stresses in beams
DESCRIPTION
stress in beamsTRANSCRIPT
3.1
Chapter 3
STRESSES IN LOADED BEAMS
3.1 PURE BENDING
3.2
3.3
Points to note:
it is called the:
note:
3.4
For a fibre JK the new length L’:
(the change in length of JK)
and the max numerical strain is given by: m
(i) Strain Due to Bending
3.5
3.6
(ii)
3.7
Common I values:
rectangular X-section, width b, height h:
Ibh
z
3
12
SI
C C
h
2
bh2
6
Circular X-section, radius r, diameter d:
Ir d
4 4
4 64
SI
C C = r
32
,4
33 dr
3.8
Example 1
Consider a beam of rectangular X-section with
load of 5 kN/m (take E = 175 GPa) determine
(a) Max. tensile and compressive stress at mid-span
(b) normal stress and strain at A
(c) radius of curvature at B.
I bh
m
of section 1
123
008 01212
1152 103
6 4. .
.
A
D B
C
Note: A is 20 mm from NA, B is on NA, D is on top fibre, E is on bottom fibre
C is on NA
3.9
(a) At mid-span, B.M. = ?
MxzkN m
10 2 522
210 .
Using I
My
x
At top fibre MPax
1.5261052.11
)06.0(31010
At bottom fibre x = 52.1 MPa
Plotting the stress distribution
3.10
Note: A positive bending moment causes compressive (-ve)
axial stresses above the neutral axis and tensile (+ve)
stresses below the N.A.
- Stress distribution is linear.
- Max. stresses are induced at top and bottom fibre i.e. fibre
furthest from N.A.
(b) At a section thro’ A
B M
KN m
. . ( )( )
. .
10 15 1
2
7 5
2
We have
AAMy
I
MPa
7 5 10 0 02
1152 10
13
3
6
. .
.
A
A
E
13 10
175 10
74 3 10
6
9
6.
3.11
(c) Since
m
y
y
x
x
269
103.74
02.06
Note:
ρ is radius of curvature of the neutral axis
x
y
(a)
3.12
Example 2
An overhanging beam of T-shaped cross-section is loaded as shown
in the Fig. Determine the max. tensile and compressive bending
stresses.
X-sectional area is divided into A1 & A2. Let distance of centroid
from bottom = .y
y A A A y A y
y
mm
1 2 1 1 2 2
20 70 60 30
20 60
50
(60) (20)
(60) (20)
Example 2
3.13
Ibh
Ad
mm
3
2
3 2
3 2
4 4
12
1
1260 20 20 60 20
1
1220 60 20 60 20
136 10
R1
= 5 kN. R2
= 10 kN
A1
A2
20
20
From equilibrium
If you have a distributed load that does not act
along the whole length of the beam, you have to
introduce another distributed load of equal
magnitude but acting in the opposite direction
beyond the right end limit of the given distributed
load.
Equivalent System
0 a Lx
0 a Lx
Represented by
W<x0>0
Represented by
W<xa>0
W
3.14
kNm
(1)
Load Intensity
Shear force:
11001433103445
xxxxxxw
kNxxxxx
wdxF
000433103445
F = -5 + 12 -10 = -3 kN
3.15
Bending Moment:
000433103445 xxxxx
FdxM
(2)
To locate position of maximum bending moment
Consider 0 < x < 3
Max. B.M. occurs at positions of zero shear force
For zero shear force, 0 = -5 + 4x, x = 1.25 m
kNxxxxxF000
433103445
M Fdx
xx x
x x kNm
5 42
43
210 3 3 4
2 2
.
Note: From the shear force plot it is noted that zero
shear force occurs at x < 3
(B.M.)
3.16
Hence the Bending Moments are :
M
kN m
5 3 43
2
3
2
.
M xx
kN m
5 42
3125
2
. .
At x = 1.25 m
At x = 3 m
Note: x < 3 m
My
Ilargest bending stress occurs at max. y (top and bottom fibre)
i.e. at y = 30 mm, -50 mm.
3.17
max. ( ).
/
.
top fibre N mm
MPa
3125 10 30
136 10
68 9
6
2
max. ( )
.
.
bottom fibre
MPa
3125 10 50
136 10
114 9
6
4
max. ( ) /
.
top fibre N mm
MPa
3 10 30
136 10
66 2
6
4
2
c bottom fibre
MPa
( )
.
3 10 50
136 10
110 3
6
4
Hence At x = 1.25 m
At x = 3 m
Hence
max. tensile stress occurs at x = 1.25 m
max. comp. stress occurs at x = 3 m.
i.e. Stresses at locations of zero shear must be investigated
Check
1 m 3 m
A D
Fxy
kN
7
B
-3
1.25 m
-5
x
Check for values of maximum
bending stresses here
3.18
3.2 SHEAR STRESSES IN BEAMS
Mxz
X
dx
Mxz
Mxz
+ (dMxz
/dx)dx
Stresses in a beam
Mxz + δMxz
3.19
Stresses in a Beam
P
Fxy
Mxz
σxx comp top, tensile bottom
xygives
P
National University of SingaporeNational University of SingaporeNational University of Singapore
Bending moment induces normal bending stresses in the direction
of the beam axis. Shear force induces shear stresses.
