# Tutorial 5 Solutions

Urban and Civil

Nov 15, 2013 (4 years and 6 months ago)

143 views

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

2) Cantilever with two l

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

= 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

= 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

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

= 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

= 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