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Question

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

The correct answer is

227 kN

Steel Plate Yielding Design Strength Explained

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.

Understanding Design Strength in Yielding

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.

Formula for Design Strength

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:

  • $A_g$ is the gross cross-sectional area of the plate.
  • $f_y$ is the characteristic yield strength of the steel material.
  • $\gamma_{m0}$ is the partial safety factor for failure in yielding.

Step-by-Step Calculation

  1. Calculate Gross Cross-sectional Area ($A_g$):

    The plate dimensions are given as 100 mm x 10 mm.

    $$ A_g = \text{Width} \times \text{Thickness} $$

    $$ A_g = 100 \text{ mm} \times 10 \text{ mm} = 1000 \text{ mm}^2 $$

  2. Identify Yield Strength ($f_y$):

    The yield strength is provided as 250 MPa.

    $$ f_y = 250 \text{ MPa} = 250 \text{ N/mm}^2 $$

  3. Determine Partial Safety Factor ($\gamma_{m0}$):

    For yielding in the gross section, the partial safety factor commonly used in structural steel design is $\gamma_{m0} = 1.10$.

  4. Calculate Design Strength ($T_{dg}$):

    Substitute the values into the formula:

    $$ T_{dg} = \frac{A_g f_y}{\gamma_{m0}} $$

    $$ T_{dg} = \frac{1000 \text{ mm}^2 \times 250 \text{ N/mm}^2}{1.10} $$

    $$ T_{dg} = \frac{250000 \text{ N}}{1.10} $$

    $$ T_{dg} \approx 227272.72 \text{ N} $$

  5. Convert to Kilonewtons (kN):

    To convert Newtons (N) to Kilonewtons (kN), divide by 1000.

    $$ T_{dg} \approx \frac{227272.72}{1000} \text{ kN} $$

    $$ T_{dg} \approx 227.27 \text{ kN} $$

Conclusion

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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Important Questions from Tension Member

  1. The following are the statements about lug angle used to connect heavily loaded tension member to gusset plates.

    (i) The length of end connection is reduced

    (ii) By using lug angles there will be saving in the gusset plate

    (iii) Cost of connection increases due to additional fasteners and angle required.

  2. A structural member subjected to tensile force in a direction parallel to its longitudinal axis is generally known as

  3. When the length of a tension member is too long:

  4. The allowable stress in axial tension is generally kept less if the thickness of the member is more than

  5. A single angle in tension is connected by one leg only. If the areas of connecting and outstanding legs are respectively a and b, then what is the net effective area of the angle?

    A) \(a-\frac{b}{1+0.35\times\frac{b}{a}}\)

    B) \(a+\frac{b}{1+0.35\times\frac{b}{a}}\)

    C) \(a-\frac{b}{1+0.20\times\frac{b}{a}}\)

    D) \(a+\frac{b}{1+0.20\times\frac{b}{a}}\)

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