4. Shear Forces And Bending Moments
4.1 Introduction
4.2 Types of Beams, Loads, and Reactions
4.3 Shear Forces and Bending Moments
4.4 Relationships Between Loads, Shear Forces, and Bending Moments
4.5 Shear-Force and Bending-Moment Diagrams
Chapter Summary & Review
Problems
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4.4 Relationships Between Loads, Shear Force, and Bending Moment
Fig. 4-10 Element of a beam used in deriving the relationships between loads, shear forces and
bending moments (all loads and stress resultants are shown in their positive directions.)
■ Distributed Load(Fig 4-10(a))
Shear Force.
Equilibrium of forces
in the vertical direction (upward forces are positive)
gives as follows.
,
(4-4)
the rate of change of the shear force =
the negative of the intensity of the distributed load
if there is no distributed load
If q=0 , then
if the distributed load is uniform
If q=constant , then
= constant
and shear force changes linearly in that part of the beam.
Fig. 4-8 Example 4-2. Shear
force and bending moment in a
cantilever beam.
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● Example
Taking the derivative
from
gives
𝑑𝑉
= −𝑞
𝑑𝑥
(a)
(4-5)
= - (Area of the loading diagram between A and B)
Bending Moment.
Let’s now consider
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the moment equilibrium in Fig. 4-10a.
Summing moments
Discarding products of differentials
𝑑𝑀
=𝑉
𝑑𝑥
the rate of change of the bending moment = the shear force
(4-6)
● Example
Again using the cantilever beam of Fig.4-8
the bending moment :
the derivative :
𝑑𝑀
=𝑉
𝑑𝑥
from
Integrating
between two points A and B
(4-7)
=(area of the shear-force diagram between A and B)
■ Concentrated Loads(Fig. 4-10(b))
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Now let’s consider
a concentrated load P acting on the beam element(Fig. 4-10(b))
From equilibrium of forces in the vertical direction,
we get
𝑉1 = −𝑃
(4-8)
the shear force decreases by the magnitude of the downward load P.
From equilibrium of moments about the left-hand face of the element(Fig. 4-10(b)),
we get
Since the length dx of the element is infinitesimally small,
the increment M1 is also infinitesimally small.
Thus,
the bending moment does not change.
At the left-hand side
the rate of change of the bending moment :
𝑑𝑀
=𝑉
𝑑𝑥
At the right-hand side
𝑑𝑀
= 𝑉 + 𝑉1 = 𝑉 − 𝑃
𝑑𝑥
It can be seen that
the concentrated load doesn’t affect the change of moment.
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■ Loads in the form of couples (Fig.4-10(c))
From equilibrium of the element
in the vertical direction
: the shear force doesn’t change
From equilibrium of moments about the left-hand side of the element
gives
Disregarding terms that contain differentials
(4-9)
It can be seen that
the bending moment does affect the change of moment.
The bending moment changes abruptly by 𝑀0
4.5 Shear-force and bending-moment diagrams
Diagrams showing the variation of N,V,M are very useful.
Because these diagrams quickly identify
locations and values of maximum N, V, M needed for design.
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■ Concentrated loads
Fig. 4-11 Shear-force and bending moment diagrams for a simple beam with a concentrated load.
with a concentrated load.
1)
,
(4-10 a, b)
(
)
(4-11a,b)
(
)
(4-12a)
(4-12b)
(4-13)
from
=0
the slope is zero
=V
the slope is equal to V
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The maximum shear forces and bending moments in a beam may occur at the following place s:
A cross section where a concentrated loads is applied and
the shear force changes sign
A cross section where the shear force equals zero.
A point of support where a vertical reaction is present
A cross section where a couple is applied
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■ Uniform load
Fig. 4-12 Shear-force and bending moment diagrams
for a simple beam with a uniform load.
=
=
(4-14b)
The slope is -q
𝑞𝐿
− 𝑞𝑥 = 𝑉
2
the slope is V
The maximum moment occurs
where the shear force equals zero.
(4-15)
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(4-14a)
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■ Several Concentrated loads
,
( 0 < 𝑥 < 𝑎1 )
(4-16a b)
(𝑎1 < 𝑥 < 𝑎2 ) (4-17a b)
Fig. 4-13 Shear-force and bending moment diagrams for a simple beam with several
concentrated loads.
(4-20a,b,c)
(4-18b)
(𝑎3 < 𝑥 < 𝐿 )
(4-18b)
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𝑥 = 𝑎1 , 𝑥 = 𝑎2 , 𝑥 = 𝑎3
(𝑎2 < 𝑥 < 𝑎3 )
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(4-18a)
■ Ex 4-4
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Draw the shear-force and bending-moment diagrams for a simple beam
with a uniform load of intensity q acting over part of the span(Fig. 4-14a).
Fig. 4-14 Example 4-4. Simple beam with a uniform load over part of the span.
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Solution
1) Reaction.
2)
( 0<𝑥<𝑎
)
(4-21a,b)
(4-22a,b)
(
(
)
)
(4-23a,b)
(4-24a,b)
3) maximum bending moment occurs
where the shear force equals zero
from V = 0
*(
)
(4-25)
This point can be found by setting the shear force V equal to zero and
solving for the value of x1.
☆ Special case :
If
,
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Now we substitute
----->
(4-26)
from (4-25) and (4-26)
(4-27a,b)
■ Example 4-5
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Draw the shear-force and bending-moment diagrams for a cantilever beam
with two concentrated loads(Fig. 4-15a)
1)
Fig. 4-15 Example 4-5. Cantilever beam with two concentrated loads
(a,b)
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Solution
2)
( 0 < 𝑥 < 𝑎)
(c,d)
( 𝑎 < 𝑥 < 𝐿)
(e,f)
■ Example 4-6
A cantilever beam supporting a uniform load of constant intensity ｑ
is shown in Fig. 4-16a.
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Draw the shear-force and bending-moment diagrams for this beam.
Fig. 4-16 Example 4-6. Cantilever beam with a uniform load.
(4-28a,b)
2)
(4-29a,b)
3)
𝑉𝑚𝑎𝑥 = −𝑞𝐿
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Solution
1)
(4-30a,b)
☆¤
(g)
(h)
■ Example 4-7
A beam ABC with an overhang at the left-hand end is shown in Fig. 4-17a.
The beam is subjected to a uniform load of intensity 𝑞 = 1.0 𝑘/𝑓𝑡 on the overhang AB
and a counterclockwise couple 𝑀0 = 12.0 𝑘 − 𝑓𝑡
acting midway between the supports at B and C.
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Draw the shear-force and bending-moment diagrams for this beam.
Fig. 4-17 Example 4-7. Beam with an overhang.
Solution
1)
𝑅𝐵 = 5.25 𝑘
2)
𝑅𝐶 = 1.25 𝑘
𝑞𝑏2 1
𝑀𝐵 = −
= (1.0 𝑘/𝑓𝑡)(4.0𝑓𝑡)2 = −8.0 𝑘 ∙ 𝑓𝑡
2
2
The bending moment just to the left of the couple 𝑀0 is
−8.0 𝑘 ∙ 𝑓𝑡 + 1.25 𝑘 8.0 𝑓𝑡 = 2.0𝑘 ∙ 𝑓𝑡
The bending moment just to the right of the couple 𝑀0 is
2.0𝑘 ∙ 𝑓𝑡 − 12.0𝑘 ∙ 𝑓𝑡 = −10.0𝑘 ∙ 𝑓𝑡
The bending moment at the support C is as expected.
−10.0𝑘 ∙ 𝑓𝑡 + 1.25 𝑘 8.0 𝑓𝑡 = 0