EXAMPLE 6.12
The simply supported beam in Fig. 6–26a has the cross-sectional area shown in Fig. 6–26b. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.
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EXAMPLE 6.12 (cont.)
• The maximum internal moment in the beam, 22.5 kNm, occurs at the center.
• By reasons of symmetry, the neutral axis passes through the centroid C at the mid-height of the beam, Fig. 6–26b.
Solutions
2
3 2 31 112 12
6 4
3
max max 6
2 0.25 0.02 0.25 0.02 0.16 0.02 0.3
301.3 10 m
22.5 10 0.17; 12.7 MPa (Ans)
301.3 10
I I Ad
McI
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<Parallel-Axis Theorem>
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2 2
2 2
2 2
2'
2'
( ' )
( ' 2 ' )
' 2 '
2 '
y y
y y
y y
y y y
y y
I z dA z d dA
z z d d dA
z dA d z dA d dA
I d z A Ad
I Ad
0
z
y0
z'
y'
dz
dy
z'
y' dA
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' '
' '
( ' )( ' )
( ' ' ' ' )
' ' ' '
' '
yz z y
z y y z
z y y z
y z z y y z
y z y z
I yzdA y d z d dA
y z z d y d d d dA
y z dA d z dA d y dA d d dA
I d z A d y A Ad d
I Ad d
0
2 2
2 2
2 2
2'
2'
( ' )
( ' 2 ' )
' 2 '
2 '
z z
z z
z z
z z z
z z
I y dA y d dA
y y d d dA
y dA d y dA d dA
I d y A Ad
I Ad
0
0
EXAMPLE 6.12 (cont.)
• The maximum internal moment in the beam, 22.5 kNm, occurs at the center.
• By reasons of symmetry, the neutral axis passes through the centroid C at the mid-height of the beam, Fig. 6–26b.
Solutions
2
3 2 31 112 12
6 4
3
max max 6
2 0.25 0.02 0.25 0.02 0.16 0.02 0.3
301.3 10 m
22.5 10 0.17; 12.7 MPa (Ans)
301.3 10
I I Ad
McI
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 6.12 (cont.)
• A three-dimensional view of the stress distribution is shown in Fig. 6–26d.
• At point B,
Solutions
MPa 2.11103.301
15.0105.22 ; 6
3
BB
B IMy
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=0.01592 m
6.5 UNSYMMETRICAL BENDING
Moment applied along principal axis
dAyMM
dAzMM
dAFF
AZZR
AyyR
AxR
0 ;
0 ;
0 ;
If y and z are the principal axes, (The integral is called the product of inertia)
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0 yzyzdA I
(1) DOUBLY SYMMETRIC BEAMS with INCLINED LOADS
• Moment arbitrarily applied
yzx
z y
M zM yI I
= +
= +
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• Orientation of neutral axis:**
*
*
0
tan tan
yzx
z y
yz z
y z y
M zM yI I
MI Iyz I M I
Occurs at the point farthest from the neutral axis.max
(2) ARBITRARY-AXIS METHOD
0 1 2 X a a y a z
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0 1 2 0
21 2 1 2
21 2 1 2
( ) 0 0, 0
( )
( )
X
y X yz y
z X z yz
F x dx a dA a ydA a zdA y z a
M x z dx a yzdA a z dA a I a I
M x y dx a y dA a yzdA a I a I
0
1 22 2
2
0( ) ( )
,
( ) ( )
z y y yz y z z yz
y z yz y z yz
z y y yz y z z yzx
y z yz
aM I M I M I M I
a aI I I I I I
y M I M I z M I M II I I
* *
2
*
*
( ) ( )0
tan
z y y yz y z z yzx
y z yz
y z z yz
z y y yz
y M I M I z M I M II I I
M I M Iyz M I M I
• Orientation of neutral axis:
occurs at the point farthest from the neutral axis.max
(3) PRINCIPAL-AXIS METHOD
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2
( ) ( )z y y yz y z z yzx
y z yz
z y y z yzx
y z z y
y M I M I z M I M II I I
yM I zM I M zM yI I I I
• If product of inertia Iyz=0, y and z axes are called “principal axes of inertia.”
• Transformation of moments of inertia
' ' ' '2' '
' ' ' '2' '
' '' '
cos 2 sin 22 2
cos 2 sin 22 2
sin 2 cos 22
y z y zy y z
y z y zz y z
y zyz y z
I I I II z dA I
I I I II y dA I
I II yzdA I
' '
' '
2' ' ' ' 2
max,min ' '
tan 2
2
2 2
y zP
y z
y z y zy z
II I
I I I II I
