Steel Compression Member Design - IS 800:2007

🧱 Steel Compression + Bending Design (IS 800:2007)















Beam-Column Design Calculation (IS 800:2007)

Problem Statement:
Check the safety of an ISHB200 column carrying a factored axial load $P_u = 150\text{ kN}$ and a factored bending moment $M_u = 15\text{ kNm}$. The effective length $L_{\text{eff}}$ of the column is $3000\text{ mm}$. Steel grade is Fe 250 ($f_y = 250\text{ MPa}$) with Buckling Class c.

Step 1: Input Data & Cross-Section Properties

  • Section: ISHB200
  • Area ($A$): $40.9\text{ cm}^2 = 4090\text{ mm}^2$
  • Elastic Section Modulus ($Z_x$): $287.3\text{ cm}^3 = 287300\text{ mm}^3$
  • Governing Radius of Gyration ($r_y$): $2.14\text{ cm} = 21.4\text{ mm}$
  • Yield Stress ($f_y$): $250\text{ MPa}$
  • Partial Safety Factor ($\gamma_{m0}$): $1.10$
  • Elastic Modulus ($E$): $2 \times 10^5\text{ MPa}$

Step 2: Calculate Slenderness Ratio ($\lambda_{sl}$)

$$\lambda_{sl} = \frac{L_{\text{eff}}}{r_y} = \frac{3000}{21.4} = 140.187$$

Step 3: Compressive Strength Calculation (Clause 7.1.2)

  1. Euler Buckling Stress ($f_{cr}$): $$f_{cr} = \frac{\pi^2 \cdot E}{\lambda_{sl}^2} = \frac{\pi^2 \times 200000}{(140.187)^2} = 100.44\text{ MPa}$$
  2. Non-dimensional Slenderness Ratio ($\lambda$): $$\lambda = \sqrt{\frac{f_y}{f_{cr}}} = \sqrt{\frac{250}{100.44}} = 1.5777$$
  3. Imperfection Factor ($\alpha$): For Buckling Class c, $\alpha = 0.49$.
  4. Intermediate Value ($\phi$): $$\phi = 0.5 \times \left[1 + \alpha(\lambda - 0.2) + \lambda^2\right]$$ $$\phi = 0.5 \times \left[1 + 0.49(1.5777 - 0.2) + (1.5777)^2\right] = 2.0821$$
  5. Stress Reduction Factor ($\chi$): $$\chi = \frac{1}{\phi + \sqrt{\phi^2 - \lambda^2}} = \frac{1}{2.0821 + \sqrt{(2.0821)^2 - (1.5777)^2}} = 0.2906$$
  6. Design Compressive Stress ($f_{cd}$): $$f_{cd} = \frac{\chi \cdot f_y}{\gamma_{m0}} = \frac{0.2906 \times 250}{1.10} = 66.05\text{ MPa}$$
  7. Design Axial Capacity ($P_d$): $$P_d = \frac{f_{cd} \cdot A}{1000} = \frac{66.05 \times 4090}{1000} = 270.14\text{ kN}$$

Step 4: Bending Moment Capacity (Clause 8.2)

$$M_d = \frac{Z_x \cdot f_y}{\gamma_{m0} \times 10^6} = \frac{287300 \times 250}{1.10 \times 10^6} = 65.30\text{ kNm}$$

Step 5: Interaction Ratio & Safety Check (Clause 9.3)

$$\text{Interaction Ratio} = \frac{P_u}{P_d} + \frac{M_u}{M_d}$$ $$\text{Interaction Ratio} = \frac{150}{270.14} + \frac{15}{65.30} = 0.555 + 0.230 = 0.785$$

Since $\text{Interaction Ratio} = 0.785 \le 1.0$, the design section is safe.

Summary Table

Property / Parameter Calculated Value
Selected Section ISHB200
Area ($A$) $4090.0\text{ mm}^2$
Section Modulus ($Z_x$) $287300.0\text{ mm}^3$
Radius of Gyration ($r$) $21.4\text{ mm}$
Slenderness Ratio ($\lambda_{sl}$) $140.2$
Buckling Class C
Reduction Factor ($\chi$) $0.291$
Design Compressive Stress ($f_{cd}$) $66.0\text{ MPa}$
Design Axial Capacity ($P_d$) $270.14\text{ kN}$
Design Moment Capacity ($M_d$) $65.30\text{ kNm}$
Interaction Ratio $0.785$
Status ✔ SAFE