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Stress Analysis I ME 205 – Fall 2023
Chapter (6) - Pure Bending of Beams
Beams
Members that are slender and support loadings applied perpendicular to their longitudinal axis
are called beams (see the figure).
Objectives
➢ Determine the stress in members caused by bending
➢ Derive and understand the bending theory, its limitations and its applications for strength
based design and analysis of symmetric bending of beams
➢ Determine the normal stress developed in the symmetric bending of beams.
1
2
✓ It depends on the beam cross-section
✓ It depends on the properties of the cross section (i.e., location of centroid)
How to calculate the bending stress ?
Does the stress caused by bending is normal or shear stress ?
Cross-section
What the Bending Moment does to the Beam?
✓ It causes compression on one face and tension on the other
✓ It causes the beam to deflect
3
Double symmetric cross-sections
(Two axes of symmetry)
Single symmetric cross-sections
(One axis of symmetry)
Asymmetric cross-sections
(No axis of symmetry)
Various Beam Cross-sections
Axis of symmetry
Axis of
symmetry
Axis of
symmetry
4
At least One axis of symmetry
5
Other Loading Types
• Principle of Superposition: The normal
stress due to pure bending may be
combined with the normal stress due
to axial loading and shear stress due to
shear loading to find the complete state
of stress.
• Eccentric Loading: Axial loading which
does not pass through section centroid
produces internal forces equivalent to
an axial force and a moment.
• Transverse Loading: Concentrated or
distributed transverse load produces
internal forces equivalent to a shear
force and a moment.
Beam sign convention
The sign convention for bending moment (M) and shear force (V) in beams are shown:
6
7
Theory of Simple Bending of Straight Beams
When a bending moment is applied to a straight prismatic beam, the longitudinal lines become
curved and vertical transverse lines remain straight and undergo a rotation.
Basic Assumptions
1. Beam material is homogenous and isotropic and obeys Hook’s law,
2. Longitudinal axis x (within the neutral surface) does not experience any change in the length,
3. All cross sections remain plane and perpendicular to longitudinal axis during the deformation.
8
➢ Member remains symmetric, i.e., y-axis is the axis of symmetry,
➢ Bending moment (M) causes beam to bends uniformly to form a
circular arc,
➢ Length of top decreases and length of bottom increases,
➢ A neutral surface must exist that is parallel to the upper and lower
surfaces and for which the length does not undergo any change,
➢ In this case. stresses and strains are negative (compressive) above
the neutral plane and positive (tension) below it for the given M
direction.
Beam with a plane of symmetry in pure bending
9
( )
( )
max max
max
max
max
max
(strain varies linearly)
or
x
x
L y
L L y y
y y
L
y y
ρ
y
y
 
    
 

 

 
 
= −

= − = − − =−
= =− =−
= =
=−
Strain due to Bending of a Straight Beam
➢ A contraction will occur (-ε) in the fibers located above the
neutral axis (+y).
➢ The elongation will occur (+ε) in the fibers located below the
axis (-y).
Consider a beam segment of length L. After deformation, the length
of the neutral surface remains L’ (=JK). At other sections,
where ρ is the radius of curvature.
L
ymax
ymax
10
Stress Due to Bending
• For a linearly elastic material,
max max
max max
(stress varies linearly)
x x
y y
E E
y y
   
= =− =−
• For static equilibrium,
max
max
max max
0
0
x x
y
F dA dA ydA y
y
dA
y

 
= = = − =− → =
   
First moment with respect to neutral plane is zero. Therefore,
the neutral surface must pass through the section centroid.
• For static equilibrium,
where S= I/ymax is the section modulus (‫المقطع‬ ‫)معامل‬ of the beam.
( ) ( ) 2
max max
max
max max max
max
max
x
I
y
M y dA y dA y dA
y y y
M y M
I S
 
 

 
= − = − − = =
 
 
= =
  
x
NA
M y
I
 =− The maximum stress σmax due to bending is: max
max max
x
NA
M y
I
 =
x
dF dA

=
ymax
Hollow circular section
11
Moment of Inertia for Regular Shapes
4 4
2
4 64
R d
I
A R
 

= =
=
N. A.
C
Circular cross section
N. A.
C
( )
( )
4 4
2 2
4
o i
o i
R R
I
A R R


