2. Stress due to bending
2
Stresses in straight beam
Stresses in curved bars
3. Assumptions
a) Cross-sectional area is constant and has an
axis of symmetry (y) that is perpendicular to
the direction of the applied moment.
3
4. a) Material is homogeneous and isotropic,
and it behaves in a linear-elastic manner
4
5. a) Cross sectional plane remains plane
b) In-plane distortion of cross section is
negligible
6. • The moment is positive when it tends to straighten out
the member. Defines the location of the centroid for
the cross-sectional area. R defines the yet unspecified
location of the neutral axis, and r locates the arbitrary
point or area element dA on the cross section.
6
r
7. The strain and the stress are non-linearly function of
the radius r
7
r
r
-
R
Ek
E
/d
is
k
where
r
r
R
k
rd
r
R
8. Location of the neutral axis
A
A
A
A
A
x
r
dA
A
R
0
dA
r
dA
R
0
dA
r
r
R
k
E
dA
0
F
m
Equilibriu
Applying
8
9. Normal Stress
r
-
R
y
R),
-
r
(
e
where
y
R
Ae
My
R
r
Ar
r
R
M
R
r
EkA
r
A
RA
2
AR
Ek
M
rdA
dA
R
2
r
dA
R
Ek
dA
r
r
R
k
E
dA
y
M
0
M
m
Equilibriu
Applying
A A
A
2
2
A
A
9
11. Example 1
The curved bar has a cross-sectional area shown. If it is
subjected to bending moments of 4 kN.m, determine the
maximum normal stress developed in the bar.
11
12. Example 1 (cont.)
Solution
M= - 4 kN.m
Area (A) = 502+0.5 50 30 = 3250 mm2
The radius at the centrooid (r) =
Position of the neutral
mm
08
.
233
3250
260
30
50
5
.
0
225
50
A
A
r
~
r
2
mm
05
.
14
50
250
280
ln
30
280
50
200
250
ln
50
r
dA
dr
r
50
r
dA
...
where
mm
42
.
231
r
dA
A
R
280
250
250
200
A
A
12
13. Example 1 (cont.)
Solution
Applying the curved-beam formula to calculate the normal
stress at B and A
(Ans)
MPa
129
66
.
1
280
3250
280
42
.
231
10
4
R
r
Ar
r
R
M
MPa
116
66
.
1
200
3250
200
42
.
231
10
4
R
r
Ar
r
R
M
6
A
B
A
6
B
B
B
13
