Question 1 |

Real matrices [A]_{3 \times 1},[B]_{3 \times 3},[C]_{3 \times 5} , [D]_{5 \times 3},[E]_{5 \times 5}, [F]_{5\times 1} are given. Matrices [B] and [E] are symmetric.

Following statements are made with respect to these matrices.

I. Matrix product \left [ F \right ]^{T} \left [ C \right ]^{T} [B] [C] [F] is a scalar.

II. Matrix product \left [ D \right ]^{T} [F] [D] is always symmetric.

With reference to above statements, which of the following applies ?

Following statements are made with respect to these matrices.

I. Matrix product \left [ F \right ]^{T} \left [ C \right ]^{T} [B] [C] [F] is a scalar.

II. Matrix product \left [ D \right ]^{T} [F] [D] is always symmetric.

With reference to above statements, which of the following applies ?

Statement I is true but II is false | |

Statement I is false but II is true | |

Both the statement are true | |

Both the statement are false |

Question 1 Explanation:

Statement 1 is true.

Consider statement 2

\left [ D \right ]^{T}\left [ F \right ]\left [ D \right ] is always symmetric.

Now since

\begin{aligned} \left [ {D}'FD \right ]{}'&= {\left [ FD \right ]}'{\left ( {D}' \right )}' \\ &={\left ( FD \right )}'D \\ &={D}'{F}'D \\ &={D}'FD \end{aligned}

(Since F may or may not be symmetric)

\therefore \;\; {D}'FD is not always symmetric is true.

\therefore 1 is true and 2 is false.

Consider statement 2

\left [ D \right ]^{T}\left [ F \right ]\left [ D \right ] is always symmetric.

Now since

\begin{aligned} \left [ {D}'FD \right ]{}'&= {\left [ FD \right ]}'{\left ( {D}' \right )}' \\ &={\left ( FD \right )}'D \\ &={D}'{F}'D \\ &={D}'FD \end{aligned}

(Since F may or may not be symmetric)

\therefore \;\; {D}'FD is not always symmetric is true.

\therefore 1 is true and 2 is false.

Question 2 |

The summation of series S=2+\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+....\infty is

4.5 | |

6 | |

6.75 | |

10 |

Question 2 Explanation:

S is an arithmetico-geometric series and can be summed up as follows,

\begin{aligned} S&=2+\left ( \frac{5}{2}+\frac{8}{2^{2})}+\frac{11}{2^{3}}+...\infty \right ) \\ &=2+x\;\; (say)\\ \text{where } x&=\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+...\infty \;\; ...(i)\\ \text{Multiplying by }&\text{1/2, we get, } \\ \frac{1}{2}x&=\frac{5}{2^{2}}+\frac{8}{2^{3}}+\frac{11}{2^{4}}+...\infty \;\;...(ii) \\ \text{Substracting (ii) } & \text{from (i), we get, }\\ x-\frac{1}{2}x&=\frac{5}{2}+\frac{\left ( 8-5 \right )}{2^{2}} \\ &+\frac{\left ( 11-8 \right )}{2^{3}} +\frac{\left ( 13-11 \right )}{2^{4}}+...\infty \\ \frac{x}{2}&=\frac{5}{2}+\left ( \frac{3}{2^{2}}+\frac{3}{2^{3}}+\frac{3}{2^{4}}+...\infty \right ) \\ \frac{x}{2}&=\frac{5}{2}+\frac{\frac{3}{2^{2}}}{\left [ 1-\frac{1}{2} \right ]} \\ \frac{x}{2}&=\frac{5}{2}+\frac{3}{2}=4 \\ \therefore \;\; x&=8\\ \text{and } S&=2+x=10\end{aligned}

\begin{aligned} S&=2+\left ( \frac{5}{2}+\frac{8}{2^{2})}+\frac{11}{2^{3}}+...\infty \right ) \\ &=2+x\;\; (say)\\ \text{where } x&=\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+...\infty \;\; ...(i)\\ \text{Multiplying by }&\text{1/2, we get, } \\ \frac{1}{2}x&=\frac{5}{2^{2}}+\frac{8}{2^{3}}+\frac{11}{2^{4}}+...\infty \;\;...(ii) \\ \text{Substracting (ii) } & \text{from (i), we get, }\\ x-\frac{1}{2}x&=\frac{5}{2}+\frac{\left ( 8-5 \right )}{2^{2}} \\ &+\frac{\left ( 11-8 \right )}{2^{3}} +\frac{\left ( 13-11 \right )}{2^{4}}+...\infty \\ \frac{x}{2}&=\frac{5}{2}+\left ( \frac{3}{2^{2}}+\frac{3}{2^{3}}+\frac{3}{2^{4}}+...\infty \right ) \\ \frac{x}{2}&=\frac{5}{2}+\frac{\frac{3}{2^{2}}}{\left [ 1-\frac{1}{2} \right ]} \\ \frac{x}{2}&=\frac{5}{2}+\frac{3}{2}=4 \\ \therefore \;\; x&=8\\ \text{and } S&=2+x=10\end{aligned}

Question 3 |

The value of the function f(x)=\lim_{x\rightarrow 0}\frac{x^{3}+x^{2}}{2x^{3}-7x^{2}} is

0 | |

-\frac{1}{7} | |

\frac{1}{7} | |

\infty |

Question 3 Explanation:

f\left ( x \right )=\lim_{x\rightarrow 0}\left [ \frac{x^{3}+x^{2}}{2x^{3}-7x^{2}} \right ]

Since this has \frac{0}{0} form, limit can be found by repeated application of L'Hospitals rule.

\begin{aligned}f\left ( x \right )&=\lim_{x\rightarrow 0}\left [ \frac{3x^{2}+2x}{6x^{2}-14x} \right ] \\ &=\lim_{x\rightarrow 0}\left [ \frac{6x+2}{12x-14} \right ] \\ &=\left [ \frac{6\times 0+2}{12\times 0-14} \right ] \\ &=-\frac{1}{7} \end{aligned}

Since this has \frac{0}{0} form, limit can be found by repeated application of L'Hospitals rule.

\begin{aligned}f\left ( x \right )&=\lim_{x\rightarrow 0}\left [ \frac{3x^{2}+2x}{6x^{2}-14x} \right ] \\ &=\lim_{x\rightarrow 0}\left [ \frac{6x+2}{12x-14} \right ] \\ &=\left [ \frac{6\times 0+2}{12\times 0-14} \right ] \\ &=-\frac{1}{7} \end{aligned}

Question 4 |

For the plane truss shown in the figure, the number of zero force members for the
given loading is

4 | |

8 | |

11 | |

13 |

Question 4 Explanation:

If three members meet at a joint and two of them are collinear, then the third member will carry zero force provided that there is no external load at the joint.

Thus using above statement we arrive at 8 zero force members which are highlighted by 'o' sign.

Question 5 |

The unit load method used in structural analysis is

applicable only to statically indeterminate structures | |

another name for stiffness method | |

an extension of Maxwell's reciprocal theorem | |

derived from Castigliano's theorem |

Question 5 Explanation:

Unit load method is used to find deflection at any point of structure. It is derived from Castigliano's theorem.

\Delta=\frac{\partial U}{\partial Q}=\int M \frac{\partial M}{\partial Q} \frac{d x}{E I}=\int \frac{M m d x}{E I}

\Delta=\frac{\partial U}{\partial Q}=\int M \frac{\partial M}{\partial Q} \frac{d x}{E I}=\int \frac{M m d x}{E I}

There are 5 questions to complete.