Question Bank • QB_FM_ADV_078 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints

Question Statement:

A combine harvester straw-chopper drive shaft is subjected to a completely reversed bending moment of $M = 400\text{ N}\cdot\text{m}$ and a steady torsional moment of $T = 300\text{ N}\cdot\text{m}$. The shaft is made of forged steel with an endurance limit in reversed bending of $S_e = 200\text{ MPa}$ and a yield strength in shear of $S_{sy} = 180\text{ MPa}$. Using the ASME design code for transmission shafting (with shock and fatigue factors $k_b = 1.5$ and $k_t = 1.0$), the equivalent twisting moment $T_e = \sqrt{(k_b M)^2 + (k_t T)^2}$ in $\text{N}\cdot\text{m}$ is ________ (round off to 1 decimal place).

Answer Key:

670.8

Step-by-Step Solution & Explanation

``` ASME Transmission Shaft Design: Reversed Bending: M = 400 N*m, k_b = 1.5 Steady Torsion: T = 300 N*m, k_t = 1.0 ``` **Step-by-step Calculation:** 1. Design bending moment ($M_d$): $M_d = k_b \times M = 1.5 \times 400 = 600\text{ N}\cdot\text{m}$ 2. Design torsional moment ($T_d$): $T_d = k_t \times T = 1.0 \times 300 = 300\text{ N}\cdot\text{m}$ 3. Equivalent twisting moment ($T_e$): $T_e = \sqrt{M_d^2 + T_d^2} = \sqrt{(600)^2 + (300)^2} = \sqrt{360000 + 90000} = \sqrt{450000}$ $T_e = 300 \sqrt{5} = 670.82\text{ N}\cdot\text{m} \approx 670.8\text{ N}\cdot\text{m}$
← Previous (QB_FM_ADV_075) Next (QB_FM_GATE_005) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test