Question 1 |
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 1 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 2 |
The ratio of the plastic moment capacity of a beam section to its yield moment capacity
is termed as
aspect ratio | |
load factor | |
shape factor | |
slenderness ratio |
Question 2 Explanation:
Ratio of \frac{M_p}{M_y}= Shape factor
Question 3 |
If the section shown in the figure turns from fully-elastic to fully-plastic, the depth of neutral axis (N.A.), \bar{y}, decreases by
10.75 mm | |
12.25 mm | |
13.75 mm | |
15.25 mm |
Question 3 Explanation:
\bar{y}=\frac{A_1y_1+A_2y_2}{A_1+A_2}=\frac{300\times 2.5+300 \times 35}{300+300}=18.75m
Question 4 |
A prismatic propped cantilever beam of span L and plastic moment capacity M_{p} is subjected to a concentrated load at its mid-span. If the collapse load of the beam is \alpha \frac{M_{p}}{L}, the value of \alpha is ______
2 | |
4 | |
6 | |
8 |
Question 4 Explanation:
P_{u}=\frac{6M_{P}}{l} So, \alpha =6
Question 5 |
The dimensions of a symmetrical welded I-section are shown in the figure.
The plastic section modulus about the weaker axis (in cm^{3}, up to one decimal place) is ______
The plastic section modulus about the weaker axis (in cm^{3}, up to one decimal place) is ______
60.4 | |
89.9 | |
120.6 | |
30.8 |
Question 5 Explanation:
\begin{aligned} Z_p&= \frac{A}{2}(\bar{x_1}+\bar{x_2})\\ &=\left [ \frac{140\times 9}{2}\left ( 35+35 \right ) \right ] \\ & \times 2+\left [ \frac{\left ( 200-18 \right )\times 6.1}{2} \right ] \\ &\times \left ( \frac{3.05}{2}+\frac{3.05}{2} \right ) \\ &=88200+1693.055 \\ &=89893.055\, mm^{3} \\ & \simeq 89.9\, cm^{3} \end{aligned}
Question 6 |
A fixed-end beam is subjected to a concentrated load (P) as shown in the figure. The beam has two different segments having different plastic moment capacities (M_{p},2M_{p}) as shown.
The minimum value of load (P) at which the beam would collapse (ultimate load) is
The minimum value of load (P) at which the beam would collapse (ultimate load) is
7.5M_{p}/L | |
5.0M_{p}/L | |
4.5M_{p}/L | |
2.5M_{p}/L |
Question 6 Explanation:
D_{S}=2
\therefore Number of plastic hinge required for complete collapse =D_{S}+1=2+1=3
Mechanism 1:
\begin{aligned} \Delta &=\frac{2L}{3\theta }=\frac{4L}{3} \\ \Rightarrow \;\; \theta &=2\theta \end{aligned}
For Principal of virtual work done,
\begin{aligned} -2M_{P}\theta -2M_{P}\theta -2M_{P}\theta -M_{P}\theta +P\left ( \frac{2L}{3}\theta \right )&=0 \\ 4M_{P}\theta +3M_{P}\phi &=\frac{2PL}{3}\theta \\ 8M_{P}\phi +3M_{P}\phi &=\frac{4PL}{3}\phi \\ 11M_{P}&=\frac{4P_{u}L}{3}\phi \\ P_{u}&=\frac{33}{4}\frac{M_{P}}{L} \\ \therefore \;\; P_{u}&=8.25\frac{M_{P}}{L} \end{aligned}
Mechanism 2:
\begin{aligned} 2M_{P}\theta +M_{P}\theta +M_{P}\theta +M_{P}\theta &=P\left ( \frac{2L}{3} \right ) \\ 5M_{P}&=\frac{2PL}{3} \\ P_{u}&=\frac{15M_{P}}{2L} \\ \therefore \;\; P_{u}&=7.5\frac{M_{P}}{L} \end{aligned}
\therefore Number of plastic hinge required for complete collapse =D_{S}+1=2+1=3
Mechanism 1:
