Mechanics of Solids
Chapter 5
1 / 19
STRESSES IN BEAMS
5.1 INTRODUCTION
In previous chapter concern was with shear forces and bending moment in beams. Focus in this
chapter is on the stresses and strains associated with those shear forces and bending moments.
Loads on a beam will cause it to bend or flex.
P
x
B
A
xy – plane = plane of bending
y
ν
z
x
P
A
B
y
Deflection Curve
y
Cross –Section assumed symmetric about xy - plane
5.2 PURE BENDING AND NONUNIFORM BENDING
Pure Bending = flexure of a beam under constant bending moment
 shear force = 0
( V = 0 = dM / dx ); no change in moment.
Nonuniform Bending = flexure of a beam in the presence of shear forces
 bending moment is no longer constant
Moment Diagram example:
M
Pa
a
a
PURE
BENDING
NONUNIFORM
BENDING
Mechanics of Solids
Chapter 5
2 / 19
5.3 CURVATURE OF A BEAM
A beam in NONUNIFORM BENDING (V  0) will have a varying
curvature.
ds = curve length
dx
x
A
y
O’
O’ = center of curvature
 = radius of curvature
 = curvature = -1 = 1
ds
P

B
A beam in PURE BENDING ( V = 0) will have will have constant
curvature = circle
ds = arc length of circular segment
d
 d = ds
For small deflections: ds  dx

Sign Convention:
1


d
dx
(1)
x
O
+ curvature
y
- curvature
opposite from text because we
use +  y
5.4 NORMAL STRAINS
a
Mo
b
x
O
z
Mo
c
d
y
y
a
b
Lengthened  in tension
Mo c
d
Mo
shortened  in compression
Mechanics of Solids
Chapter 5
3 / 19
Somewhere between the top and bottom of the beam is a place
where the fibers are neither in tension or compression.
neutral axis of the cross section
dashed line = neutral surface of the beam
when bent:
a b lengthens
c d shortens
The normal strain is:  x  
causes normal strains, x
y

  y ___________________ ( 2 )
Where, y = distance from neutral axis
From Eqn ( 2 ):
- y = + εx (elongation)
for + 
+ y = - εx (shortening)
+y
OMIT
Transverse Strains:
 z   x    y
Where  = Poisson’s Ratio
Mechanics of Solids
Chapter 5
4 / 19
5.5 NORMAL STRESSES IN BEAMS
If material is elastic with linear stress-strain diagram, THEN:
 = E (Hooke’s Law)
varies
linearly
with y
x = Ex = - Ey ____________ ( 3 )
Where x is longitudinal axis of beam and x is the normal stresses in this direction acting on the
cross section. These stresses varies linearly with the distance y from the neutral surface.
REF: + CURVATURE = + STRESSES above neutral surface
x
x
Mo
c2
z
y
c1
dA
y
y

x
dA    E y dA  0
 y dA  0
must equal ZERO because there is NO
resultant normal force that acts on the
ENTIRE cross section
(4)
Eqn ( 4 ) is the 1st Moment of the Area of the cross section w.r.t. z-axis and it is zero
 z-axis must pass thru the centroid of the cross section.
 z-axis is also the neutral axis
 neutral axis passes thru the centroid of the cross section
Limited to beams where y-axis is the axis of symmetry.
y, z –axes are the PRINCIPAL CENTROIDAL AXES.
Consider the Moment Resultant of x :
Mechanics of Solids
Chapter 5
5 / 19
dM O   x y dA
RECALL Eqn ( 3 ):
 x   E y
M O     x y dA  E  y 2 dA
M = -M0
M   E I
I   y 2 dA

1


 E 
M
EI
 EI = FLEXURAL RIGIDITY
M
I
M
 I
x 
x 
 where, I = Moment of Inertia of
cross sectional area w.r.t.
z-axis ( neutral axis )
 substitute into Eqn ( 3)

