Types of beams, loadings and supports. Shear force and ... - FKM

reelingripebeltUrban and Civil

Nov 15, 2013 (3 years and 8 months ago)

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Bending


Shear and Moment Diagram,

Graphical method to construct shear

and moment diagram, Bending deformation of a

straight member, The flexure formula

1

2

Shear and moment diagram

3

Axial load diagram

Torque diagram

Both of these diagrams show the internal forces acting on the members.
Similarly, the shear and moment diagrams show the internal shear and moment
acting on the members

Type of Beams

Statically Determinate

4

Simply Supported Beam

Overhanging Beam

Cantilever Beam

Type of Beams

Statically Indeterminate

5

Continuous Beam

Propped Cantilever Beam

Fixed Beam

6

Example 1

Equilibrium equation for 0


x


3m:

A

B

* internal V and M should be assumed +ve

kN
V
V
F
F
y
9
0
0







)
(
9
0
0
kNm
x
M
M
Vx
M






M

V

F

x

Shear Diagram

Lecture 1

8

Sign convention: V=
-
9kN


M

V

F

x

Shear Diagram

9

Sign convention: M=
-
9x kNm

X=0: M= 0

X=3: M=
-
27kNm

M=
-
9x

M =
-
9x kN.m

V =
-
9 kN

F

x

10

V=9kN

M=X

At cross section A
-
A

X

At section A
-
A

Example 2

11

kN
5
0
10
5
0
F
kN
5
0
)
2
(
)
1
(
10
;
0
y











y
y
y
y
A
A
A
C
C
M
1) Find all the external forces

12

)
(
5
0
0
1
0
down
kN
V
V
F
F
m
x
y







)
(
5
0
0
1
0
ccw
kNm
Vx
M
Vx
M
M
m
x
A








)
(
5
0
10
0
2
1
up
kN
V
V
F
F
m
x
y








)
(
)
5
10
(
10
0
)
1
(
10
0
2
1
ccw
kNm
x
M
Vx
M
M
Vx
M
m
x
A











Force equilibrium

Force equilibrium

Moment equilibrium

Moment equilibrium

)
(
5
)
(
5
1
0
ccw
x
M
down
kN
V
m
x




)
(
5
10
)
(
5
2
1
ccw
x
M
up
kN
V
m
x





M=5x

M=10
-
5x

Boundary cond for V and M

Solve it

Draw the shear and moment diagrams for
simply supported beam.

14

15

Distributed Load

16

For calculation purposes, distributed load can be represented as a single
load acting on the center point of the distributed area.


Total force = area of distributed load (W : height and L: length)

Point of action: center point of the area

Example

17

Example

18

Solve it

Draw the shear and moment diagrams the
beam:

19

Solving all the external loads

kN
Wl
F
48
)
6
(
8



Distributed load will be

Solving the FBD

0
12
36
4
)
3
(
48
0
)
3
(
4
0








x
y
y
x
A
A
kN
A
kN
B
F
B
M
21

Boundary Condition

4
0


x
x
V
V
x
F
Y
8
12
0
8
12
0







2
2
2
4
12
0
4
12
0
4
)
8
12
(
0
4
0
x
x
M
x
x
M
x
x
x
M
x
Vx
M
M
A














Equilibrium eq

22

Boundary Condition

6
4


x
x
V
V
x
F
Y
8
48
0
8
36
12
0








144
48
4
0
4
144
48
0
4
)
4
(
36
)
8
48
(
0
)
2
/
(
8
0
2
2
2


















x
x
M
x
x
M
x
x
x
M
x
x
Vx
M
M
A
Equilibrium eq

23

4
0


x
x=0 V= 12 kN

x=4 V=
-
20 kN

x
V
8
12


2
4
12
x
x
M


x=0 M= 0 kN

x=4 V=
-
16 kN

6
4


x
x
V
8
48


x=4 V= 16kN

x=6 V= 0 kN

144
48
4
2




x
x
M
x=4 V=
-
16kN

x=6 V= 0 kN

Graph based on equations

24

y = c

Straight horizontal line

y = mx + c

y=3x + 3

y=
-
3x + 3

y=3x
2

+ 3

y=
-
3x
2

+ 3

y = ax
2

+ bx +c

25

Graphical method

26





x
x
w
V
V
V
x
x
w
V
F
y














0
)
(
:
0

Relationship between load and
shear:













2
0
:
0
x
k
x
w
x
V
M
M
M
x
k
x
x
w
M
x
V
M
o


















Relationship between shear and
bending moment:

26

27

27

Dividing by

x and taking the limit as

x

0, the
above two equations become:

Regions of distributed load:

Slope of shear
diagram at
each point

Slope of
moment
diagram
at each
point

=


distributed
load intensity
at each point

= shear at each
point

)
(
x
w
dx
dV


V
dx
dM

Example

28

29

30




dx
x
w
V
)
(



dx
x
V
M
)
(
The previous equations become:

change in
shear

=

Area under
distributed load

change in
moment

=

Area under shear
diagram

31

+ve area under shear diagram

32

33

Bending deformation of a straight
member

34

Observation:

