Stresses in Machine Elements
Lecture Number - 03
Venkat S
Mechanical Engineering
SINHGAD COLLEGE OF ENGG ,vadgaon
Strength of Materials
C-Section
1.Fig-1 shows the section of an inverted steel channel used as a
beam .The beam is simply supported over a span of 3 meters
and carries two equal concentrated loads at points distant
0.30 m from each support.
Find the value of each of these loads if the maximum tensile
stress is not to exceed 95 N/mm2.Find also the corresponding
maximum compressive stress.
Strength of Materials
Solution
Let each point load be W Newton
Each vertical reaction = W Newton
Max B.M
=M=Wx300Nmm.
Distance of the neutral axis from the top edge =
280  100  50  250  87.5  56.25
y
mm
280  100  250  87.5
1695.3

 27.67 mm
612.5
Strength of Materials
Continuation
Distance of the neutral axis from the bottom edge
=100-27.67=72.33mm
When the maximum tensile stress is 95 N/mm2 corresponding
maximum compressive stress
27.67
 c 
 95 N mm2
72.33
 c  36.34 N mm2
Strength of Materials
Continuation
Moment of inertia about the neutral axis
27.67  12.5  2 15  72.333

280  27.67
I
 250 

mm4
3
3
3
3
3
 5470100mm4
M 

I
y
Strength of Materials
Continuation
W  300
95

5470100 72.33
W
95  5470100
N  23949N  23.949 kN.
72.33  300
Strength of Materials
Example-1
• A beam ABC 6m long is supported at A and B 4m apart with an
overhang of 2m. The beam carries a uniformly distributed
load of 3 kN/m over the whole length. The proposed section
for the beam is shown in fig-4 .Determine the extreme
stresses produced in the beam.
Strength of Materials
Solution Steps
• Solution syntax:
• Taking moments about the left end A,
Vb  4  3  6  3
Vb  13.50 kN
Va  3  6  13.5  4.5 kN
Then find out:
Maximum sagging bending moment (Md)=3.375 kNm
Maximum negative bending moment (Mb)=-6 kNm
Strength of Materials
Continuation
Properties of the section :
Distance of the centroidal axis from the top edge
300  80  40  276  68  46

 18.477 mm
300  80  276  68
Find out :
Distance of the centroidal axis from the bottom edge ?
Moment of inertia of the section about the centroidal axis I ?
Maximum stresses due to maximum sagging moment :
compressive stress at top edge ?
Tensile stress at bottom edge ?
Strength of Materials
Continuation
Maximum stresses due to maximum hogging moment:
Tensile stress at the top edge ?
Compressive stress at the bottom edge ?
Strength of Materials
Example-2
• A steel channel section, ISLC 225 is used as a simply supported
beam on a span of 4m. The channel is to be designed for a
working bending stress of 100 N/mm2 . It has to carry a
uniformly distributed load on the whole span . Calculate the
permissible load when
• (i) the channel stands upright 225 mm high
• (ii) the channel lies flat with the 225 mm horizontal
• The properties of the channel ISLC 225 are,
• A=3053 mm2 Ixx =2547.9x104 mm4 , Iyy= 209.5 x 104 mm4 ,
• Cyy=24.6 mm Depth=225mm, Flange width= 90 mm
Strength of Materials
Strength of Materials
Solution Steps
(i) When the channel is placed with the 225 mm dimension
vertical
Maximum B.M that can be resisted
M 
 max
ymax
I xx  22648 Nm
Let the safe load on the beam be w N/m
w=11324 N/m
Strength of Materials
Continuation
(ii) When the channel is placed with the 225 dimension
horizontal
In this case ,maximum B.M that can be resisted
M
y
max
 max
ymax
.Iyy
 90  24.6  65.4 mm
M  3203Nm
Strength of Materials
Continuation
• Let the safe load on the beam be w N/m.
2
wl
 3203
8
w  1601.5 N / m
Strength of Materials