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Chapter 4 Shear Forces and Bending Moments
4.1 Introduction
Consider a beam subjected to transverse loads as shown in figure, the deflections occur in the plane same as the loading plane, is called the plane of bending. In this chapter we discuss shear forces and bending moments in beams related to the loads.
4.2 Types of Beams, Loads, and Reactions
Type of beams
a. simply supported beam (simple beam)
b. cantilever beam (fixed end beam)
c. beam with an overhang
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Type of loads
a. concentrated load (single force)
b. distributed load (measured by their intensity) : uniformly distributed load (uniform load) linearly varying load
c. couple
Reactions
consider the loaded beam in figure equation of equilibrium in horizontal direction Fx = 0 HA - P1 cos = 0
HA = P1 cos MB = 0 - RA L + (P1 sin ) (L - a) + P2 (L - b) + q c2 / 2 = 0
(P1 sin ) (L - a) P2 (L - b) q c2
RA = CCCCCCC + CCCC + CC
L L 2 L
(P1 sin ) a P2 b q c2
RB = CCCCC + CC + CC
L L 2 L for the cantilever beam Fx = 0 HA = 5 P3 / 13
12 P3 (q1 + q2) b Fy = 0 RA = CC + CCCCC
13 2
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12 P3 q1b q1 b MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3)
13 2 2 for the overhanging beam MB = 0 - RA L + P4 (L– a) + M1 = 0 MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1
RA = CCCCCC RB = CCCC
L L
4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a concentrated load P applied at the end
A, at the cross section mn, the shear force and bending moment are found Fy = 0 V = P M = 0 M = P x sign conventions (deformation sign conventions) the shear force tends to rotate the material clockwise is defined as positive the bending moment tends to compress the upper part of the beam and elongate the lower part is defined as positive
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Example 4-1 a simple beam AB supports a force P and a couple M0, find the shear V and bending moment M at
(a) at x = (L/2)_ (b) at x = (L/2)+
3P M0 P M0 RA = C - C RB = C + C
4 L 4 L
(a) at x = (L/2)_ Fy = 0 RA - P - V = 0
V = RA - P = - P / 4 - M0 / L M = 0 - RA (L/2) + P (L/4) + M = 0
M = RA (L/2) + P (L/4) = P L / 8 - M0 / 2
(b) at x = (L/2)+ [similarly as (a)]
V = - P / 4 - M0 / L
M = P L / 8 + M0 / 2
Example 4-2 a cantilever beam AB subjected to a linearly varying distributed load as shown, find the shear force V and the bending moment M q = q0 x / L Fy = 0 - V - 2 (q0 x / L) (x) = 0
V = - q0 x2 / (2 L)
Vmax = - q0 L / 2
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M = 0 M + 2 (q0 x / L) (x) (x / 3) = 0
M = - q0 x3 / (6 L)
Mmax = - q0 L2 / 6
Example 4-3 an overhanging beam ABC is supported to an uniform load of intensity q and a concentrated load P, calculate the shear force V and the bending moment M at D from equations of equilibrium, it is found
RA = 40 kN RB = 48 kN at section D Fy = 0 40 - 28 - 6 x 5 - V = 0
V = - 18 kN M = 0
- 40 x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0
M = 69 kN-m from the free body diagram of the right-hand part, same results can be obtained 4.4 Relationships Between Loads, Shear Forces, and Bending Moments consider an element of a beam of length dx subjected to distributed loads q equilibrium of forces in vertical direction
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Fy = 0 V - q dx - (V + dV) = 0 or dV / dx = - q integrate between two points A and B
B B
∫ dV = -∫ q dx
A A
i.e. B
VB - VA = -∫ q dx
A
= - (area of the loading diagram between A and B) the area of loading diagram may be positive or negative moment equilibrium of the element M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0 or dM / dx = V maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0 integrate between two points A and B
B B
∫ dM = ∫ V dx
A A
i.e. B
MB - MA = ∫ V dx
A
= (area of the shear-force diagram between A and B) this equation is valid even when concentrated loads act on the beam between A and B, but it is not valid if a couple acts between A and B
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concentrated loads equilibrium of force
V - P - (V + V1) = 0 or V1 = - P
i.e. an abrupt change in the shear force occurs at any point where a concentrated load acts equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0 or M1 = P (dx/2) + V dx + V1 dx j 0 since the length dx of the element is infinitesimally small, i.e. M1 is also infinitesimally small, thus, the bending moment does not change as we pass through the point of application of a concentrated load loads in the form of couples equilibrium of force V1 = 0
i.e. no change in shear force at the point of application of a couple equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0 or M1 = - M0 the bending moment changes abruptly at a point of application of a couple 8
4.5 Shear-Force and Bending-Moment Diagrams concentrated loads consider a simply support beam AB with a concentrated load P
RA = P b / L RB = P a / L for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L note that dM / dx = P b / L = V for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L note that dM / dx = - P a / L = V with Mmax = P a b / L uniform load consider a simple beam AB with a uniformly distributed load of constant intensity q
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RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2 note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2 several concentrated loads for 0 < x < a1 V = RA M = RA x
M1 = RA a1 for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1) similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4 construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L for 0 < x < a
V = RA M = RA x
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for a < x < a + b
V = RA - q (x - a)
M = RA x - q (x - a)2 / 2 for a + b < x < L
V = - RB M = RB (L - x) maximum moment occurs where V = 0
i.e. x1 = a + b (b + 2c) / 2L
Mmax = q b (b + 2c) (4 a L + 2 b c + b2) / 8L2 for a = c, x1 = L / 2
Mmax = q b (2L - b) / 8 for b = L, a = c = 0 (uniform loading over the entire span)
Mmax = q L2 / 8
Example 4-5 construct the V- and M-dia for the cantilever beam supported to P1 and P2
RB = P1 + P2 MB = P1 L + P2 b for 0 < x < a
V = - P1 M = - P1 x for a < x < L
V = - P1 - P2 M
= - P1 x - P2 (x - a)
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Example 4-6 construct the V- and M-dia for the cantilever beam supporting to a constant uniform load of intensity q
RB = q L MB = q L2 / 2 then V = - q x M = - q x2 / 2
Vmax = - q L Mmax = - q L2 / 2 alternative method x V - VA = V - 0 = V = -∫ q dx = - q x
0
x
M - MA = M - 0 = M = -∫ V dx = - q x2 / 2
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Example 4-7 an overhanging beam is subjected to a uniform load of q = 1 kN/m on AB and a couple M0 = 12 kN-m on midpoint of BC, construct the V- and M-dia for the beam
RB = 5.25 kN RC = 1.25 kN shear force diagram
V = - q x on AB
V = constant on BC
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bending moment diagram
MB = - q b2 / 2 = - 1 x 42 / 2 = - 8 kN-m the slope of M on BC is constant (1.25 kN), the bending moment just to the left of M0 is
M = - 8 + 1.25 x 8 = 2 kN-m the bending moment just to the right of M0 is
M = 2 - 12 = - 10 kN-m and the bending moment at point C is
MC = - 10 + 1.25 x 8 = 0 as expected