Question

A solid uniform shaft of circular cross section is subjected to a maximum bending moment of 3 kNm and a twisting moment of 4 kNm. The equivalent torsional moment is

• A. 1 kNm
• B. 4 kNm
• C. 5 kNm
• D. 7 kNm

Hint:

There is also another theory called the maximum principal stress theory (Guest's Theory) which gives an equivalent bending moment. However, for shaft design considering torsional strength and shear stress, the equivalent torsional moment based on the maximum shear stress theory is generally more relevant.

Answer 1

The Correct Option is C

Step 1: Understand the concept of equivalent torsional moment.
When a shaft is subjected to both bending moment and twisting moment, it experiences complex stresses. To design the shaft based on a single equivalent torque that would produce the same maximum shear stress as the combined loading, the concept of equivalent torsional moment (\( T_e \)) is used. Step 2: Recall the formula for equivalent torsional moment based on maximum shear stress theory (Rankine's Theory).
The maximum shear stress theory states that failure occurs when the maximum shear stress in a complex stress state reaches the maximum shear stress at the yield point in a simple tension test. The equivalent torsional moment based on this theory is given by: $$T_e = \sqrt{M^2 + T^2}$$ where:
\( T_e \) is the equivalent torsional moment
\( M \) is the bending moment
\( T \) is the twisting moment
Step 3: Substitute the given values into the formula.
We are given:
Maximum bending moment \( M = 3 \) kNm
Twisting moment \( T = 4 \) kNm
Substitute these values into the equivalent torsional moment formula: $$T_e = \sqrt{(3 \text{ kNm})^2 + (4 \text{ kNm})^2}$$$$T_e = \sqrt{9 \text{ kNm}^2 + 16 \text{ kNm}^2}$$$$T_e = \sqrt{25 \text{ kNm}^2}$$ $$T_e = 5 \text{ kNm}$$ Therefore, the equivalent torsional moment is 5 kNm.