Plastic Analysis

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
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   Design of Steel Structures
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
A
aspect ratio
B
load factor
C
shape factor
D
slenderness ratio
GATE CE 2020 SET-2   Design of Steel Structures
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
A
10.75 mm
B
12.25 mm
C
13.75 mm
D
15.25 mm
GATE CE 2019 SET-1   Design of Steel Structures
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 ______
A
2
B
4
C
6
D
8
GATE CE 2018 SET-2   Design of Steel Structures
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 ______
A
60.4
B
89.9
C
120.6
D
30.8
GATE CE 2018 SET-1   Design of Steel Structures
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
A
7.5M_{p}/L
B
5.0M_{p}/L
C
4.5M_{p}/L
D
2.5M_{p}/L
GATE CE 2016 SET-2   Design of Steel Structures
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}
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
A
\frac{8M_{p}}{L}
B
\frac{6M_{p}}{L}
C
\frac{4M_{p}}{L}
D
\frac{2M_{p}}{L}
GATE CE 2016 SET-1   Design of Steel Structures
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:
A
16.5 M_{p}/L
B
15.5 M_{p}/L
C
15.0 M_{p}/L
D
16.0 M_{p}/L
GATE CE 2015 SET-2   Design of Steel Structures
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 _____.
A
3.3
B
2.8
C
6.2
D
8.6
GATE CE 2015 SET-1   Design of Steel Structures
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
A
\frac{2M_{p}}{L}
B
\frac{4M_{p}}{L}
C
\frac{6M_{p}}{L}
D
\frac{8M_{p}}{L}
GATE CE 2014 SET-2   Design of Steel Structures
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}
There are 10 questions to complete.

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