# Trusses

 Question 1
An idealised bridge truss is shown in the figure. The force in Member $U_{2} L_{3}$ is _____ kN (round off to one decimal place).

 A 12.5 B 16.8 C 14.1 D 10.6
GATE CE 2023 SET-1   Structural Analysis
Question 1 Explanation:

$\sum F_{y}=0, \quad 50-20-20-F_{U_{2} L_{3}} \sin 45^{\circ}=0$
$\therefore \quad F_{U_{2} L_{3}}=\frac{10}{\sin 45^{\circ}}=14.14 \mathrm{kN}$
$\therefore$ Rounding off to one decimal.
 Question 2
Consider the pin-jointed truss shown in the figure (not to scale). All members have the same axial rigidity, $\mathrm{AE}$. Members $\mathrm{QR}, \mathrm{RS}$, and $\mathrm{ST}$ have the same length L. Angles OBT, RCT, SDT are all $90^{\circ}$. Angles BQT, CRT, DST are all $30^{\circ}$. The joint $T$ carries a vertical load $P$. The vertical deflection of joint $T$ is $k \frac{P L}{A E}$. What is the value of $k$ ?

 A 1.5 B 4.5 C 3 D 9
GATE CE 2023 SET-1   Structural Analysis
Question 2 Explanation:

Considering joint $T$,

$\sum F_{y}=0, \quad F_{Q T} \sin 60^{\circ}=P$
$\therefore \quad \mathrm{F}_{\mathrm{QT}}=\frac{2 \mathrm{P}}{\sqrt{3}}$
$\sum \mathrm{F}_{\mathrm{X}}=0, \mathrm{~F}_{\mathrm{BT}}=-\mathrm{F}_{\mathrm{QT}} \cos 60^{\circ}=-\frac{P}{\sqrt{3}}$
$\therefore \quad$ Strain energy $(U)=\left(\frac{P^{2} L}{2 A E}\right)_{Q T}+\left(\frac{P^{2} L}{2 A E}\right)_{B T}$
$=\frac{\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right)^{2}(3 \mathrm{~L})}{2 \mathrm{AE}}+\frac{\left(-\frac{\mathrm{P}}{\sqrt{3}}\right)^{2}\left(\frac{3 \mathrm{~L}}{2}\right)}{2 \mathrm{AE}}$
$=\frac{2 P^{2} \times L}{A E}+\frac{P^{2} \times L}{4 A E}$
$\frac{8 P^{2} L+P^{2} L}{4 A E}=\frac{9 P^{2} L}{4 A E}$
$\therefore \quad$ Vertical deflection of joint $T$
$=\frac{\partial \mathrm{U}}{\partial \mathrm{P}}=\frac{18 \mathrm{PL}}{4 \mathrm{AE}}=4.5 \frac{\mathrm{PL}}{\mathrm{AE}}$
$\therefore \quad \mathrm{K}=4.5$

 Question 3
The plane truss shown in the figure is subjected to an external force $P$. It is given that $P = 70 kN, a = 2 m,$ and $b = 3 m$.

The magnitude (absolute value) of force (in kN) in member $EF$ is _______. (round off to the nearest integer)
 A 12 B 30 C 87 D 92
GATE CE 2022 SET-1   Structural Analysis
Question 3 Explanation:

\begin{aligned} \Sigma M_J&=0\\ V_A \times 8-H_A \times 1-70 \times 4&=0\\ 8V_A-H_A&=280 \;\;\;(i) \end{aligned}
To find $H_A$ (Cut the truss by 1-1)
Consider left hand side
\begin{aligned} \Sigma M_E&=0\\ V_A \times 4-H_A \times 4&=0\\ V_A&=H_A \;\;\;(ii)\\ \text{using (i) and (ii)}&\\ 8V_A-V_A&=280\\ 7V_A&=280\\ V_A&=40kN\\ H_A&=40kN\\ H_J&=40kN\\ V_J&=70-40=30kN\\ \end{aligned}
To find force in member EF (Cit the truss by 2-2)
Consider right hand side
Force in member EF
$F_{EF}=V_J=30kN$
 Question 4
Refer the truss as shown in the figure (not to scale).

If load, $F=10 \sqrt{3} \mathrm{kN}$, moment of inertia, $I=8.33 \times 10^{6} \mathrm{~mm}^{4}$, area of cross-section, $A=10^{4} \mathrm{~mm}^{2}$, and length, L=2 m for all the members of the truss, the compressive stress (in $\mathrm{kN} / \mathrm{m}^{2}$,in integer) carried by the member Q-R is _________
 A 370 B 650 C 760 D 500
GATE CE 2021 SET-1   Structural Analysis
Question 4 Explanation:

$V_{P}=V_{S}=5 \sqrt{3} \mathrm{kN}$
Considering equilibrium of LHS of section (1 )-(1)

\begin{aligned} \Sigma M_{T}(C W)&=0\\ (5 \sqrt{3} \times a)+F_{Q R}\left(\frac{\sqrt{3} a}{2}\right)&=0\\ \Rightarrow \qquad \qquad \qquad \qquad \quad F_{O A}&=-10 \mathrm{kN} or 10 \mathrm{kN} (C) \\ \text{ Compressive} &\text{ stress in member QR} \left(\sigma_{\mathrm{C}}\right)\\ \sigma_{\mathrm{C}}&=\frac{F_{Q A}}{2 A} \\ &=\frac{10 \mathrm{kN}}{2 \times\left(10^{4} \times 10^{-6}\right) \mathrm{m}^{2}} \\ &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}
 Question 5
A truss EFGH is shown in the figure, in which all the members have the same axial rigidity R. In the figure, P is the magnitude of external horizontal forces acting at joints F and G.

If $R=500 \times 10^{3} \mathrm{kN}$,$P=150 \mathrm{kN}$ and L=3 m, the magnitude of the horizontal displacement of joint G (in mm,round off to one decimal place) is ________
 A 0.4 B 0.6 C 0.9 D 0.2
GATE CE 2021 SET-1   Structural Analysis
Question 5 Explanation:

Note: No need to calculate 'I< force in all members because 'P force is zero fore all members except FG
\begin{aligned} \text { 1. } \Delta_{\mathrm{HG}}=& \sum_{i=1}^{n} \frac{P_{i} k_{i} L_{i}}{A_{i} E_{i}} \\ \Delta_{\mathrm{HG}}=& \text { Horizontal deflection at joint } \mathrm{G} \\ \therefore \qquad & \qquad \qquad\Delta_{\mathrm{HG}} =\underbrace{\frac{P \times 1 \times L}{A E}}_{\text {FG-member }} \times \underbrace{Q^{\prime}}_{\text {(For all other members) }} \\ & \qquad \qquad\Delta_{\mathrm{HG}} =\frac{P L}{A E}=\left(\frac{150 \times 3}{500 \times 10^{3}} \times 10^{3}\right) \mathrm{mm} \\ & \qquad \qquad \qquad=0.90 \mathrm{~mm} \end{aligned}