STRUCTURAL MECHANICS 2
Tutorial 5: Bending stresses in beams
Solutions
For each of the beams below, find the maximum bending stress in the beam and the load factor against first yield assuming
that the yield stress is
Y
= 250MPa everywhere.
To find th
e maximum bending stress, it is necessary to draw the bending moment diagram, identify the moments that could
cause the maximum stress, and evaluate the bending stresses at appropriate locations. The load factor
against first yield
may be found by divid
ing the yield stress by the maximum bending stress.
A
B
7m
13m
9kN
Uniform section beam with
Z = 3.64
10
5
mm
3
A
B
1500mm
2100mm
105kN
60kN
C
Uniform section beam with
Z = 2.50
10
5
mm
3
1) Point loaded cantilever
2) Cantilever with two l
oads
1) Bending moment at A = 9*7 = 63 kNm
Bending stress is maximum at A = (M/Z) = 63*
10
6
/ 3.64
10
5
= 173 MPa
Load factor on first yield
= 250 / 173 = 1.44
2) Bending moment at B is 60 * 600 = 36
10
3
kNmm = 36
10
6
Nmm
Bending moment at A is 60 * 2100
–
105*1500 =

31.5
10
3
kNmm =

31.5
10
6
Nmm
Thus the largest bending moment is at B
Bending stress is maximum here = (M/Z) = 36
10
6
Nm
m / 2.50
10
5
= 144 MPa
Load factor on first yield
= 250 / 144 = 1.74
A
B
8m
3m
20kN/m
Uniform section beam with
Z = 4.40
10
5
mm
3
A
C
1.5m
2.1m
100kN
50kN
z
B
Non

uniform beam with linearly varying section modulus
Z = (1+z)
10
5
mm
3
with z in metres
3) Beam with distributed load
4) Non

uniform cantilever
3) Calculate the support reactions: RA = (20*8*4
–
20*3*1.5)/8 = 68.75 kN
RB = (20*11*5.5)/8 = 151.25 kN Check RA + RB = 68.75 + 151.25 = 220 kN checks
Sketch the
shear force diagram: shows that shear = 0 at z = 68.75 / 20 = 3.4375 m
Sketch the BMD
Maximum sagging bending moment is at z = 3.4375m, and is M = RA*z
–
wz
2
/2 = 68.75*3.4375

20*3.4375
2
/2 = 118.16
kNm = 118.16
10
6
Nmm
Maximum hogging bending moment is
at B, and is M =

wL
2
/2 where L = 3m =

20*3
2
/2 =

90 kNm = 90
10
6
Nmm
Thus the largest bending moment is in sagging at z = 3.4375m
Bending stress is maximum here = (M/Z) = 118.16
10
6
Nmm / 4.40
10
5
= 268.5 MPa
Load factor on first yield
= 250 / 268.5
= 0.93. This beam is unsafe: the loading will make it pass yield.
4) Sketch the shear force diagram: LHS of beam is +50kN, RHS of beam is
–
50kN.
Sketch the BMD: Value at A is 50*2.1
–
100*1.5 =

45kNm Value at B is 50*0.6 = +30kNm
Larges
t bending moment is at A, but the section modulus is larger there too.
Places that need checking are the points where peak moments occur at A and B
Section modulus at A = (1+2.1)
10
5
= 3.1
10
5
mm
3
Section modulus at B = (1+0.6)
10
5
= 1.6
10
5
mm
3
Bendin
g stress at A is (M/Z) = 45 / 3.1
10
5
mm
3
= 145.2 MPa
Bending stress at B is (M/Z) = 30 / 1.6
10
5
mm
3
= 187.5 MPa
Maximum is therefore at B = 187.5 MPa
Load factor on first yield
= 250 / 187.5 = 1.33
A
B
8m
3m
20kN/m
Uniform
beam o
f unsymmetrical section
Z
top
= 5.60
10
5
mm
3
Z
bottom
= 7.60
10
5
mm
3
Find the maximum tensile stress
and the maximum compressive stress
A
B
6m
14m
8kN/m
5m
Uniform beam with I = 78.4
10
6
mm
4
y = 200mm
5) Beam with unsymmetrical section
6) Beam with partial distributed load
5) Initial calculations are identical to (3)
Sketch the BMD
Maximum sagging bending moment is at z = 3.4375m, and is
M = RA*z
–
wz
2
/2 = 68.75*3.4375

20*3.4375
2
/2 = 118.16 kNm = +118.16
10
6
Nmm
Maximum hogging b
ending moment is at B, and is M =

wL
2
/2 where L = 3m =

20*3
2
/2 =

90 kNm = 90
10
6
Nmm
Maximum tensile stress:
Sagging in beam gives tension on bottom: M/Z = 118.16
10
6
/ 7.60
10
5
= 156.1 MPa
Hogging at B gives tension on top M/Z = 90.
10
6
/ 5.60
10
5
= 160.7 MPa
Maximum compressive stress:
Sagging in beam gives compression on: top M/Z = 118.16
10
6
/ 5.60
10
5
= 211.0 MPa
Hogging at B gives compression on bottom M/Z = 90.
10
6
/ 7.60
10
5
= 118.4 MPa
Thus the largest tension is at B and of valu
e 160.7 MPa with a load factor of
= 250/160.7 = 1.556
The largest compression is in beam and of value 211.0 MPa with a load factor of
= 250/211.0 = 1.185
6) This is the same beam as in the previous tutorial, reversed and with double the load.
Cal
culate the support reactions: RA = 8*5*(3+5/2) / 14 = 15.71 kN
RB = = 8*5*(6+5/2) / 14 = 24.29kN Check RA + RB = 15.71 + 24.29 = 40 kN checks
Sketch the shear force diagram: shows that shear = 0 at z = 6 + 15.71 / 8 = 7.964 m
Sketch the BMD
Maximum
sagging bending moment is at z = 7.964m, and is M = RA*z
–
w(z

6)
2
/2 = 15.71*7.964

8*(7.964

6)
2
/2 =
109.69
kNm = 109.69
10
6
Nmm
This is the largest bending moment
Value of Z = I/y = 88.4
10
6
/ 200 = 4.42
10
5
mm
3
Maximum bending stress is = (M/Z) =
109.69
10
6
Nmm / 4.42
10
5
= 248.2 MPa
Load factor on first yield
= 250 / 248.2 = 1.007. This beam is right at the point of first yield
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Comments 0
Log in to post a comment