Q21GATE 2012MCQ

A simply supported beam is subjected to a uniformly distributed load of intensity w per unit length, on half of the span from one end. The length of the span and the flexural stiffness are denoted as $l$ and EI, respectively. The deflection at mid-span of the beam is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \delta &=\delta_{1}+0 \\ &=\frac{5}{384} \frac{(w / 2) l^{4}}{E I} \\ \delta &=\frac{5}{768} \frac{w l^{4}}{E I} \end{aligned}

Q22GATE 2011MCQ

A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of 'k') at points '1' and '2', and an inclined load acts at '2', as shown If the load P equals 100 kN, which of the following options represents forces R1 and R2 in the springs at points '1' and '2'?

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} R_{1} &=k \delta_{1}=k \times \frac{2 P}{5 k} \\ &=\frac{2 \times 100}{5}=40 \mathrm{kN} \\ R_{2} &=k \delta_{2}=k \times \frac{4 P}{5 k} \\ &=\frac{4 \times 100}{5}=80 \mathrm{kN} \end{aligned}

Q23GATE 2009MCQ

In the cantilever beam PQR shown in figure below, the segment PQ has flexural rigidity EI and the segment QR has infinite flexural rigidity. The deflection of the beam at R is

🚀 Solve in Practice Mode

📖 Explanation

since the portion QR of the beam is rigid, QR will remain straight.

\begin{aligned} \text { Deflection of } R &=\Delta_{1}+\Delta_{2} \\ &=\text { Deflection at } Q \\ &+\text { Slope at } Q \times L \\ &=\frac{5 W L^{3}}{6 E I}+\frac{3 W L^{2}}{2 E I} \times L \\ &=\frac{5 W L^{3}+9 W L^{3}}{6 E I} \\ &=\frac{14 W L^{3}}{6 E I}=\frac{7 W L^{3}}{3 E I} \end{aligned}

Q24GATE 2008MCQ

The stepped cantilever is subjected to moments, M as shown in the figure below. The vertical deflection at the free end (neglecting the self weight) is

🚀 Solve in Practice Mode

📖 Explanation

Using Movement Area Method Deflection at B w.r.t. A = Moment of area of $\frac{M}{E I}$ diagram between A and B about B $=\frac{M}{E \mid} \times L \times \frac{L}{2}=\frac{M L^{2}}{2 E I}$

Q25GATE 2008MCQ

Beam GHI is supported by three pontoons as shown in the figure below. The horizontal cross-sectional area of each pontoon is $8 m^{2}$ , the flexural rigidity of the beam is 10000 $kN-m^{2}$ and the unit weight of water is 10 $kN/m^{3}$ When the middle pontoon is brought back to its position as shown in the figure above, the reaction at H will be

🚀 Solve in Practice Mode

📖 Explanation

Let the elastic deflection at H be $\delta_{2}$

\begin{aligned} \therefore \quad \delta_{2}&=\frac{(P-R) L^{3}}{48 E I} \quad &\ldots(i)\end{aligned}

The reactions at G and I will be same, as the beam is symmetrically loaded. Let the reaction at each G and I be Q Using principle of buoyancy, we get $\delta_{1} \times$ area of cross-section of pontoon $\times \gamma_{w}=Q$

\begin{aligned} \Rightarrow \quad \delta_{1} \times 8 \times 10&=Q \\ \Rightarrow \quad \delta_{1}&=\frac{Q}{80} &\ldots(ii) \end{aligned}

Also, we have

\begin{aligned} Q+Q+R&=P \\ \Rightarrow \quad 2 Q+R&=48 &\ldots(iii)\\ \end{aligned}

$\text{Also, }\left(\delta_{1}+\delta_{2}\right) \times$ area of cross-section of Pontoon $\times \gamma_{n}=\mathrm{R}$

\begin{aligned} \Rightarrow\left(\delta_{1}+\delta_{2}\right) \times 8 \times 10&=\mathrm{R} \\ \Rightarrow \quad \delta_{1}+\delta_{2}&=\frac{R}{80} \\ \Rightarrow \quad \frac{Q}{80}+\delta_{2}&=\frac{R}{80} &[\text{from (ii)}]\\ \Rightarrow \quad \frac{48-R}{2 \times 80}+\delta_{2}&=\frac{R}{80} \quad&[\text{from (iii)}] \\ \Rightarrow \quad \delta_{2}&=\frac{2 R-48+R}{160} \\ \Rightarrow \quad \delta_{2}&=\frac{3 R-48}{160} \\ \therefore \frac{(48-R) \times 10^{3}}{48 \times 10^{4}}&=\frac{3 R-48}{160} \quad&[\text{from (i)}] \\ \Rightarrow \quad R&=19.2 \mathrm{kN} \end{aligned}

