# Plastic Analysis

 Question 1
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   Design of Steel Structures
Question 1 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 2
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 2 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 3
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 3 Explanation:
Ratio of $\frac{M_p}{M_y}=$ Shape factor
 Question 4
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 4 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 5
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 5 Explanation:

$P_{u}=\frac{6M_{P}}{l}$ So, $\alpha =6$
 Question 6
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 6 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 7
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 7 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 8
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 8 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 9
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 10
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 10 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}
There are 10 questions to complete.