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 ?
1.5 | |
1.7 | |
1.3 | |
2 |
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
The values of S and M_P (rounded-off to one decimal place) are
S = 2.0, M_P= 58.9 kN-m | |
S = 2.0, M_P=100.2 kN-m | |
S = 1.5, M_P= 58.9 kN-m | |
S = 1.5, M_P=100.2 kN-m |
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}
\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
541 mm and 214 mm | |
214 mm and 541 mm | |
380 mm and 151 mm | |
151 mm and 380 mm |
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}
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
The plastic moment, M_{p} calculated for the collapse mechanism using static method and kinematic method is
M_{P, \text { static }} \gt \frac{2 P L}{9}=M_{P, \text { kinematic }} | |
M_{P, \text { static }}=\frac{2 P L}{9} \neq M_{P, \text { kinematic }} | |
M_{P, \text { static }}=\frac{2 P L}{9}=M_{P, \text { kinematic }} | |
M_{P, \text { static }} \lt \frac{2 P L}{9}=M_{P, \text { kinematic }} |
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 ________
The resultant force in bolt R (in kN,round off to one decimal place) is ________
28.2 | |
46.8 | |
22.6 | |
38.2 |
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.