−
=
= −
Rectangular cross section
b
h
N. A.
h/2
3
12
bh
I
A bh
=
=
C
N. A.
3 3
12 12
o i i i
o o i i
b h b h
I
A b h b h
= −
= −
bi
bo
hi
ho
h/2
Hollow Rectangular section
C
12
Section Moduli for Regular Shapes
3
max
3
2
max
,
12 2
12
6
2
bh h
I y
bh
I bh
S
h
y
= =
= = =
For example:
The section modulus (S) for rectangular
cross section of height h and width b can
be obtained as:
13
Deformations in a Transverse Cross Section
• Deformation due to bending moment M is quantified by
the curvature of the neutral surface
max max max
max max max
1 1
1
M y M
y E y E y I EI
M
EI
 


= = = =
=
where ρ is the radius of curvature (‫االنحناء‬ ‫قطر‬ ‫)نصف‬.
14
• Stress due to eccentric loading found by superposing the uniform
stress due to a centric load and linear stress distribution due a
pure bending moment
( ) ( )
centric bending
x x x
P My
A I
  
= + = 
Eccentric Axial Loading in a Plane of Symmetry
• Eccentric loading
F P and M P d
= = 
• Validity requires stresses below proportional limit, deformations
have negligible effect on geometry, and stresses not evaluated near
points of load application.
15
Example (1): A member having the dimensions shown is used to resist an internal bending moment of
M=90kN.m. Find the maximum stress in the member if the moment is applied about the z-axis as shown. Sketch
the stress distribution over the cross-section.
Solution
The maximum normal bending stress is given by:
The moment of inertia about the z-axis can be obtained by:
max
max max
( / 2)
2
M y M h
where y h
I I
 = = =
( )
3
3
6 4
0.2 0.15
56.25 10
12 12
bh
I m
−

= = = 
3
max 6
90 10 0.15
120
2 56.25 10
MPa
 −
 
= =
 
16
Example (2): An open-link chain is obtained by bending low-carbon steel rods
into the shape shown. For 160lb load, find: (a) the maximum tensile and
compressive stresses, and (b) distance between section centroid and neutral axis.
Solution:
• Equivalent centric load and bending moment
160
160 0.6=104
P lb
M P d lb×in
=
=  = 
( )2
2 2
0.25 0.1963in
160
815
0.1963
axial
A R
P
psi
A
 

= = =
= = =
• Normal stress due to a centric load
( )
( )
4
4 3 4
1 1
4 4
max
max 3
0.25 3.068 10 in
104 0.25
8475
0.068 10
I R
M y
psi
I
 

−
−
= = = 

= = =

• Normal stress due to bending moment
17
(1) Maximum tensile and compressive stresses
( )
( )
max max
max max
815 8475
815 8475
axial
t
axial
c
  
  
= +
= +
= −
= −
( )
max 9260
t
psi
 =
( )
max 7660
c
psi
 =−
(2) Neutral axis location
3
0
815 3.068 10
0.0240
105
o
o
M y
P
A I
P I
y in
A M
−
= −
 
=  = =
0.0240
o
y in
=
Using superposition method:
----------------------------------------------------------------------------------------------------------------------------------------------------------
18
Section properties:
The moment of inertia about the neutral axis (which passed from the
centroid of the cross section) is:
The maximum normal bending stress is given by: max
max
M y
I
 =
( )( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( )
3 2 3 2
3 2 3
1 1
4 0.5 4 0.5 3.40 0.25 2 0.5 3 0.5 3 3.40 2
12 12
1
0.5 10 0.5 10 5.5 3.4 91.73 .
12
I
in
 
= + − + + −
 
 
+ + − =
( )( ) ( )( ) ( )( )
( ) ( )( ) ( )
1
1
0.25 4 0.5 2 2 3 0.5 5.5 10 0.5
3.40 .
4 0.5 2 3 0.5 10 0.5
i i
i
i
i
y A
y in
A
=
=
+ +
 
 
= = =
+ +


( )
( )
3
t max
4 10 10.5 3.40 12
3715.12 3.715
91.73
psi ksi

  − 
= = =
( )
3
max
4 10 3.40 12
1779.1 1.779
91.73
c psi ksi

  
= = =
Example (3): Determine the maximum tensile and compressive bending stress in the beam if it is subjected
to a moment of M=4 kip.ft.
Solution
19
Example (4): A cast-iron machine part is acted upon by a 3kN.m couple. Knowing E=165GPa and
neglecting the effects of fillets, find (a) the maximum tensile and compressive stresses, and (b) the
radius of curvature.
Solution
Calculate the location of section centroid (ӯ):
3
114 10
38
3000
i i
i
y A
Y mm
A

