# Design of Steel Structures

 Question 1
Consider the horizontal axis passing through the centroid of the steel beam cross-section shown in the figure. What is the shape factor (rounded off to one decimal place) for the cross-section ? A 1.5 B 1.7 C 1.3 D 2
GATE CE 2023 SET-1      Beams
Question 1 Explanation: \begin{aligned} Z_{P} & =\frac{A}{2}\left(\bar{y}_{1}+\bar{y}_{2}\right) \\ & =\left(3 b \times \frac{b}{2}+b \times b\right)\left(2 \times \frac{11}{20} b\right) \\ & =\frac{11}{4} b^{3} \end{aligned}
$z_{e}=\frac{1}{y_{max }}=\frac{\frac{b \times(3 b)^{3}}{12}+\frac{2 b(b)^{3}}{12}}{\frac{3 b}{2}}$
$z_{e}=\frac{29}{18} b^{3}$
Shape factor $(S)=\frac{Z_{P}}{Z_{e}}=\frac{\frac{11}{4} b^{3}}{\frac{29}{18} b^{3}}=1.7$
 Question 2
For the square steel beam cross-section shown in the figure, the shape factor about $z- z$ axis is $S$ and the plastic moment capacity is $M_P$. Consider yield stress $f_y = 250 MPa$ and $a = 100 mm.$ The values of $S$ and $M_P$ (rounded-off to one decimal place) are
 A $S = 2.0, M_P= 58.9 kN-m$ B $S = 2.0, M_P=100.2 kN-m$ C $S = 1.5, M_P= 58.9 kN-m$ D $S = 1.5, M_P=100.2 kN-m$
GATE CE 2022 SET-2      Plastic Analysis
Question 2 Explanation:
Shape factor for diamond shaped section = 2
\begin{aligned} S&=\frac{M_P}{M_y}=\frac{f_yZ_P}{f_yZ_e}\\ M_P&=S.M_y=S.f_y.Z_e \end{aligned} \begin{aligned} I_{ZZ}&=\frac{a^4}{12}\\ Z_{ZZ}&=\frac{a^4 \times \sqrt{2}}{12 \times a }=\frac{\sqrt{2}a^3}{12}mm^3\\ M_P&=\left [ 2 \times 250 \times \frac{\sqrt{2} \times (100)^3}{12} \right ] \times 10^{-6} kNm\\ &=58.93kNm \end{aligned}

 Question 3
A weld is used for joining an angle section ISA 100 mm x 100 mm x 10 mm to a gusset plate of thickness 15 mm to transmit a tensile load. The permissible stress in the angle is 150 MPa and the permissible shear stress on the section through the throat of the fillet weld is 108 MPa. The location of the centroid of the angle is represented by $C_{yy}$ in the figure, where $C_{yy}=28.4mm$. The area of cross-section of the angle is 1903 $mm^2$. Assuming the effective throat thickness of the weld to be 0.7 times the given weld size, the lengths $L_1$ and $L_2$ (rounded- off to the nearest integer) of the weld required to transmit a load equal to the full strength of the tension member are, respectively A 541 mm and 214 mm B 214 mm and 541 mm C 380 mm and 151 mm D 151 mm and 380 mm
GATE CE 2022 SET-1      Structural Fasteners
Question 3 Explanation:
As per IS:800-2007 Clause no. 11.2.1
Full strength of tension member,
\begin{aligned} P&=\sigma _{at} \times A_g\\ P&=150 \times 1903\\ P&=285450N \end{aligned}
Weld strength of $L_1$,
\begin{aligned} P_1&=L_1\times (0.7 \times s) \times 108\\ P_1&=L_1\times (0.7 \times 5) \times 108\\ P_1&=378L_1 \end{aligned}
Weld strength of $L_2$,
\begin{aligned} P_2&=L_2\times (0.7 \times s) \times 108\\ P_2&=L_2\times (0.7 \times 5) \times 108\\ P_2&=378L_2 \end{aligned}
Applying moment CG equal to zero,
\begin{aligned} P_1 \times 28.4 &=P_2 \times (100-28.4)\\ \frac{P_1}{P_2}&=\frac{71.6}{28.4}\\ \frac{L_1 \times 378}{L_2 \times 378}&=\frac{71.6}{28.4}=2.521\\ L_1&=2.521L_2\\ \text{Total weld strength}&=\text{Tensile strength}\\ (L_1+L_2) \times 3.5 \times 108&=285450\\ L_1+L_2&=755.158\\ 2.521L_2+L_2&=755.152\\ L_2&=214.472mm\\ L_1&=540.685mm\\ \end{aligned}
 Question 4
A prismatic steel beam is shown in the figure. The plastic moment, $M_{p}$ calculated for the collapse mechanism using static method and kinematic method is
 A $M_{P, \text { static }} \gt \frac{2 P L}{9}=M_{P, \text { kinematic }}$ B $M_{P, \text { static }}=\frac{2 P L}{9} \neq M_{P, \text { kinematic }}$ C $M_{P, \text { static }}=\frac{2 P L}{9}=M_{P, \text { kinematic }}$ D $M_{P, \text { static }} \lt \frac{2 P L}{9}=M_{P, \text { kinematic }}$
GATE CE 2021 SET-2      Plastic Analysis
Question 4 Explanation: \begin{aligned} \text{At collapse,} \quad M_{p} \theta+M_{p} \phi&=P \Delta\\ \Rightarrow \quad \quad 3 M_{P} \frac{\Delta}{l}+\frac{3 M_{P} \Delta}{2 l} &=P \Delta \\ M_{P} &=\frac{2 P l}{9}\\ \text{Also,} \qquad\qquad \quad M_{P ,\text { static }}&=M_{P,\text{ kinematic}} \end{aligned}
 Question 5
A column is subjected to a total load (P) of 60 kN supported through a bracket connection, as shown in the figure (not to scale). The resultant force in bolt R (in kN,round off to one decimal place) is ________
 A 28.2 B 46.8 C 22.6 D 38.2
GATE CE 2021 SET-1      Structural Fasteners
Question 5 Explanation:
\begin{aligned} F_{1}&=\frac{P}{n}=\frac{60}{6}=10 \mathrm{kN}\\ F_{2}&=\frac{P \cdot e}{\sum r_{i}^{2}} \times r_{R}=\frac{60 \times 100\mathrm{~mm}}{4 \times 50^{2}+2 \times 40^{2}} \times 40\\ F_{2}&=\frac{60 \times 100 \times 40}{10000+3200}=\frac{60 \times 100 \times 40}{13200}=\frac{200}{11} \mathrm{kN}\\ F_{R}&=F_{1}+F_{2}\\ &=10+\frac{200}{11}=\frac{310}{11}=28.2 \mathrm{kN} \end{aligned}

There are 5 questions to complete.