Question 1 |

Consider two axially loaded columns, namely, 1 and 2, made of a linear elastic material with Young's modulus 2 \times 10^{5} MPa, square cross- section with side 10 mm, and length 1 m. For Column 1, one end is fixed and the other end is free. For Column 2, one end is fixed and the other end is pinned. Based on the Euler's theory, the ratio (up to one decimal place) of the buckling load of Column 2 to the buckling load of Column 1 is ________

2.5 | |

8 | |

5.2 | |

4 |

Question 1 Explanation:

\begin{aligned} P_{c r 1}&=\frac{\pi^{2} E \cdot I}{(2 l)^{2}}=\frac{\pi^{2} E I}{4 l^{2}} \\ P_{c r 2}&=\frac{\pi^{2} E \cdot I}{\left(\frac{l}{\sqrt{2}}\right)^{2}}=\frac{2 \pi^{2} E I}{l^{2}}\\ \therefore \text{Required ratio }&=\frac{P_{C r 2}}{P_{c r 1}}=\frac{\frac{2 \pi^{2} E I}{l^{2}}}{\frac{\pi^{2} E I}{4 l^{2}}}=8 \end{aligned}

Question 2 |

In a system,two connected rigid bars AC and BC are of identical length, L with pin supports at A and B. The bars are interconnected at C by a frictionless hinge. The rotation of the hinge is restrained by a rotational spring of stiffness, k. The system initially assumes a straight line configuration, ACB. Assuming both the bars as weightless, the rotation at supports, A and B, due to a transverse load, P applied at C is

\frac{PL}{4k} | |

\frac{PL}{2k} | |

\frac{P}{4k} | |

\frac{Pk}{4L} |

Question 2 Explanation:

Deflection under load P=\theta L

Work done by force P=\frac{1}{2} P(\theta L)

Strain energy stored in spring

\begin{aligned} &=\frac{1}{2}[k(2 \theta)](2 \theta) \\ \frac{1}{2} P(\theta L) &=\frac{4 k \theta^{2}}{2} \\ \Rightarrow \quad \theta &=\frac{P L}{4 k} \end{aligned}

Work done by force P=\frac{1}{2} P(\theta L)

Strain energy stored in spring

\begin{aligned} &=\frac{1}{2}[k(2 \theta)](2 \theta) \\ \frac{1}{2} P(\theta L) &=\frac{4 k \theta^{2}}{2} \\ \Rightarrow \quad \theta &=\frac{P L}{4 k} \end{aligned}

Question 3 |

The possible location of shear centre of the channel section, shown below, is

P | |

Q | |

R | |

S |

Question 3 Explanation:

For no twisting,

\begin{aligned} V \times e &=H \times h\\ \Rightarrow \;\; e&=\frac{Hh}{V}\\ \end{aligned}

Hence, possible location of shear center is P.

Question 4 |

Two steel columns P (length L and yield strength f_{y}= 250 MPa) and Q (length 2L and yield strength f_{y}= 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:

0.5 | |

1 | |

2 | |

4 |

Question 4 Explanation:

Buckling load,

\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}

\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}

Question 5 |

The sketch shows a column with a pin at the base and rollers at the top. It is subjected to an
axial force P and a moment M at mid-height. The reaction(s) at R is/are

a vertical force equal to P | |

a vertical force equal to P/2 | |

a vertical force equal to P and a horizontal force equal to M/h | |

a vertical force equal to P/2 and a horizontal force equal to M/h |

Question 5 Explanation:

\begin{aligned} \Sigma F_y &=0 \\ \Rightarrow \;\; V_R-P&=0\\ \Rightarrow \;\;V_R&=P\\ \Sigma M_Q&=0\\ \Rightarrow \;\; H_Rh-M&=0\\ \Rightarrow \;\;H_R&=\frac{M}{h} \end{aligned}

Question 6 |

The ratio of the theoretical critical buckling load for a column with fixed ends to that of another
column with the same dimensions and material, but with pinned ends, is equal to

0.5 | |

1 | |

2 | |

4 |

Question 6 Explanation:

Eulars critical load,

\begin{aligned} P_{e}&=\frac{\pi^{2}EI}{I^{2}_{\text{eft}}}\\ \frac{P_{e1}}{P_{e2}}=\left[\frac{(I_{\text{elf}})_{2}}{(I_{\text{elf}})_{1}} \right ]^{2}&=\left[\frac{I}{1/2}\right ]^{2}=4 \end{aligned}

\begin{aligned} P_{e}&=\frac{\pi^{2}EI}{I^{2}_{\text{eft}}}\\ \frac{P_{e1}}{P_{e2}}=\left[\frac{(I_{\text{elf}})_{2}}{(I_{\text{elf}})_{1}} \right ]^{2}&=\left[\frac{I}{1/2}\right ]^{2}=4 \end{aligned}

Question 7 |

The effective length of a column of length L fixed against rotation and translation
at one end and free at the other end is

0.5L | |

0.7L | |

1.414L | |

2L |

Question 8 |

The point within the cross sectional plane of a beam through which the resultant
of the external loading on the beam has to pass through to ensure pure bending
without twisting of the cross-section of the beam is called

moment centre | |

centroid | |

shear centre | |

elastic center |

Question 9 |

A rigid bar GH of length L is supported by a hinge and a spring of stiffness K as shown in the figure below. The buckling load, P_{Cr}, for the bar will be

0.5KL | |

0.8KL | |

1.0KL | |

1.2KL |

Question 9 Explanation:

Let the deflection in the spring be \delta and force in

the spring be F.

Taking moments about G, we get

\begin{aligned} P_{c r} \times \delta&=F \times L\\ \Rightarrow \quad P_{c r}&=\frac{K \delta \times L}{\delta} &\left[\because F=K\delta \right ]\\ \Rightarrow \quad P_{c r}&=K L \end{aligned}

the spring be F.

Taking moments about G, we get

\begin{aligned} P_{c r} \times \delta&=F \times L\\ \Rightarrow \quad P_{c r}&=\frac{K \delta \times L}{\delta} &\left[\because F=K\delta \right ]\\ \Rightarrow \quad P_{c r}&=K L \end{aligned}

Question 10 |

Cross-section of a column consisting to two steel strips, each of thickness t and
width b is shown in the figure below. The critical loads of the column with perfect
bond and without bond between the strips are P and P_0 respectively. The ratio P/P_0 is

2 | |

4 | |

6 | |

8 |

Question 10 Explanation:

We know that critical load for a column is proportional to moment of inertia irrespective of end conditions of the column i.e.

P_{c r} \propto 1

When the steel strips are perfectly bonded, then

f_{p b}=\frac{b \times(2 t)^{3}}{12}=\frac{8 b t^{3}}{12}

When the steel strips are not bonded, then

\begin{aligned} I_{wb}&=2 \times \frac{b t^{3}}{12}=\frac{2 b t^{3}}{12} \\ \therefore \quad \frac{P}{P_{0}}&=\frac{8 b t^{3} / 12}{2 b t^{3} / 12}=4 \end{aligned}

P_{c r} \propto 1

When the steel strips are perfectly bonded, then

f_{p b}=\frac{b \times(2 t)^{3}}{12}=\frac{8 b t^{3}}{12}

When the steel strips are not bonded, then

\begin{aligned} I_{wb}&=2 \times \frac{b t^{3}}{12}=\frac{2 b t^{3}}{12} \\ \therefore \quad \frac{P}{P_{0}}&=\frac{8 b t^{3} / 12}{2 b t^{3} / 12}=4 \end{aligned}

There are 10 questions to complete.