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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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