# Beams

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15 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)

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Beams

Beams:

Comparison with
trusses, plates

Examples:

1. simply supported beams

2. cantilever beams

L, W, t: L >> W and L >> t

L

W

t

Beams
-

shear force

and
bending moments

q

q

> 0

q

< 0

Shear Forces, Bending Moments
-

Sign
Conventions

Shear forces
:

Bending moments
:

left section

right section

positive shear:

negative shear:

positive moment

negative moment

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.

Forces, and Bending Diagram

dV dM
q V
dx dx
  
Beam
-

Normal Strain

Pure bending problem

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

M

M

Normal Strain
-

Analysis

x

neutral axis (N.A.):

Coordinate system:

longitudinal strain:

y

q

r

'
x
L x x
L x
y
y

r q rq
rq r
   
 

 
  
N.A.

Beam
-

Normal Stress

x x
Ey
E
 
r
  
Hooke’s Law:

Maximum stresses:

M

M

M

x

y

Neutral axis:

0 0 0
0 0
x x
A A
c
A
Ey
F dA dA
ydA y

r
     
   
 

Flexure Formula

M M

2
2
1

x
A A
y E M
dM dA y M E dA y dA
EI
 

r r r
       
 
2
: second moment of inertial (with respe
ct to the neutral axis)
A
I y dA

M

x

y

Moment balance:

x x
Ey
E
 
r
  
x
M y
I

 

Comparison:

Moment of Inertia
-

I

dA
y
I
A

2
Example 2:

Example 3:

w

h

4h

w

w

w

h

Design of Beams for Bending Stresses

Design Criteria:

n
u
Y
allowable
allowable

or

,

1.

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

Procedure:

Find M
max

Calculate the required section modulus

Pick a beam with the least cross
-
sectional area or weight

; : section modulus
x
M y M I
S
I S y
 

    
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

Horizontal shear stresses:

V

2

1

t
H

1
1
, : first moment
h
H
y
VQ
Q ydA
Iw
t
 

x

y

h
1

h
2

y
1

Shear Stresses inside Beams

H
VQ
Iw
t t
 
Relationship between the horizontal shear stresses and the vertical shear stresses:

Shear stresses
-

force balance

Iw
VQ

t
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

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

x

h
1

h
2

y
1

y

1
1
h
y
Q ydA

Shear Stresses inside Beams

2
L
4
L
Example 5: Find shear stresses at points A, O and B located at cross section

a
-
a.

P

a

a

4
h
4
L
4
h
4
h
4
h
w

A

B

O

Shear Stress Formula
-

Limitations

Iw
VQ

t
-

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

(1) the largest normal stress

(2) the largest shearing stress

(3) the shearing stress at point a

Deflections of Beam

1

M
EI
r

2
2
3 2
2
1

1
d y
dx
dy
dx
r

 
 

 
 
 
 
 
Deflection curve of the beam: deflection of the neutral axis of the beam.

x

y

x

y

Derivation:

2 2
2 2

d y M d y
EI M
dx EI dx
  
2
2

dM d d y
V V EI
dx dx dx
 
  
 
 
Moment
-
curvature relationship:

Curvature of the deflection curve:

Small deflection:

(1)

(2)

(3)

Equations (1), (2) and (3) are totally equivalent.

P

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

P

Deflection:
y

Deflection:
y
1

Deflection:
y
2

1 2
y y y
 
Method of Superposition

Example 9

P

F

F

q

A

B

B

C

+

d
B

Method of Superposition

Example 9

P

F

A

B

+

d
B

𝜃

=
𝜃
bending
+
𝜃
𝛿
𝐵

𝜃
bending

𝜃
𝛿
𝐵

𝜃
bending
=
4𝑃

2
81 𝐼

𝜃
𝛿
𝐵

tan
𝜃
𝛿
𝐵
=
𝛿


Method of Superposition

Example 9

F

q

B

C

C

B

C

F

q

𝛿

=
𝛿
𝐹
+
𝛿
𝑞

𝛿
𝐹
=


3
3 𝐼

𝛿
𝑞
=
𝑞

4
8 𝐼

𝜃
A
=
4𝑃

2
81 𝐼
+
2
𝑃

3
9
𝐼

+
𝑞

4
8
𝐼

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.