Question

Consider the horizontal axis passing through the centroid of the steel beam cross-section shown (a symmetric "plus" of arm width $b$). What is the shape factor (rounded off to one decimal place) for the cross-section? \includegraphics[width=0.5\linewidth]{image77.png}

• A. 1.5
• B. 1.7
• C. 1.3
• D. 2.0

Hint:

For built-up symmetric sections, compute $I_x$ by add–subtract of rectangles, and get $Z_p$ as $A\bar{y}$ with the PNA at the centroid; find $\bar{y}$ from the centroid of the half area.

Answer 1

The Correct Option is B

Step 1: Area and symmetry.
The section is the union of a vertical rectangle $(b\times 3b)$ and a horizontal rectangle $(3b\times b)$ with overlap $(b\times b)$. \[ A = (3b^2+3b^2-b^2)=5b^2. \] Depth $=3b\Rightarrow c=\dfrac{3b}{2}=1.5b$ about the centroidal horizontal axis.

Step 2: Elastic section modulus $Z=\dfrac{I_x{c}$.}
\[ I_x = I_x(\text{vert.})+I_x(\text{horiz.})-I_x(\text{overlap}) = \frac{b(3b)^3}{12}+\frac{(3b)b^3}{12}-\frac{b\cdot b^3}{12} = \frac{29}{12}b^4. \] \[ Z = \frac{I_x}{c}=\frac{\frac{29}{12}b^4}{1.5b}=\frac{29}{18}b^3\approx 1.611\,b^3. \]

Step 3: Plastic section modulus $Z_p$.
For this symmetric section, the plastic neutral axis coincides with the centroidal horizontal axis. Top half area $=A/2=2.5b^2$. Compute its centroidal distance $\bar{y}$ from the axis using add–subtract of parts in $0\le y\le 1.5b$: \[ A_1=1.5b^2,\ y_1=0.75b; A_2=1.5b^2,\ y_2=0.25b; A_3=0.5b^2,\ y_3=0.25b. \] \[ Q_{\text{top}}=A_1y_1+A_2y_2-A_3y_3=(1.5\cdot0.75+1.5\cdot0.25-0.5\cdot0.25)b^3 =1.375\,b^3. \] \[ \bar{y}=\frac{Q_{\text{top}}}{A/2}=\frac{1.375}{2.5}b=0.55b. \] \[ Z_p=A\bar{y}=5b^2\cdot0.55b=2.75\,b^3. \]

Step 4: Shape factor.
\[ \text{Shape factor }= \frac{Z_p}{Z}= \frac{2.75}{1.611}\approx 1.707 \approx 1.7. \] \[ \boxed{1.7} \]