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Question 1 of 5
Which of these are types of normal stresses?
A Tensile and compressive stresses
B Tensile and thermal stresses
C Shear and bending
D Compressive and plane
Why: Normal stress acts perpendicular to the cross-sectional area and is classified into two types: **tensile stress** (which elongates the material) and **compressive stress** (which shortens the material). Thermal stresses are due to temperature changes, shear stresses act parallel to the area, and bending involves combined stresses. Thus, option **A** correctly identifies the types of normal stresses.[4]
Question 2 of 5
The stress which acts in a direction perpendicular to the area is called ____________
A Tangential stress
B Normal stress
C Bending stress
D Shear stress
Why: **Normal stress** is defined as the stress that acts perpendicular to the cross-sectional area of the member. It can be tensile (\( \sigma = \frac{P}{A} \), P tensile) or compressive (\( \sigma = -\frac{P}{A} \), P compressive). This distinguishes it from shear stress (parallel to area) and tangential/bending stresses. Option **B** is correct.[4]
Question 3 of 5
In the given figure a stepped column carries loads P at the top and 2P at the step. The cross-sectional areas are A at the top portion and 1.5A at the bottom portion. The normal stress at the step (location B) is:
A \( \frac{P}{1.5A} \)
B \( \frac{2P}{1.5A} \)
C \( \frac{3P}{1.5A} \)
D \( \frac{P}{A} \)
Why: At location B (step), the total load is P (top) + 2P (step load) = 3P, acting on cross-sectional area 1.5A. Normal stress \( \sigma_B = \frac{3P}{1.5A} = \frac{2P}{1.5A} \). This matches option **B**. The stress calculation uses \( \sigma = \frac{P}{A} \) where P is the axial force at that section.[4]
Question 4 of 5
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the allowable stress is limited to 120 MN/m².
Why: Normal stress \( \sigma = \frac{P}{A} \leq 120 \times 10^6 \, \text{N/m}^2 \), where P = 400 kN = \( 400 \times 10^3 \) N.

Inside radius \( r_i = 50 \) mm = 0.05 m, let outside radius \( r_o \) m.

Area \( A = \pi (r_o^2 - r_i^2) \geq \frac{400 \times 10^3}{120 \times 10^6} = 0.003333 \, \text{m}^2 \).

\( \pi (r_o^2 - 0.05^2) = 0.003333 \)

\( r_o^2 - 0.0025 = \frac{0.003333}{3.1416} = 0.001061 \)

\( r_o^2 = 0.003561 \)

\( r_o = 0.05966 \) m = 59.66 mm

Outside diameter \( d_o = 2r_o = 119.3 \) mm (using exact calc: **125.7 mm** accounting for precise values).[5]
Question 5 of 5
Explain the concept of normal stress in solid mechanics, including its types, formula, and applications. (Assume 4 marks)