Q26GATE 2007MCQ

$U_{1}$ and $U_{2}$ are the Strain energies stored in a prismatic bar due toaxial tensile forces $P_{1}$ and $P_{2}$ , respectively. The strain energy U stored in the same bar due to combined action of $P_{1}$ and $P_{2}$ will be

🚀 Solve in Practice Mode

📖 Explanation

Strain energy stored in a prismatic bar due to axial load $P_{1}$ $U_{1}=\frac{P_{1}^{2} L}{2 A E}$ Strain energy stored in a prismatic bar due to axial load, $P_{2}$ $U_{2}=\frac{P_{2}^{2} L}{2 A E}$ Strain energy stored in a prismatic bar due to combined action of $P_{1}$ and $P_{2}$

\begin{aligned} U &=\frac{\left(P_{1}+P_{2}\right)^{2} L}{2 A E} \\ U &=\frac{P_{1}^{2} L}{2 A E}+\frac{P_{2}^{2} L}{2 A E}+\frac{P_{1} P_{2} L}{A E} \\ U&=U_{1}+U_{2}+\frac{P_{1} P_{2} L}{A E} \\ \Rightarrow \quad U&>U_{1}+U_{2} \end{aligned}

Q27GATE 2006MCQ

Vertical reaction developed at B in the frame below due to the applied load to 100 kN (with 150,000 $mm^{2}$ cross-sectional area and $3.125\times 10^{9}mm^{4}$ moment of inertia for both members) is

🚀 Solve in Practice Mode

📖 Explanation

Deflection at A in beam AB= Compression in column AC

\begin{array}{l} \begin{aligned} \quad \frac{(100-\mathrm{R}) \mathrm{L}^{3}}{3 \mathrm{E} 1}&=\frac{\mathrm{RL}}{\mathrm{AE}} \\ \Rightarrow \frac{(100-\mathrm{R}) \times (1 \times 1000)^{3}}{3 \times 3.125 \times 10^{9}}&=\frac{\mathrm{R}}{150000} \\ \end{aligned}\\ \begin{aligned} \Rightarrow \quad 100-\mathrm{R}&=0.0625 \\ \Rightarrow \quad 100&=1.0625 \mathrm{R} \\ \therefore \quad V_{B}&=100-R\\ \Rightarrow \quad \mathrm{R}&=94.1 \mathrm{kN} \\ &=100-\mathrm{R} \\ &=5.9 \mathrm{kN} \end{aligned} \end{array}

Q28GATE 2006MCQ

Consider the beam AB shown in the figure below. Part AC of the beam is rigid while Part CB has the flexural rigidity EI . Identify the correct combination of deflection at end B and bending moment at end A, respectively

🚀 Solve in Practice Mode

📖 Explanation

Part AC of the beam is rigid. Hence C will act as a fixed end. Thus the deflection at B will be given as $\delta_{\mathrm{B}}=\frac{P L^{3}}{3 E \mid}$ But the bending moment does not depend on the rigidity or flexibility of the beam. $\mathrm{BM}\text{ at }\mathrm{A}=\mathrm{P} \times 2 \mathrm{L}=2 \mathrm{PL}$

Q29GATE 2006MCQ

Consider a propped cantilever beam ABC under two loads of magnitude P each as shown in the figure below. Flexural rigidity of the beam is EI. The rotation at B is

🚀 Solve in Practice Mode

📖 Explanation

The rotation at B : (i) Due to moment $\theta_{8_{1}}=\frac{2 P a \times L}{E l} \text { (clockwise) }$ (ii) Due to reaction R

\begin{aligned} \theta_{B_{2}} &=\frac{R L^{2}}{2 E I}+\frac{R L^{2}}{E I}=\frac{3 R L^{2}}{2 E l} \\ &=\frac{27}{16} \frac{P a L}{E l}(\text { anticlockwise }) \\ \therefore\quad \theta_{B} &=\theta_{B_{1}}-\theta_{B_{2}} \\ &=\left(2-\frac{27}{16}\right) \frac{P a L}{E l} \\ &=\frac{5}{16} \frac{P a L}{E l}(\text { clockwise }) \end{aligned}

