Understanding Steel Plate Rupture Strength Calculation

This problem asks us to find the design strength of a steel plate when it fails by rupture across a critical section containing bolt holes. We are given the plate dimensions, the size and number of bolt holes, and the ultimate strength of the steel.

Given Data:

• Plate dimensions: 250 mm × 150 mm × 10 mm
• Bolt diameter: 16 mm
• Number of bolts (and holes): 2
• Ultimate strength of steel ($f_u$): 410 MPa

Determining the Critical Section and Net Area

The design strength in rupture depends on the net area of the plate at the critical section. The critical section is the cross-section with the minimum net area. This typically occurs where the maximum number of holes are present, perpendicular to the direction of load.

We need to assume the direction of load application. Usually, the smaller dimension is considered the width across which holes are punched for connection. Let's assume the load is applied along the 150 mm dimension. Therefore, the width of the plate considered for the critical section is $w = 150$ mm.

• Thickness of the plate, $t = 10$ mm.
• Diameter of the bolt, $d_b = 16$ mm.
• For bolts, the diameter of the hole is typically taken as the bolt diameter plus a clearance of 2 mm. So, the diameter of the hole, $d_h = d_b + 2 \text{ mm} = 16 \text{ mm} + 2 \text{ mm} = 18$ mm.
• Number of holes along the critical section, $n = 2$.
• The net width ($w_{net}$) at the critical section is calculated by subtracting the total width of the holes from the gross width of the plate: $w_{net} = w - n \times d_h$ $w_{net} = 150 \text{ mm} - 2 \times 18 \text{ mm}$ $w_{net} = 150 \text{ mm} - 36 \text{ mm}$ $w_{net} = 114 \text{ mm}$
• The net area ($A_{net}$) of the plate at the critical section is the net width multiplied by the plate thickness: $A_{net} = w_{net} \times t$ $A_{net} = 114 \text{ mm} \times 10 \text{ mm}$ $A_{net} = 1140 \text{ mm}^2$

Calculating the Design Strength in Rupture

The design strength of a plate in rupture of the critical section ($T_{dn}$) is generally calculated using the formula provided in structural design codes like IS 800:2007. The formula is:

$$ T_{dn} = \frac{0.9 A_n f_{un}}{\gamma_{m1}} $$

Where:

• $A_n$ is the net area of the cross-section = 1140 mm$^2$.
• $f_{un}$ is the ultimate tensile stress at UTS of the material = 410 MPa = 410 N/mm$^2$.
• $\gamma_{m1}$ is the partial safety factor for failure at ultimate stress. For steel, this value is typically 1.25.

Now, let's substitute the values into the formula:

$$ T_{dn} = \frac{0.9 \times 1140 \text{ mm}^2 \times 410 \text{ N/mm}^2}{1.25} $$

First, calculate the numerator:

$$ 0.9 \times 1140 \times 410 = 420540 \text{ N} $$

Now, divide by the partial safety factor:

$$ T_{dn} = \frac{420540 \text{ N}}{1.25} $$

$$ T_{dn} = 336432 \text{ N} $$

To express the strength in kilonewtons (kN), we divide by 1000:

$$ T_{dn} = \frac{336432}{1000} \text{ kN} $$

$$ T_{dn} = 336.432 \text{ kN} $$

Conclusion

The calculated design strength of the steel plate in rupture of the critical section is approximately 336.432 kN. Comparing this value with the given options, the closest and correct answer is 336 kN.