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

  2. The net area of round bars to resist the tension, is the area of the cross-section at

  3. A steel plate is 300 mm wide and 10 mm thick. It has one rivet of nominal diameter 18 mm. The net sectional area of the plate is

  4. The best tension member section will be as:

  5. For preliminary sizing, the rupture strength of net section may be approximately taken by using the below equation as per IS ∶ 800-2007,

    Tdn = αAnfu/γml

    What is the "α" value for one or two bolts along the length in the end connection or equivalent weld length.

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