y

 Flexure Formula
x = Bending Stress
My
I
2
c2
c1
x
+ M
1
MAXIMUM STRESSES:
1 
Mc1
I
2  
Mc 2
I
Mechanics of Solids
Chapter 5
6 / 19
Text defines Section Moduli as:
S1 
I
c1
S2 
I
c2
1 
M
S1
2  
M
S2
Section Modulus is handy to use when evaluating bending stress w.r.t. to moment which varies
along length of a beam.
If cross section is symmetrical w.r.t. z-axis, then:
c1 = c2 = c
Mc
I
 1   2 
Moments of Inertia to know:
b
O
h
2
h
2
d
z
O
y
I
bh 3
12
I
d4
64
NOTES:
1. Preceding analysis of normal stresses is for PURE BENDING….no shear forces.
2. Flexure Formula can be used for NONUNIFORM BENDING…..proven in “Theory of
Elasticity” by Timoshenko and Goodier.
3. Flexure Formula is NOT accurate in regions of shape irregularities – stress concentrations.
EXAMPLES: No. 2 pg 315
No. 3 pg 316
Mechanics of Solids
Chapter 5
7 / 19
EXAMPLE No. 1
Given: The beam shown.
TOP
9 kN
15 kN / m
50 mm
150 mm
1m
0.5 m
2m
0.5 m
50 mm
FIND:
a.) Stress at a point 100 mm below the neutral surface and 1.3 m from the right.
b.) Maximum tensile and compressive stresses.
SOLn:
Mechanics of Solids
Chapter 5
8 / 19
Example No. 2
GIVEN:
A high-strength steel wire of diameter d = 4 mm, modulus of elasticity E = 200 GPa, proportional
limit pl = 1200 MPa is bent around a cylindrical drum of radius R0 = 0.5 m .
FIND:
a. bending moment, M
b. maximum bending stress, max
SOLn:
R0
C
d
Mechanics of Solids
Chapter 5
9 / 19
EXAMPLE No. 3
GIVEN:
The beam shown which is constructed of glued laminated wood. The uniform load includes the
weight of the beam.
9 ft P = 12 k
q = 1.5 k / ft
A
B
L = 22 ft
FIND:
a. Maximum Tensile Stress in the beam due to bending.
b. Maximum compressive stress in the beam due to bending.
SOLn:
h = 27 in.
b = 8.75 in.
Mechanics of Solids
Chapter 5
10 / 19
5.6 DESIGN of BEAMS for BENDING STRESSES
After all factors have been considered (i.e., materials, environmental conditions ) it usually boils
down to
Allow > Beam
 Allow 
M max c
I
Here is where the section modulus is useful.
RECALL:

M
S
thus, S 
M max
 allow
Appendix E and F give properties of beams.
i.e., W8 x 15
DEPTH
Weight per foot ( lbs / ft )
Nominal Depth ( in )
WIDE
FLANGE
W Shape = wide flange
Wood Beams - 2 x 4  really is: 1.5” x 3.5” net dimensions (should always use net dims.)
Why are W Shapes and S Shapes ( I-beam ) made the way they are?
has same area, A
A
h
b
If both have the same area, which beam is better?
Why is I – Beam better?
ans. I – Beam
ans. has higher M.o.I.
Mechanics of Solids
Chapter 5
11 / 19
Ideally, we would like:
2
h
2
h
2
A
2
Ah
Ah
I     
2 2
2 2
A
2
Ah 2
I
4
2
Ah 2
I
1
S   4  Ah
h
c
2
2
Can’t do this because we need the web.
For W Shapes; S ≈ 0.35 Ah
You want as much material as possible, as far from the neutral axis as possible because this is
where the greatest stress is occurring.
However, you have to be careful because if the web is too thin, it could fail by:
1.) being overstressed in shear
2.) buckling
You don’t get something for nothing.
Mechanics of Solids
Chapter 5
12 / 19
EXAMPLE No. 1
GIVEN: A steel beam shown has an allowable bending stress allow = 60 MPa and a weight density
γ = 77.0 kN / m3.
3.5 kN / m
A
C
B
300 mm
150 mm
FIND:
Required width b of the beam disregarding the weight of the beam.
SOLn:
2b
b
Mechanics of Solids
Chapter 5
13 / 19
5.8 SHEAR STRESSES IN BEAMS ( RECTANGULAR CROSS-SECTIONS )
1.  acts parallel to V ( shear force, also, y – axis )
2.  is uniform across cross-section
Assumptions:
( pg 334 )
h
2
P
Consider a system of 2 beams
REF:
τ
h
2
P
τ
τ
Assuming there is no ( or very little ) friction, the top beam can slide w.r.t. to the bottom beam.
However, if we have a single solid beam of height “ h “ do we get any sliding between 2 planes of
the beam? ans. NO
Thus, there must be shear stresses present that prevent sliding.
b
Looking at the dotted rectangle:
h
2
y1
M + dM
M

h
2
x
z
y
y1
m
n
TOTAL F1
FORCE
F2
F1
y
dA
dx
CROSS SECTION OF BEAM AT SUBELEMENT
Normal Force:
( left side )
Total Force:
( left side )
My
 x dA 
dA
I
( right side )

