Post on 27-Oct-2020
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Strength of Materials
(GATE Aerospace and GATE Mechanical) By
Mr Dinesh Kumar
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Shear Forces and Bending Moments
A slender structural member is usually classified according to the types of loads, they support. For example, a beam is a type of structural member subjected to transverse loads, that is, forces or moments having their vectors perpendicular to the axis of the bar.
Type of supports :-
o Fixed support
o Hinge support or pinned support
→ u=v = 0
→ 휃 = 0
→ Force ≠0
→ Moment ≠0
→ Shear stress ≠0
→ Bending stress ≠0
→ u=v = 0
→ 휃 ≠ 0
→ Force ≠0
→ Moment =0
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o Roller support
Types of Beams:-
o Simply supported Beam
→ v = 0
→ u ≠ 0
→ 휃 ≠ 0
→ Force ≠0
→ Moment =0
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o Cantilever beam
o Beam with overhang
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o Propped Cantilever beam
o o Fixed beam
o Continuous beam
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Type of load o Concentrated load
o Uniformly distributed load (UDL)
o Uniformly varying load (VDL)
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o Concentrated moment
]
Sign convention of shear force and bending moment
Consider the sign conventions for shear forces and bending moments.
According to above figure, a positive shear force acts clockwise against the material and a negative shear force acts counterclockwise against the material. Also, a positive bending (sagging) moment compresses the upper part of the beam and a negative bending (hogging) moment compresses the lower part.
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Relationship between loads, shear force and bending moments Consider a small section of beam of length dx as shown in the figure, it is subjected to a udl of intensity of q(x) as shown. Free body dia for shear forces and bending moment as shown.
o Force equilibrium In Vertical direction ∑F = 0 q(x)dx – V + (V+dV) = 0 q(x)dx + dV = 0 q(x) = - 풅푽
풅풙
o Moment equilibrium
Taking Moment equilibrium about z axis passing through CG of the section
∑M = 0
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Vdx/2 + M + (V+ dV) dx/2 – M – dM = 0 2 Vdx + (dV dx)/2 – dM = 0 (here 푑푣푑푥 is very small) Vdx – dM = 0 V = 풅푴
풅풙
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Shear- force and bending moment diagrams o Cantilever beam
Cantilever with an end point load
Reaction force R = P, Moment reaction M = PL
From LHS
Shear force 푃 = P (0≤x≤ 퐿)
BM 푀 = -M +Px (0≤x≤ 퐿)
= -PL+Px
= -P (L-x)
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From RHS
Shear force 푃 = P
BM 푀 = -P (L-x)
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Cantilever with UDL
Reaction force R = ql
Moment reaction M =
From LHS
Shear force = 푃 = q퐿 – qx (0≤x≤L)
BM 푀 = - + q퐿 x- (0≤x≤L)
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From RHS
Shear force = 푃 = q(L-x)
BM = - ( )
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Cantilever with point load at mid-span and UDL
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Reaction force R = qL + P
Reaction moment M = +
From LHS
Shear force 푃 = qL+ P-qx
= P+q(L-x) (0≤x≤L/2)
푃 = p+qL-qx-p
= q(L-x) (L/2≤x≤L)
Bending moment 푀 = - ( + )+(qL+P)x-
= - ( + + ) + (qL+P)x (0≤x≤L/2)
At x = L/2
푀 / = −
Bending moment 푀 = - ( + ) + (qL+P)x- –P(x-L/2)
= - ( + )+qLx
= - ( ) (L/2≤x≤L)
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Cantilever with moment at mid-span and UDL
Reaction force R = qL
Reaction momet M = -M
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Here M =
From LHS
Shear force 푃 = qL-qx
= q(L-x) (0≤x≤L)
Bending Moment 푀 = - ( -M) + qLx -
= -( + ) + M +qLx (0≤x≤L/2)
= -( + ) + M +qLx-M (L/2≤x≤L)
푀 / = M- =
푀 = -
Simply supported beam o Simply supported beam with point load
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Reaction 푅 = 푅 = p/2
From LHS
Shear force 푃 = p/2 (0≤x≤L/2)
= P/2-P
= -P/2 (L/2≤x≤L)
Bending moment 푀 = x (0≤x≤L/2)
= x – p(x-L/2) (L/2≤x≤L)
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At x = L/2
푀 =
Simply supported beam with eccentricity point load
Reaction force
푅 + 푅 = P
Moment equilibrium about A
∑푀 = 0
푅 ×L = P×q
푅 = , 푅 =
From LHS
Shear force 푃 = (0≤x≤L/2)
푃 = -P (L/2≤x≤L)
Bending moment 푀 = (0≤x≤L/2)
= - P(x-a) (L/2≤x≤L)
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At x = a
푀 =
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Simply supported beam with point moment
Reaction force 푅 + 푅 = 0
Moment equilibrium about A
∑푀 = 0
푅 ×L = M
푅 = M/L, 푅 =- M/L
From LHS
Shear force 푃 =- M/L (0≤x≤L)
BM 푀 = - x (0≤x≤a)
= - x + M (a≤x≤L)
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o Simply supported beam with UDL
Reaction 푅 = 푅 =
From LHS
Shear force 푃 = −qx (0≤x≤L)
Bending moment 푀 = 푥 −
At x = L/2
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푀 / =
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Problems
1.
Ans. A
Degree of static indeterminacy = No. of reactions – No. of equilibrium equations
From first figure,
Degree of static indeterminacy = 3-3 = 0
From second figure
Degree of static indeterminacy = 6-3 = 3
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2.
Ans. A
R + R = 0
∑M = 0 = R ×1 – 1
So, R = 1KN
R = -1 KN
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3.
Ans. B
Total load ∫푃 = ∫ sin dx =
R + R =
R = R =
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4.
Ans. A
R + R =
∑M = 0 = R ×L – ( ). .
R =
R =
From LHS
BM at midpoint of AB
푀 / = . - = Sagging
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5.
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6.
Ans. D
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R + R = 20
∑M = 0 = R × 8-20× 2 − 80
R = 40 KN
R = 20-40 = -20 KN
At section AB
M = R .x
M = 0 KN-m
M = -20×4 = -80 KN-m
At section BC
M = R .x +80
M = -20×4 + 80 = 0KN-m
M = -20×8 +80 = -80 KN-m
At section CD
M = R .x +R .(x-8) + 80
M = -20×8+40× 0 + 80 = -80KN-m
M = 0 KN-m
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7.
Ans. B
Reaction force 푅 = 10KN
Moment 푀 = 10×2 +5 = 25 KN-m
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At PQ
M = 푅 x-푀 = 10.x-25
푀 = -25KN-m
푀 = 10×2 -25 = -5 KN-m
At QR
M = 푅 x-푀 -10× (푥 − 2) +5
푀 = 0KN-m
푀 = 0KN-m
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8.
Ans. B
R + R = 0
∑M = 0 = R .L-M
R = M/L
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R = -M/L
At AB
M = -R x
M = 0
M = -M
At BC
M = -R x +R .(x-L)
M = -M
M = -M
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9.
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Ans. A
R + R = 0
∑M = M = R .L +M +2M
R = -
R =
At AB
M = R x + M
M = M
M = 2M
At BC
M = R x + M-2M
M = 0
M = M
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10.
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11.
Ans.
At BC
Reaction 푅 = P
푅 + 푅 = P
푅 = 0
At AB
Reaction 푅 = P
푀 = PL = 25× 2 = 50KN-m
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12.
Ans.
At PQ
푅 + 푅 = 10
푅 = 푅 = 5KN
At OP
BM = 푀 = 푅 .2
= 5× 2 = 10KN-m
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