= = =


( )
( ) ( )
3
2 2
3 3
2 2 9 4
12
90 20 30 40
1800 12 1200 18 868 10
12 12
x
bh
I I Ad Ad
m

−
 
= + = +
 
 
   
   
 
= +  + +  = 
   
   
   
   
   
 
The maximum tensile and compressive stresses:
The radius of curvature (ρ):
----------------------------------------------------------------------------------------------------------------------------------------------
20
( )
( )
( )
max
max
3
max 9
3
max 9
3 10 0.022
76.0
868 10
3 10 0.038
131.3
868 10
A
tensile
B
compressive
M y
I
M y
MPa
I
M y
MPa
I



−
−
=
 
= = =

  −
= = =−

9 9
3
1 165 10 868 10
47.7
3 10
M EI
m
EI M


−
  
= → = = =

21
Example (5): The largest allowable stresses for the cast iron link are 30MPa in tension and 120MPa
in compression. Determine the largest force P which can be applied to the link.
Solution
From Sample Example 4:
3 2
9 4
3 10 m
0.038m
868 10 m
A
Y
I
−
−
= 
=
= 
• Determine equivalent centric and bending loads.
0.038 0.010 0.028m
= 0.028
d
P centric load
M bending moment P d P
= − =
=
=  =
22
• Superpose stresses due to centric and bending loads
( )( )
( )( ) P
P
P
I
Mc
A
P
P
P
P
I
Mc
A
P
A
B
A
A
1559
10
868
022
.
0
028
.
0
10
3
377
10
868
022
.
0
028
.
0
10
3
9
3
9
3
−
=

−

−
=
−
−
=
+
=

+

−
=
+
−
=
−
−
−
−


• Evaluate critical loads for allowable stresses.
kN
0
.
77
MPa
120
1559
kN
6
.
79
MPa
30
377
=
−
=
−
=
=
=
+
=
P
P
P
P
B
A


max 77.0
P kN
=
• The largest allowable load:
-------------------------------------------------------------------------------------------------------------------------------------------------------------
23
Solve the problems
in Sheet (6)
Try to solve
Problem (1): An eccentric force P is applied as shown to a steel
bar of 25x90 mm cross-section. The strains at A and B have been
measured and found to be: εA =+350μ and εB=-70μ. Knowing that
E=200GPa, determine (a) the distance d, (b) the magnitude of the
force P. (Answer : (a) 5.00mm, (b) 243kN)
Problem (2): A vertical force P of magnitude 20kips is applied at
point C located on the axis of symmetry of the cross-section of a
short column. Knowing that y=5 in, determine (a) the stress at
point A, (b) the stress at point B, (c) the location of the neutral
axis. (Answer: (a) 0.576ksi, (b) -1.498ksi , (c) 2.411in)

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ME 205- Chapter 6 - Pure Bending of Beams.pdf

