Question 1 |
The number of boundary conditions required to solve the differential equation \frac{\partial^2 \phi }{\partial x^2}+\frac{\partial^2 \phi }{\partial y^2}=0 is
2 | |
0 | |
1 | |
4 |
Question 2 |
Value of the integral I=\int_{0}^{\pi /4}cos^{2}xdx is
\frac{\pi }{8}+\frac{1}{4} | |
\frac{\pi }{8}-\frac{1}{4} | |
-\frac{\pi }{8}-\frac{1}{4} | |
-\frac{\pi }{8}+\frac{1}{4} |
Question 2 Explanation:
\begin{aligned} I&=\int_{0}^{\pi /4}\cos ^{2}xdx \\ &=\int_{0}^{\pi /4}\frac{1+\cos 2x}{2}dx \\ &=\int_{0}^{\pi /4}\frac{dX}{2}+\int_{0}^{\pi /4}\frac{\cos 2x}{2}dx \\ &=\frac{1}{2}\left [ x \right ]_{0}^{\pi /4}+\frac{1}{2}\left [ \frac{\sin 2x}{2} \right ]_{0}^{\pi /4} \\ &=\frac{\pi }{8}+\frac{1}{4}\left [ \sin \frac{\pi }{2}-\sin x \right ] \\ &=\frac{\pi }{8}+\frac{1}{4}\times 1 =\frac{\pi }{8}+\frac{1}{4} \end{aligned}
Question 3 |
Limit of the following series as x approaches \frac{\pi }{2} is f(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}
\frac{2\pi }{3} | |
\frac{\pi }{2} | |
\frac{\pi }{3} | |
1 |
Question 3 Explanation:
\lim_{x\rightarrow \frac{\pi }{2}}f\left ( x \right )=\lim_{x\rightarrow \frac{\pi }{2}}\sin x=1
Question 4 |
The degree of static indeterminacy, N_s, and the degree of kinematic indeterminacy, N_k , for the plane frame shown below, assuming axial deformations to be negligible, are given by
N_s = 6 and N_k = 11 | |
N_s = 6 and N_k = 6 | |
N_s = 4 and N_k = 6 | |
N_s = 4 and N_k = 4 |
Question 4 Explanation:
Degree of static indeterminacy,
\begin{aligned} N_{s} &=3 m+r_{e}-3 j-r_{r} \\ &=3 \times 5+(3+2+2)-3 \times 6-0 \\ &=4 \end{aligned}
Degree of kinematic indeterminacy,
\begin{aligned} N_{k} &=3 j-r_{e}-m \\ &=3 \times 6-(3+2+2)-5=6 \\ N_{s} &=4 \text { and } N_{k}=6 \end{aligned}
\begin{aligned} N_{s} &=3 m+r_{e}-3 j-r_{r} \\ &=3 \times 5+(3+2+2)-3 \times 6-0 \\ &=4 \end{aligned}
Degree of kinematic indeterminacy,
\begin{aligned} N_{k} &=3 j-r_{e}-m \\ &=3 \times 6-(3+2+2)-5=6 \\ N_{s} &=4 \text { and } N_{k}=6 \end{aligned}
Question 5 |
The bending moment (in kNm units) at the mid span location X in the beam
with overhangs shown below is equal to
0 | |
-10 | |
-15 | |
-20 |
Question 5 Explanation:
\begin{aligned} \mathrm{R}+\mathrm{V}&=30 &\ldots(i)\\ \text { Now, } \mathrm{V} &\times 2-20 \times 3+10 \times 1=0 \\ \mathrm{V}&=\frac{60-10}{2}=25 \mathrm{kN}\\ \therefore \quad \mathrm{R}&=5 \mathrm{kN}\\ \mathrm{BM} \text { at }\quad X&=25 \times 1-20 \times 2=-15 \mathrm{kNm} \end{aligned}
There are 5 questions to complete.