LEC 06 Shear Stresses for Beams in Bending(pdf) - KFUPM Open ...

Chapter 4
Load and Stress Analysis
Lec. 6
Shear Stresses for Beams in Bending
A. Bazoune

4.8 Shear Stresses for
Beams in Bending
Most beams have both shear forces and bending moment
present.
It is only occasionally that we encounter beams subjected
to pure bending, that is to say having zero shear force.
to pure bending, that is to say having zero shear force.
The flexural formula was developed on the assumption of
pure bending in order to eliminate complicating effects of
shear force.
For engineering purposes, the flexure formula is valid no
matter whether the shear force is present or not.

Fig. 4.20 shows
a beam of
constant cross
section
subjected to a
shear force V
and a bending
moment M
Figure 4-20 Beam section isolation.
Note: Only forces in x-direction are
shown on the dx element.

Associate the hollow
vector with your
right hand.
If you place the
thumb of your right
hand in the negative
z- direction, then
your fingers, when
bent, will indicate
the direction of
moment M.
Figure 4-20 Beam section isolation.
Note: Only forces in x-direction are
shown on the dx element.

Shear force and
bending moment are
related by:
V
=
dM
dx
(a)
Figure 4-20 Beam section isolation.
Note: Only forces in x-direction are
shown on the dx element.

At some point
along the beam,
we cut transverse
section dx at a
distance y 1 above
the neutral axis.
We remove this
section to study
the forces that act
on it.
Figure 4-20 Beam section isolation.
Note: Only forces in x-direction are
shown on the dx element.

Because shear
force is present,
the B.M. is
changing as we
move along the x-
axis.
We can designate
the B.M. as M on
the Near Side of
the section and as
(M + dM) on the
Far Side.
Figure 4-20 Beam section isolation.
Note: Only forces in x-direction are
shown on the dx element.

The moment M
produces a normal
stress σ, and the
moment (M + dM),
a normal stress
(σ + dσ).
These normal
stresses produce
normal forces on
the vertical faces of
the element.
The compressive force on the far side being
greater than on the near side.
The Resultant of these two would cause the
section to tend to slide in the negative x-
direction.

This resultant must
be balanced by a
shear force acting
in the positive x-
direction on the
bottom of the
section.
This shear force
results in a shear
stress.
For the near face
F
N
=∫
c
y
1
σ dA
(b)
where the limits indicate that we integrate from y = y 1 to y = c.

σ = M y I
For ,
Eq. (b) becomes
F
N
M
=
I
∫
c
I ∫ ( )
y
1
(b)
y dA
(c)
For the Far side
c M+
dM c
F = ∫ σ+ dσ
dA = ∫ y dA
y1 I y1
(d)

The force on the
bottom face is the
shear stress τ times
the area of the
bottom face
FB
=τ bdx
(e)

In this equation, the integral
is the first moment of the
area of the isolated vertical
face about the neutral axis.
∫
c
y y dA
1
=
Q

The moment is usually designated Q
Q
∫ y dA=
= c y
1
y'
A'
(4-30)
where, for the isolated area from y 1 to c, is the distance from the
y-axis to the area centroid and A’ is the area.
Finally, we write
V Q
τ =
(4-31)
I b

Shear Stresses in Standard-Section Section
Beams
The shear stress distribution in a beam depends on how Q/b
varies as a function of y 1.
Fig. 4-22 shows a portion of a beam with a rectangular crosssection.
It is subjected to shear force V and a bending Moment M.

Figure 4-22
Shear Stress in a rectangular beam

A normal stress σ is developed on a cross-section such as A-A due to M.
Section A-A is in compression above the N.A. and in tension below.

Consider an element of area dA at a distance
y above the N.A. such that
Eq. (4-30) becomes
dA = bdy
2
c
c by b
Q y dA b y dy ⎡ ⎤
= ∫ = c y
y ∫ =
1 y
⎢ = −
1 2
⎥
⎣ ⎦ 2
c
y1
2 2
( 1 )

Substitute this value into Eq.(4-31)
From Table A-18 and for a rectangular cross-section, we have
Substitute A = bh and h = 2c
gives
Using this value for I in Eq. (4-32) gives
V
(
2 2
τ = c − y )
(4-32)
1
2I
I =
Ac
3
3V
⎛
τ = ⎜1−
2A
⎝
2
I =
bh
12
(b)
(4-33)
The maximum shear stress exists when y=0, which is at the bending
neutral axis. Thus
3V
τ
max
=
2A
y
c
2
1
2
⎞
⎟
⎠
3
(4-34)

As we move away from the neutral axis, the shear stress
decreases parabolically until it is zero at the outer surface
where y = ± c .
Notice that the shear
stress is maximum at
the bending neutral
axis, where the
normal stress due to
bending is zero and
that the shear stress
is zero at the outer
surfaces, where the
bending stress is a
maximum.

Particular Case:
Shear Stresses in Standard Section Beams

Example 4-7 Textbook
A beam 12 in long is to support a load of 488 lb.f acting 3 in from the left
support as shown in Fig- 4-21
a. Basing the design only on bending stress, a
designer has
selected a 3-in
column
channel with
the cross-sectional
sectional
dimensions as shown. If the direct shear is neglected, the stress beam may be
actually higher than the designer thinks. DETERMINE the principal stresses
considering bending and direct shear and compare then with that considering
bending only.
Figure 4-21.

Example 4-7 Cont’d
The loading, shear force and
bending moment diagram are
shown in Figure4-21(b)
(b).
If the direct shear force is
included in the analysis, the
maximum stresses at the top
and bottom of the beam will
be the
same
as
if only
bending were
considered.
The maximum stresses are:
Mc
σ = ±
I
1098( 1.5)
= ± = ± 992psi
1.66
Figure 4-21 (b).

Example 4-7 Cont’d

Example 4-7 Cont’d

y Q s t s ,s 1 2
in
in 3
psi
psi
psi
0 0.653 0 847 847, -847
0.25 0.648 165 840 762, -928
0.5 0.631 331 818 670, -1000
0.75 0.605 496 785 575, -1071
1 0.568 661 737 477, -1138
1.227 0.525 812 681 387, -1200
1.5 0 992 0 0, -992