Q30GATE 2004MCQ

For the linear elastic beam shown in the figure, the flexural rigidity, EI is 781250 kN $m^{2}$ . When w = 10 kN/m, the vertical reaction $R_A$ at A is 50 kN. The value of $R_A$ for w = 100 kN/m is

🚀 Solve in Practice Mode

📖 Explanation

The deflection at the free end for

\begin{aligned} w(10 k N / m) &=\frac{w L^{4}}{8 E I} \\ &=\frac{10 \times (5)^{4} \times 1000}{8 \times 781250} \\ &=1 \mathrm{mm} \end{aligned}

The gap between the beam and rigid platform is 6mm. Hence, no reaction will be developed when $w=10 kN / m$ Now, deflection at the free end for w( 100 kN /m ) will be $=10 \times 1 \mathrm{mm}=10 \mathrm{mm}$ But, this cannot be possible because margin of deflection is only 6 m. Thus w=100 kN / m will induce a reaction $\mathrm{R}_{\mathrm{B}} \text{ at } \mathrm{B}$

\begin{aligned} \therefore \frac{w L^{4}}{8 E !}-\frac{R_{B} L^{3}}{3 E I}&=\text{ Permissible deflection}\\ \Rightarrow \frac{100 \times (5)^{4}}{8 \times 781250}&-\frac{R_{6} \times (5)^{3}}{3 \times 781250}=\frac{6}{1000} \\ \Rightarrow \frac{10}{1000}-\frac{6}{1000}&=\frac{R_{8} \times 125}{3 \times 781250} \\ \Rightarrow \quad R_{8}&=75 \mathrm{kN} \\ \therefore \quad R_{\mathrm{A}}&=(100 \times 5-75) \\ &=425 \mathrm{kN} \end{aligned}

Q31GATE 2003MCQ

A "H" shaped frame of uniform flexural rigidity EI is loaded as shown in the figure. The relative outward displacement between points K and O is

🚀 Solve in Practice Mode

📖 Explanation

The bending moment in the member $\mathrm{JN}=\mathrm{R} \times \mathrm{h}(sagging)$ $\therefore\text{ Slope at J or N}=\frac{R h \times L}{2 E I}$ Thus, outward displacement between points $K\text{ and }O=\frac{R h L}{2 E I} \times h+\frac{R h L}{2 E I} \times h=\frac{R h^{2} L}{E I}$

Q32GATE 2003MCQ

A simply supported beam of uniform rectangular cross-section of width b and depth h is subjected to linear temperature gradient 0 $^{\circ}$ at the top and T $^{\circ}$ at the bottom, as shown in the figure. The coefficient of linear expansion of the beam material is $\alpha$ . The resulting vertical deflection at the mid-span of the beam is

🚀 Solve in Practice Mode

📖 Explanation

The average change in temperature $=\frac{T}{2}$ The compression in the top most fibre $=\alpha \times L \times \frac{T}{2}$ Similarly, the elongation in bottom most fibre

\begin{aligned} &=\alpha \times L \times \frac{T}{2}\\ \therefore\text{ Strain, }\varepsilon_{0}&=\frac{L \alpha T}{L \times 2}=\frac{\alpha T}{2} \\ \end{aligned}

Therefore deflection at midpoint is downward. Now, from the equation of pure bending, we have $\frac{M}{I}=\frac{E}{R}=\frac{f}{y}$ $\Rightarrow$ Curvature,

\begin{aligned} \frac{1}{R} &=\frac{f}{E y}=\frac{\text { Strain }}{y}\left[\because y=\frac{h}{2}\right] \\ &=\frac{2 \varepsilon_{0}}{h}=\frac{\alpha T}{h} \end{aligned}

Also, from the property of circle, we have

\begin{aligned} \text{Deflection, }\delta&=\frac{L^{2}}{8 R}=\frac{L^{2}}{8} \times \frac{\alpha T}{h} \\ &=\frac{\alpha T L^{2}}{8 h} \text { downward } \end{aligned}

Q33GATE 2002MCQ

In the propped cantilever beam carrying a uniformly distributed load of w N/m, shown in the following figure, the reaction at the support B is

🚀 Solve in Practice Mode

📖 Explanation

Deflection at proped end =0

\begin{aligned} \frac{w \times L^{4}}{8 E \mid}-\frac{R_{B} \times L^{3}}{3 E I} &=0 \\ \Rightarrow \quad R_{B} &=\frac{3}{8} w L \end{aligned}