-

bottom line : longer

-

top line: shorter

-

Middle line: remain the same but rotate (neutral line)


35

Strain

s
s
s
s








'
lim
0

Before deformation

x
s



After deformation,

x

has a radius of
curvature
r
,
with center of curvature at
point O’


r





x
s
Similarly


r




)
(
'
y
s
r


r

r

r

y
y
s











)
(
lim
0
Therefore

36

r

c


max
Maximum strain will be

max
max
)
(
)
/
/
(


r
r


c
y
c
y




max
)
(


c
y


-
ve: compressive state

+ve: tension

The Flexure Formula

37

The location of neutral axis is
when the resultant force of the
tension and compression is equal
to zero.




0
F
F
R
Noting

dA
dF












A
A
A
ydA
c
dA
c
y
dA
dF
max
max
)
(
0



38

Since , therefore

Therefore, the neutral axis should
be the centroidal axis

0
max

c

0


A
ydA
39



Z
Z
R
M
M
)
(








A
A
A
A
dA
y
c
dA
c
y
y
dA
y
ydF
M
2
max
max
)
(



I
Mc

max

Maximum normal stress


Normal stress at y distance

I
My


Line NA: neutral axis

Red Line: max normal stress


c = 60 mm

Yellow Line: max compressive stress


c = 60mm

I
Mc

max

I
Mc


max

Line NA: neutral axis

Red Line: Compressive stress


y
1

= 30 mm

Yellow Line: Normal stress


y
2

= 50mm

I
My
1
1



I
My
2
2


Refer to Example 6.11 pp 289

I: moment of inertial of the cross
sectional area

12
3
bh
I
x
x


64
4
4
4
D
r
I
x
x





Find the stresses at
A and B

I: moment of inertial of the cross
sectional area

Locate the centroid (coincide
with neutral axis)

mm
A
A
A
y
A
y
A
A
y
y
n
i
i
n
i
i
i
5
.
237
)
300
)(
50
(
)
300
)(
50
(
)
300
)(
50
(
325
)
300
)(
50
(
150
2
1
2
2
1
1
1
1












I: moment of inertial of the cross
sectional area

Profile I

4
6
3
3
)
10
(
5
.
112
12
)
300
(
50
12
mm
bh
I
I



A

A

4
6
2
6
2
3
)
10
(
344
.
227
)
5
.
87
)(
300
)(
50
(
)
10
(
5
.
112
12
)
(
mm
Ad
bh
I
A
A
I






I about Centroidal axis

I about Axis A
-
A using parallel axis
theorem

Profile II

4
6
2
3
2
3
)
10
(
969
.
117
)
5
.
87
)(
50
)(
300
(
12
)
50
)(
300
(
12
)
(
mm
Ad
bh
I
A
A
II






Total I

4
6
4
6
6
)
10
(
313
.
345
)
10
(
969
.
117
)
10
(
344
.
227
)
(
)
(
mm
mm
I
I
I
A
A
II
A
A
I
A
A








* Example 6
-
12 to 6
-
14 (pp 290
-
292)

Solve it

If the moment acting on the cross section of the beam is M = 6 kNm,
determine the maximum bending stress on in the beam. Sketch a
three dimensional of the stress distribution acting over the cross
section

If M = 6 kNm, determine the resultant force the bending stress
produces on the top board A of the beam

4
6
3
2
3
)
10
(
8
.
786
12
)
300
(
40
]
)
170
)(
40
)(
300
(
12
)
40
(
300
[
2
mm
I
I




Total Moment of Inertia

Max Bending Stress at the top and bottom

MPa
I
I
Mc
M
top
45
.
1
)
190
(
)
10
(
6000
3






MPa
M
bottom
45
.
1

Bottom of the flange

MPa
I
I
Mc
M
top
f
14
.
1
)
150
(
)
10
(
6000
3
_






MPa
M
bottom
f
14
.
1
_

1.45MPa

1.14MPa

6kNm

Resultant F = volume of the trapezoid

1.45MPa

1.14MPa

40 mm

300 mm

kN
N
F
R
54
.
15
15540
)
300
)(
40
(
2
)
14
.
1
45
.
1
(




Solve it

The shaft is supported by a smooth thrust load at A and smooth
journal bearing at D. If the shaft has the cross section shown,
determine the absolute maximum bending stress on the shaft

Draw the shear and moment diagram

kN
F
kN
F
F
M
A
D
D
A
3
3
)
25
.
2
(
3
)
75
.
0
(
3
)
3
(
0






External Forces

Absolute Bending
Stress

M
max

= 2.25kNm

MPa
I
Mc
8
.
52
)
25
40
(
4
)
40
(
)
10
(
2250
4
4
3
max