Tensile Strength of Steel Plates

When designing steel structures, it's important to consider how the members will behave under tensile forces. One critical aspect is determining the design strength, which is the maximum load a member can safely carry in tension. Steel plates with holes, often used for bolted connections, are susceptible to failure in different ways when subjected to tension. Two primary failure modes are yielding of the gross section and rupture of the net section at the holes.

Rupture Strength at Holes

The question specifically asks about the design strength related to the rupture of the net critical cross-sectional area (A) at the holes. When a tensile load is applied to a plate with holes, the stress concentrates around the holes. If the load increases, the material around the holes can eventually fracture or rupture. The net area is the cross-sectional area of the plate minus the area removed by the holes in the critical section.

IS 800 : 2007 Provisions

According to IS 800 : 2007, the Indian standard code for general construction in steel, the design strength of a tension member, $T_{dn}$, based on the rupture of the net section is given by a specific formula. This formula takes into account the net area, the ultimate tensile stress of the steel material, and a partial safety factor for the material.

The formula provided in IS 800 : 2007 for calculating the design strength due to rupture of the net section is:

\[ {T_{dn}} = \frac{{0.9{A_{net}}{f_u}}}{{{\gamma_{m1}}}} \]

Where:

• ${T_{dn}}$ is the design strength of the tension member based on rupture of net section.
• ${A_{net}}$ is the net effective area of the total cross-section. In the question, this is denoted by ${A}$.
• ${f_u}$ is the ultimate tensile strength of the material. This matches ${f_u}$ in the question.
• ${\gamma_{m1}}$ is the partial safety factor for failure at ultimate stress. According to Table 5 of IS 800 : 2007, the value of ${\gamma_{m1}}$ is 1.25.

Applying the Formula

Substituting the given notation from the question ($A$ for $A_{net}$) and the value of ${\gamma_{m1}}$ (1.25) into the IS 800:2007 formula, we get the expression for the design strength:

\[ \text{Design Strength} = \frac{{0.9 \times A \times {f_u}}}{{1.25}} \]

\[ \text{Design Strength} = \frac{{0.9A{f_u}}}{{1.25}} \]

Comparing with Options

Let's compare this derived expression with the given options:

1. \(\frac{{0.87A{f_u}}}{{1.25}}\) - This uses 0.87 instead of 0.9.
2. \(\frac{{1.25A{f_u}}}{{0.9}}\) - This has the factors inverted in the denominator and numerator.
3. \(\frac{{0.9A{f_u}}}{{1.10}}\) - This uses 1.10 in the denominator, which is ${\gamma_{m0}}$ (partial safety factor for yielding), not ${\gamma_{m1}}$.
4. \(\frac{{0.9A{f_u}}}{{1.25}}\) - This exactly matches the derived expression from IS 800 : 2007 for rupture strength.

Therefore, the correct expression for the design strength in tension of a plate due to rupture of the net critical cross-sectional area at the holes, as per IS 800 : 2007, is $\frac{{0.9A{f_u}}}{{1.25}}$.