Question 1 |

A perfectly flexible and inextensible cable is shown in the figure (not to scale). The external loads at F and G are acting vertically.

The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______

The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______

4.68 | |

3.78 | |

5.64 | |

8.25 |

Question 1 Explanation:

\begin{aligned} \Sigma M_F=0\\ \left ( 10.667+\frac{H}{6} \right )2-H \times 3=0\\ H=8\; kN\\ V_E=10.667+\frac{8}{6}\\ V_E=12\; kN\\ V_H=10\; kN \end{aligned}

Consider left side of section (1)-(1)

\begin{aligned} \Sigma H=0\\ T \cos \theta =H\\ T \cos \theta =8\\ T \sin \theta =2\\ \therefore \;\; T^2 \cos ^2\theta + T^2 \sin ^2 \theta =8^2+2^2\\ T=8.246\; kN \end{aligned}

Tension in segment GF is 8.246 kN.

Question 2 |

A planar elastic structure is subjected to uniformly distributed load, as shown in the figure

Neglecting self-weight, the maximum bending moment generated in the structure (in kNm, round off to the nearest integer), is ___________ .

Neglecting self-weight, the maximum bending moment generated in the structure (in kNm, round off to the nearest integer), is ___________ .

12 | |

192 | |

96 | |

48 |

Question 2 Explanation:

V_A=V_B=\frac{wL}{2}=\frac{12 \times 8}{2}=48 kN

As horizontal thrust is zero so it behaves like a beam (curved beam)

M_{max}=\frac{wL^2}{8} \; (\text{ At crown})

=\frac{12 \times 8^2}{8}=96KNm

Question 3 |

The figure shows a two-hinged parabolic arch of span L subjected to a uniformly distributed load of intensity q per unit length

The maximum bending moment in the arch is equal to

The maximum bending moment in the arch is equal to

\frac{qL^{2}}{8} | |

\frac{qL^{2}}{12} | |

Zero | |

\frac{qL^{2}}{10} |

Question 3 Explanation:

If a two hinged or three hinged parabolic arch is
subjected to UDL throughout its length, bending
moment is zero everywhere.

Question 4 |

A three hinged parabolic arch having a span of 20m and a rise of 5m carries a
point load of 10 kN at quarter span from the left end as shown in the figure. The
resultant reaction at the left support and its inclination with the horizontal are
respectively

9.01 kN and 56.31^{\circ} | |

9.01 kN and 33.69^{\circ} | |

7.50 kN and 56.31^{\circ} | |

2.50 kN and 33.69^{\circ} |

Question 4 Explanation:

Let the vertical reactions at left and right support be V_{L} and V_{R} upwards respectively.

Taking moments about right support, we get,

\begin{aligned} V_{L} \times 20-10 \times 15&=0 \\ \Rightarrow \quad V_{L}&=7.5 \mathrm{kN} \\ \therefore \quad V_{R}&=10-7.5=2.5 \mathrm{kN} \end{aligned}

Let the horizontal reaction at left support be H from left to right.

Taking moments about the crown from left, we get,

\begin{aligned} 7.5 \times 10-10 \times 5-H \times 5 &=0 \\ \Rightarrow \quad H&=5 \mathrm{kN} \end{aligned}

Resultant reaction,

\begin{aligned} R &=\sqrt{H^{2}+V_{L}^{2}} \\ &=\sqrt{5^{2}+(7.5)^{2}} \\ &=9.01 \mathrm{kN} \end{aligned}

Let the resultant reaction at the left support makes an angle \theta with the horizontal.

\begin{aligned} & \tan \theta=\frac{7.5}{5} \\ \Rightarrow & \theta=\tan ^{-1}(1.5) \\ \Rightarrow & \theta=56.31^{\circ} \end{aligned}

Question 5 |

A three hinged parabolic arch ABC has a span of 20 m and a central rise of 4 m.
The arch has hinges at the ends and at the centre. A train of two point loads of
20 kN and 10 kN, 5m apart, crosses this arch from left to right, with 20 kN load
leading. The maximum thrust induced at the supports is

25kN | |

28.13kN | |

31.25kN | |

32.81kN |

Question 5 Explanation:

In case of a three hinged parabolic arch, the influence line diagram for horizontal thrust is linear. Maximum thrust will be induced at the supports when 20 kN load is at the crown.

Ordinate of the ILD at a distance of 5 m from

A=\frac{5}{10} \times \frac{L}{4 h}=\frac{1}{2} \times\left(\frac{20}{4 \times 4}\right)=0.625 \mathrm{kN}

also \frac{L}{4 h}=\frac{20}{4 \times 4}=1.25 \mathrm{kN}

Thus, horizontal thrust.

H=10 \times 0.625+20 \times 1.25=31.25 \mathrm{kN}

Ordinate of the ILD at a distance of 5 m from

A=\frac{5}{10} \times \frac{L}{4 h}=\frac{1}{2} \times\left(\frac{20}{4 \times 4}\right)=0.625 \mathrm{kN}

also \frac{L}{4 h}=\frac{20}{4 \times 4}=1.25 \mathrm{kN}

Thus, horizontal thrust.

H=10 \times 0.625+20 \times 1.25=31.25 \mathrm{kN}

There are 5 questions to complete.