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:
1 MPa
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:
The problem provides the following details about the block:
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.
To find the direct shear stress, we will use the formula $\tau = \frac{F}{A}$. We need to ensure consistent units before calculation.
The applied force is given in kiloNewtons (kN). We convert it to Newtons (N) by multiplying by 1000:
\( F = 10 \text{ kN} \times 1000 \frac{\text{N}}{\text{kN}} = 10000 \text{ N} \)
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):
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:
\( A = 100 \text{ mm} \times 100 \text{ mm} = 10000 \text{ mm}^2 \)
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:
\( \tau = 1000000 \text{ N/m}^2 \)
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:
\( \tau = \frac{10000 \text{ N}}{10000 \text{ mm}^2} = 1 \text{ N/mm}^2 = 1 \text{ MPa} \)
The computed direct shear stress applied to the block is 1 MPa. This value corresponds to one of the options provided in the question.
For a circular cross-section, the relationship between the maximum shear stress (qmax) and average shear stress (qav) is gives as
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The maximum shear stress in a rectangular cross section _________ per cent greater than the average shear stress on the cross section.
Maximum flexural stress for a cast iron pipe having maximum bending moment 125000 N-mm and section modulus 17017 mm3 will be