_B = 0 => R_A = -R_B Taking moments about A: R_B * 6 - 300 = 0 => R_B = +50 kN (Upward), R_A = -50 kN (Downward) Step 2: Macaulay's Equation EI * (d²y / dx²) = -50*x | + 300*(x - 4)^0 Integrate once: EI * (dy/dx) = -25*x² + C1 | + 300*(x - 4)^1 Integrate twice: EI * y = (-25/3)*x³ + C1*x + C2 | + 150*(x - 4)² Step 3: Boundary Conditions At x = 0, y = 0 => C2 = 0 At x = 6, y = 0: 0 = (-25/3)*(6)³ + C1*(6) + 150*(6 - 4)² 0 = -1800 + 6*C1 + 600 6*C1 = 1200 => C1 = 200 Slope & Deflection Equations: EI * y = -8.333*x³ + 200*x | + 150*(x - 4)² Step 4: Deflection at Couple Point (x = 4 m) EI * y_C = -8.333*(4)³ + 200*(4) = -533.31 + 800 = +266.69 kN-m³ y_C = 266.69 / 40000 = 0.00667 m = 6.67 mm (Upward) Step 5: Maximum Deflection Max deflection occurs where dy/dx = 0 in segment AC (x < 4): 0 = -25*x² + 200 => x² = 8 => x = 2.828 m from A EI * y_max = -8.333*(2.828)³ + 200*(2.828) = -188.56 + 565.6 = 377.04 kN-m³ y_max = 377.04 / 40000 = 0.00942 m = 9.42 mm (Upward) Final Answers: (i) Deflection at point of couple = 6.67 mm (ii) Maximum Deflection = 9.42 mm (at 2.83 m from A)
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