  • 1. Stress Analysis I ME 205 – Fall 2023 Chapter (6) - Pure Bending of Beams Beams Members that are slender and support loadings applied perpendicular to their longitudinal axis are called beams (see the figure). Objectives ➢ Determine the stress in members caused by bending ➢ Derive and understand the bending theory, its limitations and its applications for strength based design and analysis of symmetric bending of beams ➢ Determine the normal stress developed in the symmetric bending of beams. 1
  • 2. 2 ✓ It depends on the beam cross-section ✓ It depends on the properties of the cross section (i.e., location of centroid) How to calculate the bending stress ? Does the stress caused by bending is normal or shear stress ? Cross-section What the Bending Moment does to the Beam? ✓ It causes compression on one face and tension on the other ✓ It causes the beam to deflect
  • 3. 3 Double symmetric cross-sections (Two axes of symmetry) Single symmetric cross-sections (One axis of symmetry) Asymmetric cross-sections (No axis of symmetry) Various Beam Cross-sections Axis of symmetry Axis of symmetry Axis of symmetry
  • 4. 4 At least One axis of symmetry
  • 5. 5 Other Loading Types • Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. • Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a moment. • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a moment.
  • 6. Beam sign convention The sign convention for bending moment (M) and shear force (V) in beams are shown: 6
  • 7. 7 Theory of Simple Bending of Straight Beams When a bending moment is applied to a straight prismatic beam, the longitudinal lines become curved and vertical transverse lines remain straight and undergo a rotation. Basic Assumptions 1. Beam material is homogenous and isotropic and obeys Hook’s law, 2. Longitudinal axis x (within the neutral surface) does not experience any change in the length, 3. All cross sections remain plane and perpendicular to longitudinal axis during the deformation.
  • 8. 8 ➢ Member remains symmetric, i.e., y-axis is the axis of symmetry, ➢ Bending moment (M) causes beam to bends uniformly to form a circular arc, ➢ Length of top decreases and length of bottom increases, ➢ A neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not undergo any change, ➢ In this case. stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it for the given M direction. Beam with a plane of symmetry in pure bending
  • 9. 9 ( ) ( ) max max max max max max (strain varies linearly) or x x L y L L y y y y L y y ρ y y                  = −  = − = − − =− = =− =− = = =− Strain due to Bending of a Straight Beam ➢ A contraction will occur (-ε) in the fibers located above the neutral axis (+y). ➢ The elongation will occur (+ε) in the fibers located below the axis (-y). Consider a beam segment of length L. After deformation, the length of the neutral surface remains L’ (=JK). At other sections, where ρ is the radius of curvature. L ymax ymax
  • 10. 10 Stress Due to Bending • For a linearly elastic material, max max max max (stress varies linearly) x x y y E E y y     = =− =− • For static equilibrium, max max max max 0 0 x x y F dA dA ydA y y dA y    = = = − =− → =     First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid. • For static equilibrium, where S= I/ymax is the section modulus (‫المقطع‬ ‫)معامل‬ of the beam. ( ) ( ) 2 max max max max max max max max x I y M y dA y dA y dA y y y M y M I S        = − = − − = =     = =    x NA M y I  =− The maximum stress σmax due to bending is: max max max x NA M y I  = x dF dA  = ymax
  • 11. Hollow circular section 11 Moment of Inertia for Regular Shapes 4 4 2 4 64 R d I A R    = = = N. A. C Circular cross section N. A. C ( ) ( ) 4 4 2 2 4 o i o i R R I A R R   − = = − Rectangular cross section b h N. A. h/2 3 12 bh I A bh = = C N. A. 3 3 12 12 o i i i o o i i b h b h I A b h b h = − = − bi bo hi ho h/2 Hollow Rectangular section C
  • 12. 12 Section Moduli for Regular Shapes 3 max 3 2 max , 12 2 12 6 2 bh h I y bh I bh S h y = = = = = For example: The section modulus (S) for rectangular cross section of height h and width b can be obtained as:
  • 13. 13 Deformations in a Transverse Cross Section • Deformation due to bending moment M is quantified by the curvature of the neutral surface max max max max max max 1 1 1 M y M y E y E y I EI M EI     = = = = = where ρ is the radius of curvature (‫االنحناء‬ ‫قطر‬ ‫)نصف‬.
  • 14. 14 • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment ( ) ( ) centric bending x x x P My A I    = + =  Eccentric Axial Loading in a Plane of Symmetry • Eccentric loading F P and M P d = =  • Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application.
