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SOLUTION 1:
We are given:
First, we find the position of centroid:
We take the reference axis as MN.
y is the distance of the C.G. of the T – section from the bottom MN.
The given T – section is symmetrical about Y – Y axis as shown in the above figure.
1a is the area of rectangle ABCD
3 2100 10 10 mm  
1y is the distance of C.G. of area 1a from bottom line MN
10
100 105
2
mm  
2a is the area of rectangle AEMN
3 2100 10 10 mm  
2y is the distance of C.G. of area 2a from bottom line MN
100
50
2
mm 
Thus, 1 1 2 2
1 2
77.5
a y a y
y mm
a a

 


Now, the second moment of area of the T – section.
The given section is symmetrical about the axis Y – Y and hence the C.G. of the section will
lie on Y – Y axis.
Let:
1IG  Moment of inertia of rectangle (1) about the horizontal axis and passing through C.G.
2IG  Moment of inertia of rectangle (2) about the horizontal axis and passing through C.G.
of rectangle (2).
1h  The distance between C.G. of the given section and C.G. of the rectangle (1).
1 1
105 77.5
27.5
h y y
mm
 
 


2h  The distance between C.G. of the given section and C.G. of the rectangle (2).
2 2
77.5 50
27.5
h y y
mm
 
 


3
4
1
100 10
8333.33
12
IG mm

 
3
4
2
10 100
33333.33
12
IG mm

 
Using theorem of parallel axis,
For Rectangle (1)   2 3 2 41 1 1 8333.33 10 27.5 764583.33IG a h mm    
For Rectangle (2)   3 2 433333.33 10 27.5 1589583.333mm  
Now,
4
764583.33 1589583.33
2354166.663
zz
zz
I
I mm
 


The moment of inertia of the given section about the vertical axis passing through the C.G. of
the given section:
3 3
4
10 100 100 10
12 12
841666.6667
yyI
mm
 
 


Now, we will find the maximum compressive and tensile stress.
Maximum bending moment at support
3 230 10 2.5
93750 .
2
N m
 
 
From Bending equation:
max
max
max
fM
I y
M Z f

 

Polar moment of inertia  xx zz yyI I I
Therefore, 4319583333xxI mm
Tensile stress top
M
y
I

Substituting the values, tensile stress  
93750
52.5 1000 9.53
319583333
MPa   
Compressive stress bottom
M
y
I

Substituting the values, tensile stress ...