3.20
Pure bending without shear induces only normal stresses
in the direction of the beam axis. When a shear force is
present (i.e. the bending moment varies along the length
of the beam), shear stresses are also induced. (In practice,
it is very uncommon to encounter pure bending in a
beam).
Consider an elemental length of a beam where the shear
force is constant but there is a variation in the bending
moment. e.g. a simply-supported beam with a central
point load.
3.21
Consider the portion of the beam element above a vertical
distance y from the centroid of the cross-section; i.e. we
have made an imaginary horizontal cut at y and chosen
the upper element which has a surface exposed by the cut.
For equilibrium in the axial direction;
xx yxA
xxA
dA F dA1 2 0
Using xxxz
z
M
Iy
3.22
F
M M y
IdA
M
Iy dAyx
xz xz
zA
xz
zA
Az
xzyx dAy
I
MF
Dividing by x and letting x 0
A
xz
z
yxdAy
dx
dM
Idx
dF 1
Now y dA A yA
(first moment of area A about the Z-axis)
A - area of the cross-section isolated by the horizontal
cut; i.e. above the location of the shear stress being
determined (i.e. above y)
y - vertical distance between the centroidal axis and the
CG of A
Since dM
dxF
xz
xy
dF
dx
F
IA y
yx xy
z
-this term is known as the
‘shear force per unit
length’ or ‘shear flow’
3.23
yAI
F
b
dx
dF
b
z
xy
yxxy
.1
.1
dx
dF
b
Area
yx
yx
.1
forceshear
xyyx Since
We have,
shear stress
3.24
Example 1
Determine the shear stress distribution in a beam of
rectangular cross-section (b x h) subjected to the loading
shown below.
Consider the cross-section at y-y.
3.25
xy yx
xy
z
F A y
I b
F Pxy at section y-y
Ahy b
2
y
hyy
22
1
I bhz 1
123
b b
bbh
yh
ybyh
P
xy3
12
1
22
1
2
xyP
bh
hy
6
23
2
2
At y = 0, bh
Pxy
2
3max
3.26
The shear stress distribution is parabolic; maximum at the
centroidal axis and zero at the top and bottom. (Contrast
this with the normal stress distribution caused by bending,
where the maximum stresses occur furthest from the
centroidal axis).
3.27
3.3 RELATIVE MAGNITUDES OF BENDING
AND SHEAR STRESSES
For a rectangular cross-section (bxh) simply-supported beam
with a central point load,
The maximum bending moment and shear force (numerical
values) are:
M P
LPL
F P
xz
xy
(max)
(max)
12 2
14
12
Max. bending stress:
xx
xz
z
M
Iy
PL
bh
h(max)
(max)
max
141
122
3
xxPLbh(max)
32 2
3.28
Max. shear stress:
Using the previous example with y = 0 (for max. shear
stress) and P replaced by 12P;
xyPbh(max)
34
xx
xy
Lh
(max)
(max) 2
Hence, if the beam length is much greater than its depth
(e.g. L 10h), the maximum shear stress will be at least
an order of magnitude smaller than the max. bending
(normal) stress.
3.29
Example 1
A box beam is loaded as shown in the Fig., I about N.A. =
10.5 x 10-6m4. Draw shear force and B.M. diagrams and
calculate
a) Shearing and bending stresses at point E,
b) Max. shear and bending stresses.
From equilibrium
RA = 3 kN, RB = 11 kN
Note: Point E is 40 mm from the top fibre
Beam cross-section
3.30
The shear force and bending moment diagrams can be
constructed as follows:
1.5
-4
7
x
Fxy
(kN)
-3
-8
6
4.5
1.5
Mxz
(kNm)
x
3.31
To determine ty
2
201008020
2
6040)6020(2204080)604080120( ty
mmy
y
t
t
7.567200
10408
101761016810647200
3
333
At pt. E Fxy = - 3 kN
yt
yt
Beam cross-section
3.32
36
39
104.117
10)207.56(4080
)207.56(04.008.0
m
m
yA
xyxyF A y
Ib
N m
MPa
3 10 117 4 10
10 5 10 2 0 02
0 839
3 6
6
2.
. ( . )
.
Bending stress, at E, Mxz = 4.5 kN.m
xxMy
I
MPa
4 5 10 56 7 40 10
10 5 10
7 16
3 3
6
. .
.
.
For max. shear stress,
Max. shear force = 7 kN , 2 x 4
3.33
A y
m
m
80 40 56 7 20 2 20 16 756 7 40
2
80 40 36 7 2 20 16 7 8 4 10
123 10
9 3
6 3
( . ) .( . )
. . .
MPa
Ib
yAFxy
xy
05.202.02105.10
101231076
63
3.34
For max. bending stress,
Max. B.M. = - 8 kN.m at x = 4
max
max
.
.
.
M y
I
MPa
8 10 63 3 10
10 5 10
48 2
3 3
6
(compression)
Note: xy (max) occurs at N.A.
xx (max) occurs at bottom fibre (at support B).