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**Q.1 A long uniformly distributed load of 10 kN/m and a concentrated load of 60 kN are moving together on the beam ABCD shows in the figure (not drawn to scale). The relative positions of the two loads are not fixed. The maximum shear force (in kN, round off to the nearest integer) caused at the internal hinge B due to the two loads is______ [2019 : 2 Marks, Set-I I]**

ILD for V

Max. shear

= 70KN

For a moving concentrated load of 50 kN on the beam, the magnitude of the maximum bending moment (in kN-m) obtained at the support C will be equal to _________ . [2017 : 2 Marks, Set-I]

ILD for BM at C

Max. BM occurs at C, when 50 kN load is at B.

Max. BM at C = 50 x (-4) = -200 kNm

C.G. of system = 2.5 m from any load

For maximum bending moment, system of load should be,

Maximum BM will occur below load at C. Ordinate of ILD at C

Ordinate of ILD at D

Maximum BM = 5.935 x 3 + 3.695 x 3

= 28.89 kN-m

Maximum bending stress,**Method-II (By Statics)**

Let the section at which absolute max. BM ocurs be at x distance from left support.

Max. BM at section

For absolute max. BM,

⇒

⇒ x = 10.75 m

Absolute max BM

Absolute max bending stress**AVOID MISTAK****(I)**

Max. bending stress**(II)**

Max. bending stress

This is the maximum bending stress at mid= span and not the absolute max. bending stress in the beam.**Q.4 In a beam of length L, four possible influence line diagrams for shear force at a section located at a distance of L/4 from the left end support (marked as P, Q, R and S) are shown below. The correct influence line diagram is [2014 : 1 Mark, Set-I](a) P(b) Q(c) R(d) SAns.** (A)

(a) 30 kN

(b) 40 kN

(c) 45 kN

(d) 55 kN

Ans.

Drawing the ILD of shear force just to right of Q by using

If moving distributed load is present over span PR then we get maximum shear force just to the right of Q.

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