REF:
F2  
n
m
M  dM  y dA ____________ ( 2 )
I
z
x
My
F1  
dA __________________ ( 1 )
I
similarly,
Total Force:
y = distance at which
shear stress acts.
dx
b
y

Mechanics of Solids
Chapter 5
Shear Force:
F
x
14 / 19
F3 =  b dx ______________ ( 3 )
0
F2  F1  F3  0
F3  F2  F1
(4)
Sub Eqns (1), (2), and (3) into (4):
( M  dM ) y
My
dA  
dA
I
I
dM y
bdx  
dA
I
dM  1 

  y dA
dx  Ib  
V
   y dA
Ib
bdx  
RECALL:
dM
V
dx
y = distance at which the shear force acts
b = thickness of the cross-sectional area where the stress is to be evaluated.
∫ y dA = first moment of that portion of the cross-sectional area between the transverse line (m – n)
where the stress is to be evaluated AND the extreme fiber of the beam.
LET:
Q   y dA

VQ
Ib
SHEAR FORMULA
Mechanics of Solids
Chapter 5
15 / 19
For RECTANGULAR cross-sections, ( see text pg 339 for derivation )

V
2I
 h2

  y12 
 4

Max Shear occurs at the neutral axis, y1 = 0
WHERE:
τ
 max
τmax
V h 2 3V


8I
2A
A = bh
= TOTAL CROSS-SECTIONAL AREA
(5)
and
I=
bh3
12
NOTE:
Shear Eqn ( 5 ) is limited to Cross-Sectional shapes that have sides parallel to the y-axis.
READ:
Shear Strain Effects, pg 340
5.9 SHEAR STRESSES IN BEAMS ( CIRCULAR
CROSS – SECTIONS
)
What about circular cross-sections? How do we handle beams with circular cross-sections?
Largest Shear Stresses occur at neutral axis.
r
We can assume with good accuracy that:
1. τ acts parallel to V ( shear force, also, y – axis )
2. τ is uniform across cross-sections
z
These assumptions are the same used when we
developed the shear formula:
VQ

Ib
y
Therefore, we can use this to find τ at the N.A. (Neutral Axis) which is τmax
I
Q
 r4
4
A
y
2
 r2  4 r 



2  3  
Q
2 r3
3
b  2r

A  r2

2
2
and
y
4r
3
Mechanics of Solids
Chapter 5
 max
16 / 19
 2 r3 

V 
3
VQ

  4V


4
Ib   r 
3 r 2

2 r 
 4 

 max 
4V
3A
WHERE:
A = πr2
= TOTAL CROSS-SECTIONAL AREA
HOLLOW CIRCULAR CROSS SECTIONS
r1
r2
z
I
4
(r24  r14 )
b  2(r2  r1 )
Q
y

2 3
(r2  r13 )
3
 max 
4V  r22  r2 r1  r12

3 A  r22  r12
WHERE:






A   r22  r12
= TOTAL CROSS-SECTIONAL AREA
NOTE:
If r1 = 0 we get our previous equation for solid circular cross-section.
Mechanics of Solids
Chapter 5
17 / 19
5.10 SHEAR STRESSES IN THE WEB OF BEAMS WITH FLANGES
The shear formula  
VQ
still applies because the same assumptions are made.
Ib
READ:
Pg 346 – 350 for derivation of following eqns.
y1
z
h1
t
h
τmin
h
2
h
τmax @ 2
h
2
or at , neutral axis,
where, y1 = 0
τmin
y
b
We will use:  
VQ
It
where: t = web thickness
Q
I
b 2
t
(h  h12 )  (h12  4 y12 )
8
8

1
bh 3  bh13  th13
12

where: b = width of flange
h = height of beam
h1 = web height (inside flanges)
y1 = distance from N.A.
For Wide-Flange Beams, we use the AVERAGE Shear Stress in the web:
 aver 
V
th1
where: t = web thickness
h1 = web height (inside flanges)
Mechanics of Solids
Chapter 5
18 / 19
EXAMPLE No. 1
A beam is to be made of steel that has an allowable bending stress of allow = 24 ksi and an
allowable shear stress of τallow = 14.5 ksi. Select an appropriate W shape that will carry the loading
shown.
20 kip
40 kip
A
B
6’
6’
6’
h
2
Mechanics of Solids
Chapter 5
19 / 19
EXAMPLE No. 2
The laminated beam shown supports a uniform load of 12 kN/m. If the beam is to have
a height – to – width ratio of 1.5, determine the smallest width. allow = 9 MPa, τallow = 0.6 MPa.
Neglect the weight of the beam.
12 kN / m
1m
1.5b
B
A
3m
b