Past Year Questions: Arches | Structural Analysis - Civil Engineering (CE) PDF Download

Q1: A semi-circular bar of radius R m, in a vertical plane, is fixed at the end G, as shown in the figure. A horizontal load of magnitude P kN is applied at the end H. The magnitude of the axial force, shear force, and bending moment at point Q for θ = 45º, respectively, are    [2022, Set-1]

(a)

(b)

(c)

(d)

Ans:(a)
Sol:
(F_Q) = P sin⁡ θ = P/√2 (at θ = 45º) 
(S_Q) = P cos ⁡θ = P/√2 (at θ = 45º) 
(M_Q) = PR sin⁡ θ = PR/√2 (at θ = 45º) 

Q1: 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    [2021, Set-2]
Ans: 8.1 to 8.4
Sol:

VE = 12 kN
VH = 10 kN
Consider left side of section (1)-(1)
ΣH = 0 
T cos ⁡θ = H 
T cos ⁡⁡θ = 8 
T sin θ ⁡= 2
∴ T² cos²θ ⁡+ T² sin^2⁡ θ = 8² + 2²
T = 8.246 kN
Tension in segment GF is 8.246 kN.

Q1: 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 ___________ .    [2020, Set-1]
Ans: 95 to 97
Sol: 
As horizontal thrust is zero so it behaves like a beam (curved beam)
M_max = wL²/8 (At crown) 
= 12 x 8² / 8 = 96 KN m

Q.1 The figure shows a two-hinged parabolic arch span L subjected to a uniformly distributed load of intensity q per unit length.

The maximum bending moment in the arch is equal to    [2017 : 1 Mark, Set-I]
(a) qL²/8
(b) qL²/12
(c) zero
(d) qL²/10
Ans.(C)
Solution:
If a two hinged or three hinged parabolic arch is subjected to UDL throughout its length, bending moment is zero everywhere.

Q.2 A three hinged parabolic arch having a span of 20 m and a rise of 5 m 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    [2010 : 2 Marks]

(a) 9.01 kN and 56.31⁰
(b) 9.01 kN and 33.69⁰
(c) 7.50 kN and 56.31⁰
(d) 2.50 kN and 33.69⁰
Ans. (A)
Solution:

Let the vertical reactions at left and right support be V_L and V_R upwards respectively.
Taking moments about right support, we get,

Let the horizontal reaction at left support be Hfrom left to right.
Taking moments about the crown from left, we get,

⇒ H = 5kN
Resultant reaction,


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