\begin{aligned} \Delta &=\frac{2L}{3\theta }=\frac{4L}{3} \\ \Rightarrow \;\; \theta &=2\theta \end{aligned}
For Principal of virtual work done,
\begin{aligned} -2M_{P}\theta -2M_{P}\theta -2M_{P}\theta -M_{P}\theta +P\left ( \frac{2L}{3}\theta \right )&=0 \\ 4M_{P}\theta +3M_{P}\phi &=\frac{2PL}{3}\theta \\ 8M_{P}\phi +3M_{P}\phi &=\frac{4PL}{3}\phi \\ 11M_{P}&=\frac{4P_{u}L}{3}\phi \\ P_{u}&=\frac{33}{4}\frac{M_{P}}{L} \\ \therefore \;\; P_{u}&=8.25\frac{M_{P}}{L} \end{aligned}
Mechanism 2:
\begin{aligned} 2M_{P}\theta +M_{P}\theta +M_{P}\theta +M_{P}\theta &=P\left ( \frac{2L}{3} \right ) \\ 5M_{P}&=\frac{2PL}{3} \\ P_{u}&=\frac{15M_{P}}{2L} \\ \therefore \;\; P_{u}&=7.5\frac{M_{P}}{L} \end{aligned}
Question 7 |
A propped cantilever of span L carries a vertical concentrated load at the mid-span. If the plastic moment capacity of the section is M_{p}, the magnitude of the collapse load is
\frac{8M_{p}}{L} | |
\frac{6M_{p}}{L} | |
\frac{4M_{p}}{L} | |
\frac{2M_{p}}{L} |
Question 7 Explanation:
D_{s}\, =\, 1
\therefore No. of plastic hinge required for complete collapse =1+1 =2
\begin{aligned} -M_{P}\theta -M_{P}\theta -M_{P}\theta +\frac{PL}{2}\theta &=0 \\ 3M_{P}&=\frac{PL}{2} \\ P&=\frac{6M_{P}}{L} \end{aligned}
Question 8 |
A fixed end beam is subjected to a load, W at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is:
16.5 M_{p}/L | |
15.5 M_{p}/L | |
15.0 M_{p}/L | |
16.0 M_{p}/L |
Question 9 |
For formation of collapse mechanism in the following figure, the minimum value of P_{u} is cM_{p}/L. M_{p} and 3M_{p} denote the plastic moment capacities of beam sections as shown in this figure. The value of c is _____.
3.3 | |
2.8 | |
6.2 | |
8.6 |
Question 9 Explanation:
\begin{aligned} L\phi&=3L\theta \\ \phi&=3\theta \\4M_{P}\theta +2M_{P}\phi &=PL\phi \\ \frac{4M_{P}\phi }{3}+2M_{P}\phi &=PL\phi \\ \frac{10M_{P}\phi }{3}&=PL\phi \\ P=\frac{10M_{P}}{3L} \; \Rightarrow \; C&=\frac{10}{3}=3.33 \end{aligned}
Question 10 |
A prismatic beam (as shown below) has plastic moment capacity of M_{p}, then the collapse load P of the beam is
\frac{2M_{p}}{L} | |
\frac{4M_{p}}{L} | |
\frac{6M_{p}}{L} | |
\frac{8M_{p}}{L} |
Question 10 Explanation:
Here degree of static indeterminacy =0
\therefore Number of plastic hinges required for mechanical equlibrium
=D_{s}+1=0+1=1
From principal of virtual work
\begin{aligned} -M_{P}\theta -M_{P}\theta +P\frac{L}{2}\theta -\frac{P}{2}\times \frac{L}{3}\theta &=0 \\ -2M_{P}\theta +\frac{PL}{2}\theta -\frac{PL}{6}\theta &=0 \\ 2M_{P}=\frac{PL}{2}-{PL}{6}=\frac{\left ( 3-1 \right )PL}{6}&=\frac{1}{3}PL \\ P&=\frac{6M_{p}}{L} \end{aligned}
\therefore Number of plastic hinges required for mechanical equlibrium
=D_{s}+1=0+1=1
From principal of virtual work
\begin{aligned} -M_{P}\theta -M_{P}\theta +P\frac{L}{2}\theta -\frac{P}{2}\times \frac{L}{3}\theta &=0 \\ -2M_{P}\theta +\frac{PL}{2}\theta -\frac{PL}{6}\theta &=0 \\ 2M_{P}=\frac{PL}{2}-{PL}{6}=\frac{\left ( 3-1 \right )PL}{6}&=\frac{1}{3}PL \\ P&=\frac{6M_{p}}{L} \end{aligned}
There are 10 questions to complete.