Structural Mechanics and Digital Communication Problems

Shear Stress in Beams

• 16.1 Rectangular beam: b = 100 mm, d = 250 mm, L = 3 m, w = 40 kN/m → Find τmax + distribution
• 16.2 Triangular section: b = 100 mm, h = 150 mm, F = 13.5 kN → Find τmax + distribution
• 16.3 Circular section: d = 100 mm, F = 30 kN → Find τmax + distribution
• 16.4 I-section: Flange = 150×20 mm, Web = 300×10 mm, F = 50 kN → Find τmax
• 16.5 I-section: D = 350 mm, B = 200 mm, web = 12.5 mm, flange = 25 mm, F = 200 kN → Shear stress distribution
• 16.6 T-section: F = 100 kN, I = 113.4 × 10⁶ mm⁴ → Shear at points + distribution
• 16.7 Unequal I-section: σc = 17.5 MPa, F = 100 kN → Bending moment + shear distribution
• 16.8 Irregular section: F = 20 kN → Shear at points + distribution
• 16.9 Square (diagonal): Diagonal = 2b, shear force = F → τmax + distribution
• 16.10 RSJ: D = 200 mm, B = 160 mm, tf = 22 mm, web = 12 mm → Shear share (flange vs web)

Torsion of Shafts

• 27.1 Solid shaft: D = 50 mm, τ = 40 MPa → Find torque
• 27.2 Solid shaft: T = 10 kN·m, τ = 45 MPa → Find diameter
• 27.3 Hollow shaft: D = 80 mm, d = 50 mm, τ = 45 MPa → Find torque
• 27.4 Solid shaft: D = 60 mm, N = 150 rpm, τ = 50 MPa → Find power
• 27.5 Hollow shaft: D = 100 mm, d = 40 mm, N = 120 rpm, τ = 50 MPa → Find power
• 27.6 Solid shaft: D = 100 mm, P = 120 kW, N = 150 rpm → Find shear stress
• 27.7 Hollow shaft: P = 200 kW, N = 80 rpm, τ = 60 MPa, d = 0.6D → Find diameters
• 27.8 Solid shaft: P = 100 kW, N = 160 rpm, τ = 70 MPa, Tmax = 1.2T → Find diameter
• 27.9 Solid shaft: D = 125 mm, θ = 1°, L = 1.5 m, C = 70 GPa → Find torque
• 27.10 Hollow shaft: D = 100 mm, d = 60 mm, τ = 35 MPa, C = 85 GPa → Find angle of twist
• 27.11 Solid shaft: D = 120 mm, P = 200 kW, N = 100 rpm, θ = 2°, C = 90 GPa → Find length
• 27.12 Solid shaft: D = 80 mm, θ = 1.5°, L = 5 m, τ = 42 MPa, C = 84 GPa → Find max torque
• 27.13 Solid shaft: T = 1.6 kN·m, τ = 60 MPa, θ = 1° per 20D, C = 80 GPa → Find diameter
• 27.14 Solid vs Hollow: Solid D = 200 mm, hollow d = 150 mm (same area) → Find (a) power ratio (b) twist ratio
• 27.15 Replace shaft: Solid D = 60 mm, hollow d = 0.5D → Find hollow diameters + material saving
• 27.16 Replace shaft: Solid D = 80 mm, hollow D = 100 mm → Find internal diameter
• 27.17 Replace shaft: Solid (Al): D = 50 mm, Hollow (Steel): same D, C₁ = 28 GPa, C₂ = 85 GPa → Find internal diameter
• 27.18 Hollow vs Solid: Hollow D = 300 mm, d = 200 mm, Cₛ = 2.4 Cₐ → Find solid diameter + rigidity ratio

Q. Digital Communication Systems

• Q1 State and explain Sampling Theorem. Derive Nyquist rate.
• Q2 Define sampling frequency, Nyquist rate, and Nyquist interval with examples.
• Q3 A signal of bandwidth 5 kHz is sampled → Find: (a) Minimum sampling frequency (b) Nyquist interval
• Q4 A 2 kHz signal is sampled using PCM with 8 bits/sample → Find: (a) Bit rate (b) Bandwidth required
• Q5 Explain Pulse Code Modulation (PCM) with block diagram.
• Q6 Define quantization and explain quantization error.
• Q7 A 3 kHz signal is encoded using 4-bit PCM → Find: (a) Sampling rate (b) Bit rate (c) Bandwidth
• Q8 Explain ASK, FSK, and PSK with waveforms.
• Q9 Compare BPSK and QPSK in terms of bandwidth and efficiency.
• Q10 Differentiate between coherent and non-coherent detection.
• Q11 A 3 kHz signal is encoded using 4-bit PCM → Find: (a) Sampling rate (b) Bit rate (c) Bandwidth
• Q12 1 kHz input signal sampled at 25% above Nyquist rate, 12-bit PCM system → Find: Bandwidth required & total bits transmitted
• Q13 Explain QPSK modulation and demodulation with block diagram.
• Q14 Explain QAM (Quadrature Amplitude Modulation) and its advantages.