  • 15. 15 Example (1): A member having the dimensions shown is used to resist an internal bending moment of M=90kN.m. Find the maximum stress in the member if the moment is applied about the z-axis as shown. Sketch the stress distribution over the cross-section. Solution The maximum normal bending stress is given by: The moment of inertia about the z-axis can be obtained by: max max max ( / 2) 2 M y M h where y h I I  = = = ( ) 3 3 6 4 0.2 0.15 56.25 10 12 12 bh I m −  = = =  3 max 6 90 10 0.15 120 2 56.25 10 MPa  −   = =  
  • 16. 16 Example (2): An open-link chain is obtained by bending low-carbon steel rods into the shape shown. For 160lb load, find: (a) the maximum tensile and compressive stresses, and (b) distance between section centroid and neutral axis. Solution: • Equivalent centric load and bending moment 160 160 0.6=104 P lb M P d lb×in = =  =  ( )2 2 2 0.25 0.1963in 160 815 0.1963 axial A R P psi A    = = = = = = • Normal stress due to a centric load ( ) ( ) 4 4 3 4 1 1 4 4 max max 3 0.25 3.068 10 in 104 0.25 8475 0.068 10 I R M y psi I    − − = = =   = = =  • Normal stress due to bending moment
  • 17. 17 (1) Maximum tensile and compressive stresses ( ) ( ) max max max max 815 8475 815 8475 axial t axial c       = + = + = − = − ( ) max 9260 t psi  = ( ) max 7660 c psi  =− (2) Neutral axis location 3 0 815 3.068 10 0.0240 105 o o M y P A I P I y in A M − = −   =  = = 0.0240 o y in = Using superposition method: ----------------------------------------------------------------------------------------------------------------------------------------------------------
  • 18. 18 Section properties: The moment of inertia about the neutral axis (which passed from the centroid of the cross section) is: The maximum normal bending stress is given by: max max M y I  = ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 3 2 3 2 3 2 3 1 1 4 0.5 4 0.5 3.40 0.25 2 0.5 3 0.5 3 3.40 2 12 12 1 0.5 10 0.5 10 5.5 3.4 91.73 . 12 I in   = + − + + −     + + − = ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( ) 1 1 0.25 4 0.5 2 2 3 0.5 5.5 10 0.5 3.40 . 4 0.5 2 3 0.5 10 0.5 i i i i i y A y in A = = + +     = = = + +   ( ) ( ) 3 t max 4 10 10.5 3.40 12 3715.12 3.715 91.73 psi ksi    −  = = = ( ) 3 max 4 10 3.40 12 1779.1 1.779 91.73 c psi ksi     = = = Example (3): Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M=4 kip.ft. Solution
  • 19. 19 Example (4): A cast-iron machine part is acted upon by a 3kN.m couple. Knowing E=165GPa and neglecting the effects of fillets, find (a) the maximum tensile and compressive stresses, and (b) the radius of curvature. Solution Calculate the location of section centroid (ӯ): 3 114 10 38 3000 i i i y A Y mm A  = = =   ( ) ( ) ( ) 3 2 2 3 3 2 2 9 4 12 90 20 30 40 1800 12 1200 18 868 10 12 12 x bh I I Ad Ad m  −   = + = +               = +  + +  =                       
  • 20. The maximum tensile and compressive stresses: The radius of curvature (ρ): ---------------------------------------------------------------------------------------------------------------------------------------------- 20 ( ) ( ) ( ) max max 3 max 9 3 max 9 3 10 0.022 76.0 868 10 3 10 0.038 131.3 868 10 A tensile B compressive M y I M y MPa I M y MPa I    − − =   = = =    − = = =−  9 9 3 1 165 10 868 10 47.7 3 10 M EI m EI M   −    = → = = = 
  • 21. 21 Example (5): The largest allowable stresses for the cast iron link are 30MPa in tension and 120MPa in compression. Determine the largest force P which can be applied to the link. Solution From Sample Example 4: 3 2 9 4 3 10 m 0.038m 868 10 m A Y I − − =  = =  • Determine equivalent centric and bending loads. 0.038 0.010 0.028m = 0.028 d P centric load M bending moment P d P = − = = =  =
  • 22. 22 • Superpose stresses due to centric and bending loads ( )( ) ( )( ) P P P I Mc A P P P P I Mc A P A B A A 1559 10 868 022 . 0 028 . 0 10 3 377 10 868 022 . 0 028 . 0 10 3 9 3 9 3 − =  −  − = − − = + =  +  − = + − = − − − −   • Evaluate critical loads for allowable stresses. kN 0 . 77 MPa 120 1559 kN 6 . 79 MPa 30 377 = − = − = = = + = P P P P B A   max 77.0 P kN = • The largest allowable load: -------------------------------------------------------------------------------------------------------------------------------------------------------------
  • 23. 23 Solve the problems in Sheet (6) Try to solve Problem (1): An eccentric force P is applied as shown to a steel bar of 25x90 mm cross-section. The strains at A and B have been measured and found to be: εA =+350μ and εB=-70μ. Knowing that E=200GPa, determine (a) the distance d, (b) the magnitude of the force P. (Answer : (a) 5.00mm, (b) 243kN) Problem (2): A vertical force P of magnitude 20kips is applied at point C located on the axis of symmetry of the cross-section of a short column. Knowing that y=5 in, determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis. (Answer: (a) 0.576ksi, (b) -1.498ksi , (c) 2.411in)