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Question

A block is of dimensions of the upper surface 100 mm x 100 mm. The height of the block is 10 mm. A tangential force of 10 kN is applied at the centre of the upper surface. The block is displaced by 1 mm with respect to the lower face. Direct shear stress in the element is:

The correct answer is

1 MPa

Understanding Direct Shear Stress Concept

Direct shear stress is a measure of the internal forces that cause an object to deform by sliding layers past each other. It arises when a force acts parallel to the cross-sectional area of the material. The calculation involves dividing the applied shear force by the area over which it acts.

The fundamental formula used to calculate direct shear stress ($\tau$) is:

\( \tau = \frac{F}{A} \)

In this formula:

  • F represents the tangential force applied to the object.
  • A is the cross-sectional area resisting this force.

Analyzing Given Parameters for Shear Stress Calculation

The problem provides the following details about the block:

  • Dimensions of the upper surface: 100 mm x 100 mm
  • Height of the block: 10 mm
  • Tangential force applied (F): 10 kN
  • Displacement of the block: 1 mm

Note: While the height and displacement are crucial for calculating shear strain or modulus, only the force and the area of application are needed for direct shear stress.

Calculating Direct Shear Stress Value

To find the direct shear stress, we will use the formula $\tau = \frac{F}{A}$. We need to ensure consistent units before calculation.

Step 1: Convert Force to Standard Units (Newtons)

The applied force is given in kiloNewtons (kN). We convert it to Newtons (N) by multiplying by 1000:

\( F = 10 \text{ kN} \times 1000 \frac{\text{N}}{\text{kN}} = 10000 \text{ N} \)

Step 2: Calculate the Area in Consistent Units (Square Meters)

The force is applied on the upper surface with dimensions 100 mm x 100 mm. This surface area (A) needs to be calculated in square meters ($m^2$).

First, convert the dimensions from millimeters (mm) to meters (m):

  • 100 mm = 0.1 m

Now, calculate the area:

\( A = \text{Length} \times \text{Width} = (0.1 \text{ m}) \times (0.1 \text{ m}) = 0.01 \text{ m}^2 \)

Alternatively, we can work with millimeters and convert the final stress unit:

\( A = 100 \text{ mm} \times 100 \text{ mm} = 10000 \text{ mm}^2 \)

Step 3: Compute Shear Stress

Substitute the force and area values into the shear stress formula $\tau = \frac{F}{A}$.

Using the area in square meters:

\( \tau = \frac{10000 \text{ N}}{0.01 \text{ m}^2} \)

Performing the division yields:

\( \tau = 1000000 \text{ N/m}^2 \)

Step 4: Convert Stress to Megapascals (MPa)

Stress is commonly expressed in Megapascals (MPa). Recall that 1 Pascal (Pa) equals 1 N/m$^2$, and 1 MPa equals $1 \times 10^6$ N/m$^2$.

Convert the calculated stress:

\( \tau = \frac{1000000 \text{ N/m}^2}{1 \times 10^6 \text{ N/m}^2/\text{MPa}} = 1 \text{ MPa} \)

Using the area in square millimeters provides a shortcut, as 1 N/mm$^2$ is equivalent to 1 MPa:

\( \tau = \frac{10000 \text{ N}}{10000 \text{ mm}^2} = 1 \text{ N/mm}^2 = 1 \text{ MPa} \)

Final Result for Direct Shear Stress

The computed direct shear stress applied to the block is 1 MPa. This value corresponds to one of the options provided in the question.

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Important Questions from Shear Stress and Bending Stress

  1. For a beam to be classified as a beam of uniform strength, which of the following conditions must be met?
  2. For a circular cross-section, the relationship between the maximum shear stress (qmax) and average shear stress (qav) is gives as

  3. The ratio of moment carrying capacity of a square cross-section beam of dimension D to the moment carrying capacity of a circular cross-section of diameter D is:

  4. The maximum shear stress in a rectangular cross section _________ per cent greater than the average shear stress on the cross section.

  5. Maximum flexural stress for a cast iron pipe having maximum bending moment 125000 N-mm and section modulus 17017 mm3 will be

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