Question 1 |
A prismatic fixed-fixed beam, modelled with a total lumped-mass of 10 kg as a single degree of freedom (SDOF) system is shown in the figure.
If the flexural stiffness of the beam is 4 \pi^{2} \mathrm{kN} / \mathrm{m}, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
If the flexural stiffness of the beam is 4 \pi^{2} \mathrm{kN} / \mathrm{m}, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
8 | |
16 | |
10 | |
18 |
Question 1 Explanation:
Given Data:
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
Question 2 |
Employ stiffness matrix approach for the simply supported beam as shown in the figure to calculate unknown displacements/rotations. Take length, L=8 m; modulus of elasticity, E=3 \times 10^{4} \mathrm{~N} / \mathrm{mm}^{2}; moment of inertia, I=225 \times 10^{6} \mathrm{mm}^{4}.
The mid-span deflection of the beam (in mm,round off to integer) under P=100 kN in downward direction will be ___________
The mid-span deflection of the beam (in mm,round off to integer) under P=100 kN in downward direction will be ___________
186 | |
91 | |
119 | |
192 |
Question 2 Explanation:
By stiffness matrix method
Step-1 : Generation of stiffness matrix
Column-1
\begin{aligned} \mathrm{k}_{11}&=\frac{3 E(2 I)}{(L / 2)^{3}}+\frac{3 E(I)}{(L / 2)^{3}}=\frac{72 \mathrm{E} I}{L^{3}}\\ k_{21}&=-\frac{3 E(2 I)}{(L / 2)^{2}}+\frac{3 E(I)}{(L / 2)^{2}}=-\frac{12 E I}{L^{2}} \end{aligned}
Column-2
\begin{aligned} k_{22}&=\frac{3 E(2 I)}{(L / 2)}+\frac{3 E(I)}{(L / 2)}=18 \frac{E I}{L} \\ \text { Stiffness matrix }[k]&=\left[\begin{array}{cc} 72 \frac{E I}{L^{3}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{E I}{L} \end{array}\right] \end{aligned}
Step-2: Calculation of unknown Nodal displacements \left(\Delta_{\mathrm{B}}, \theta_{\mathrm{B}}\right)
\begin{aligned} \text{Using} \qquad \qquad \qquad \qquad [P]&=[k][\Delta]\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \quad & \quad\left[\begin{array}{l} P \\ O \end{array}\right]=\left[\begin{array}{cc} \frac{72 E I}{L^{2}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{8 E I}{L} \end{array}\right]\left[\begin{array}{l} \Delta_{B} \\ \theta_{B} \end{array}\right]\\ \text{On solving} \qquad \qquad \qquad \quad\\\Delta_{\mathrm{B}}&=\frac{P L^{3}}{64 E I}(\downarrow) \\ \therefore \qquad \qquad \qquad \qquad \qquad \theta_{\mathrm{B}}&=\frac{P L^{2}}{96 E I}(\mathrm{CW}) \\ \Delta_{\mathrm{B}}&=\frac{\left(100 \times 10^{3}\right) \times(8000)^{3}}{64 \times 3 \times 10^{4} \times 225 \times 10^{6}}\\ &=118.519 \mathrm{~mm} \simeq 119 \mathrm{~mm} \end{aligned}
Question 3 |
For the beam shown below, the stiffness coefficient K_{22} can be written as
\frac{6EI}{L^{2}} | |
\frac{12EI}{L^{3}} | |
\frac{3EI}{L} | |
\frac{EI}{6L^{2}} |
Question 3 Explanation:
By giving unit displacement in 2^{\text {nd }} direction without giving displacement in any other direction, the force developed R_{6}, is,
K_{22}=\frac{12 E I}{L^{3}}
Question 4 |
The stiffness coefficient k_{ij} indicates
force at i due to a unit deformation at j | |
deformation at j due to a unit force at i | |
deformation at i due to a unit force at j | |
force at j due to a unit deformation at i |
Question 4 Explanation:
Stiffness(k) is the force required to produce unit deformation.
Thus k_{ij} denotes force required in direction i due to unit deformation (displacement) in direction j.
Thus k_{ij} denotes force required in direction i due to unit deformation (displacement) in direction j.
Question 5 |
For a linear elastic frame, if stiffness matrix is doubled with respect to the existing
stiffness matrix, the deflection of the resulting frame will be
Twice the existing value | |
Half the existing value | |
The same as existing value | |
Indeterminate value |
Question 5 Explanation:
Stiffness matrix and deflection are related as,
P=K \delta
Thus, when stiffness matrix is doubled, then deflection will reduce to half of the existing value.
P=K \delta
Thus, when stiffness matrix is doubled, then deflection will reduce to half of the existing value.
There are 5 questions to complete.