# 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.