A plate of size 100 mm x 10 mm having yield strength of 250 MPa, the design strength of plate in yielding of the cross-section is
227 kN
This solution explains how to calculate the design strength of a steel plate based on its cross-section yielding, a critical concept in structural engineering.
The design strength of a steel member against yielding is determined by its gross cross-sectional area and the material's yield strength. It represents the maximum stress the plate can withstand before permanent deformation occurs, considering safety factors.
The design strength of a plate in yielding of the gross section ($T_{dg}$) is calculated using the following formula, typically based on structural design codes like IS 800:
$$ T_{dg} = \frac{A_g f_y}{\gamma_{m0}} $$
Where:
The plate dimensions are given as 100 mm x 10 mm.
$$ A_g = \text{Width} \times \text{Thickness} $$
$$ A_g = 100 \text{ mm} \times 10 \text{ mm} = 1000 \text{ mm}^2 $$
The yield strength is provided as 250 MPa.
$$ f_y = 250 \text{ MPa} = 250 \text{ N/mm}^2 $$
For yielding in the gross section, the partial safety factor commonly used in structural steel design is $\gamma_{m0} = 1.10$.
Substitute the values into the formula:
$$ T_{dg} = \frac{A_g f_y}{\gamma_{m0}} $$
$$ T_{dg} = \frac{1000 \text{ mm}^2 \times 250 \text{ N/mm}^2}{1.10} $$
$$ T_{dg} = \frac{250000 \text{ N}}{1.10} $$
$$ T_{dg} \approx 227272.72 \text{ N} $$
To convert Newtons (N) to Kilonewtons (kN), divide by 1000.
$$ T_{dg} \approx \frac{227272.72}{1000} \text{ kN} $$
$$ T_{dg} \approx 227.27 \text{ kN} $$
The calculated design strength of the steel plate in yielding of the cross-section is approximately 227.27 kN. This value closely matches one of the provided options.
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