Beams
Beams:
t
L
L, W, t: L >> W and L >> t
W
Comparison with trusses, plates
Examples:
1. simply supported beams
2. cantilever beams
Beams - loads and internal loads
Loads: concentrated loads, distributed loads, couples (moments)
q
q>0
q<0
Internal loads: shear force and bending moments
Shear Forces, Bending Moments - Sign
Conventions
Shear forces:
positive shear:
negative shear:
Bending moments:
positive moment
negative moment
left section
right section
Shear Forces, Bending Moments - Static
Equilibrium Approach
Procedure:
1. find reactions;
2. cut the beam at a certain cross section, draw F.B.D. of one piece of the beam;
3. set up equations;
4. solve for shear force and bending moment at that cross section;
5. draw shear and bending moment diagrams.
Example 1: Find the shear force and bending diagram at any cross section of
the beam shown below.
Relationship between Loads, Shear
Forces, and Bending Diagram
dV
 q
dx
dM
V
dx
Beam - Normal Strain
M
M
no transverse load
Pure bending problem
no axial load
no torque
Observations of the deformed beam under pure bending
Length of the longitudinal elements
Vertical plane remains plane after deformation
Beam deforms like an arc
Normal Strain - Analysis
neutral axis (N.A.):
x
radius of curvature:
Coordinate system:
q
longitudinal strain:
r
y
N.A.
L x ' x
x 

L
x
r  y  q  rq

y


rq
r
Beam - Normal Stress
Hooke’s Law:
 x  E x  
M
M
Ey
r
y
M
x
Maximum stresses:
Neutral axis:
Fx  0


  x dA  0 

A
A
 ydA  0
A
 yc  0
Ey
r
dA  0
Flexure Formula
y
Moment balance:
M
M  M
x
dM    x dA  y  M    E
A
y2
r
dA 
E
r
2
 y dA 
A
1
r

M
EI
I   y 2dA: second moment of inertial (with respect to the neutral axis)
A
 x  E x  
x  
M y
I
Ey
r
Comparison:
Axially loaded members
Torsional shafts:
Moment of Inertia - I
I   y 2 dA
A
Example 2:
h
w
Example 3:
h
w
w
w
4h
Design of Beams for Bending Stresses
Design Criteria:
1.
   allowable,
 allowable 
 Y or  u
n
2. cost as low as possible
Design Question:
Given the loading and material, how to choose the shape and the size
of the beam so that the two design criteria are satisfied?
Design of Beams for Bending Stresses
x  
M y
M
  ;
I
S
S
I
: section modulus
y
Procedure:
• Find Mmax
• Calculate the required section modulus
• Pick a beam with the least cross-sectional area or weight
• Check your answer
Design of Beams for Bending Stresses
Example 4: A beam needs to support a uniform loading with density of
200 lb /ft. The allowable stress is 16,000 psi. Select the shape and the size
of the beam if the height of the beam has to be 2 in and only rectangular and
circular shapes are allowed.
6 ft
Shear Stresses inside Beams
shear force: V
V
y
Horizontal shear stresses:
h1
y1
h2
2
1
tH
x
VQ
tH 
,
Iw
h1
Q   ydA: first moment
y1
Shear Stresses inside Beams
Relationship between the horizontal shear stresses and the vertical shear stresses:
y
y1
h1
h2
VQ
t tH 
Iw
x
Shear stresses - force balance
VQ
t
Iw
V: shear force at the transverse cross section
Q: first moment of the cross sectional area above the level at which
the shear stress is being evaluated
h1
Q   ydA
y1
w: width of the beam at the point at which the shear stress is being
evaluated
I: second moment of inertial of the cross section
Shear Stresses inside Beams
Example 5: Find shear stresses at points A, O and B located at cross section
a-a.
P
a
h4
A
O
B
a
L
4
L
4
L
2
h4
h4
h4
w
Shear Stress Formula - Limitations
VQ
t
Iw
- elementary shear stress theory
Assumptions:
1. Linearly elastic material, small deformation
2. The edge of the cross section must be parallel to y axis, not applicable for
triangular or semi-circular shape
3. Shear stress must be uniform across the width
4. For rectangular shape, w should not be too large
Shear Stresses inside Beams
Example 6: The transverse shear V is 6000 N. Determine the vertical shear stress
at the web.
Beams - Examples
Example 7: For the beam and loading shown, determine
(1) the largest normal stress
(2) the largest shearing stress
(3) the shearing stress at point a
Deflections of Beam
Deflection curve of the beam: deflection of the neutral axis of the beam.
y
P
y
x
x
Derivation:
1
Moment-curvature relationship:
Curvature of the deflection curve:
r
1
r


M
EI
d2y
dx 2
  dy  2 
1    
  dx  
32
Small deflection:
d2y M

dx 2 EI
V
dM
dx
d2y
M
dx 2
(1)
d  d2y 
 EI 2 
dx 
dx 
(2)
 EI
 V
(3)
Equations (1), (2) and (3) are totally equivalent.
Deflections by Integration of the
Moment Differential Equation
Example 8 (approach 1):
Deflections by Integration of the Load
Differential Equation
Example 8 (approach 2):
Method of Superposition
P
q
Deflection: y
P
Deflection: y2
Deflection: y1
y  y1  y2
Method of Superposition
Example 9
F
q
P
A
B
B
C
F
+
dB
Method of Superposition
Example 9
P
A
B
𝜃𝐴 = 𝜃bending + 𝜃𝛿𝐵
F
+
𝜃bending
𝜃bending
4𝑃𝑎2
=
81𝐸𝐼
dB
𝜃𝛿𝐵
𝜃𝛿𝐵 ≈ tan𝜃𝛿𝐵
𝛿𝐵
=
𝑎
Method of Superposition
Example 9
F
q
C
B
q
F
C
C
B
𝛿𝐵 = 𝛿𝐹 + 𝛿𝑞
𝐹𝑏3
𝛿𝐹 =
3𝐸𝐼
𝑞𝑏 4
𝛿𝑞 =
8𝐸𝐼
4𝑃𝑎2 2𝑃𝑏 3 𝑞𝑏 4
𝜃A =
+
+
81𝐸𝐼 9𝐸𝐼𝑎 8𝐸𝐼𝑎
Statically Indeterminate Beam
Number of unknown reactions is larger than the number of independent
Equilibrium equations.
Propped cantilever beam
Clamped-clamped beam
Continuous beam
Statically Indeterminate Beam
Example 10. Find the reactions of the propped beam shown below.