W kN
A
C
w kN/ length
B
W = 1.5 kN
w= 1.5 kN/m
D
l (AC)= 1.5 m
l (CB)= 2.25m
l (BD)= 1.5m
SFY = 0 = RA + RB – W – w x 1.5 = 0
SMA = 0 = RB x 3.75 – W x 1.5 – w x 1.5 x 4.5 = 0
RA = 0.45 kN
Mechanics of Solids
RB = 3.3 kN
38
1. Consider section x-x is taken in between
A and C (0 < x < 1.5)
Shear Force
Bending Moment
Vx + RA = 0
Mx – RAx = 0
VA= - 0.45 kN
At x= 0
VC= - 0.45 kN
At x= 1.5m ; MC= 0.675 kNm
Mechanics of Solids
; MA= 0
39
2. Consider section x-x is taken in between
C and B (1.5 < x < 3.75)
Shear Force
Bending Moment
Vx + RA – W = 0
Mx + W (x – 1.5) – RAx = 0
VC= 1.05 kN
At x= 1.5m
VB= 1.05 kN
At x= 3.75m ; MB= - 1.6875 kNm
Mechanics of Solids
; MC= 0.675 kNm
40
3. Consider section x-x is taken in between
B and D (3.75 < x < 5.25)
Shear Force
Vx + RA + RB – W – w (x – 3.75) = 0
At x= 3.75m ; VB= - 2.25 kN
At x= 5.25m ; VD= 0
Bending Moment
Mx – RAX – RB (x – 3.75) + W (x- 1.5) + w (x – 3.75)2/ 2
At x= 3.75m
; MB= -1.6875 kNm
At x= 5.25m
; MD= 0 kNm
Mechanics of Solids
41
W kN
A
C
w kN/ length
B
D
1.05 kN
0
- 0.45 kN
Shear Force
Diagram
- 2.25 kN
0.675 kNm
0
Bending Moment
Diagram
- 1.6875 kNm
Mechanics of Solids
42
Example 3.4
A beam is subjected to Varying distributed load.
Calculate internal forces and moments and draw
shear force and bending moment diagrams.
In given problem varying distributed load is given.
For calculating reactions at support the distributed
load has to be replaced by a single resultant force
at the location x. (Maximum intensity is w0 )
Mechanics of Solids
43
Fig.12 Example 3.4. A distributed loading is replaced
by its resultant.
Mechanics of Solids
44
The external supports RB and MB are now easily
obtained by applying the conditions of equilibrium.
Mechanics of Solids
45
It is not permissible to use the above resultant R to
calculate shear force and bending moments within
the beams.
We can, however, use general method to find the
internal forces and bending moments
Mechanics of Solids
46
(a)
(b)
(c)
Mechanics of Solids
47
At x= 0 ; V = 0
At x= L ; V = woL / 2
At x= 0 ; M = 0
At x= L ; M = - woL2 / 6
Mechanics of Solids
48
Mechanics of Solids
49
Parabolic distribution
woL
2
+ ve Shear
Hogging
Cubic distribution
Mechanics of Solids
50
DIFFERENTIAL EQUILIBRIUM
RELATIONSHIPS
1. It is an alternative procedure for obtaining internal
forces and moments for the slender members
2. Instead of cutting a beam in two and applying the
equilibrium conditions to one of the segments, very
small differential element of the beam will be considered
3. The conditions of equilibrium combined with a limiting
conditions will lead us to differential equations connecting
the load, the shear force, and the bending moment
Mechanics of Solids
51
4. Integration of these relationships for particular
cases furnishes us with an alternative method for
evaluating shear forces and bending moments.
Mechanics of Solids
52
Mechanics of Solids
53
If the variation of q(x) is smooth and if Dx is very
small then R is very nearly given by q Dx and the
line of action of R will very nearly pass through the
midpoint ‘o’ of the element.
The conditions of equilibrium applied to Fig. 14c
are then
Mechanics of Solids
54
Integrating above equations with appropriate
conditions will give values of shear forces and
bending moments
Mechanics of Solids
55
Example 3.5
In Fig.15a a beam carrying a uniformly distributed load
of intensity q = - wo is supported by a pinned joint at A
and a roller support at B. We shall obtain shear-force
and bending-moment diagrams by integration of the
differential relationships (3.11) and (3.12).
Mechanics of Solids
56
RA = R B = wo L / 2
Mechanics of Solids
57
We have two boundary conditions available to find
C1 and C2.
External moments are absent at either end of the beam,
hence
Mb= 0
at x = 0
Mb= 0
at x = L
Inserting these boundary conditions values of C1 and
C2 yield
Mechanics of Solids
58
The shear force and bending moment are then
given by
The bending moment is maximum when the shear
force is zero.
Mechanics of Solids
59
Mechanics of Solids
60
Example 3.6
Consider the beam shown in Fig. 3.16a with simple
transverse supports at A and B and loaded with a
uniformly distributed load q = - w0 over a portion of the
length. It is desired to obtain the shear-force and
bending-moment diagrams.
Mechanics of Solids
62
Part AC
Shear Force
dV1
 wO  0
dx
V1  wO x  C1
Part CB
Shear Force
dV2
0
dx
V2  C3
Mechanics of Solids
Bending Moment
dM b1
dM b1
V  0

 wO x  C1  0
dx
dx
1
M b1  wO x 2  C1 x  C2
2
Bending Moment
dM b 2
V  0
dx

dM b 2
 C3  0
dx
M b 2  C3 x  C4
63
For 4 Constants i.e. C1, C2, C3 and C4 we need
to have 4 boundary conditions
At x = 0;
At x = a;
MA = 0
At x = L;
V1 = V2 = VC
MB = 0
and
Mb1 = Mb2 = MC
By applying these BC we get values of C1, C2,
C3 and C4 as follows
1
a
C1  wO a (  2)
2
L
C2  0
Mechanics of Solids
1 wO a 2
C3 
2 L
1
C4  wO a 2
2
64
Part AC
Shear Force
1
a
V1  wO x  wO a (  2)
2
L
1
a
1
( L  b)
VA  wO a(  2)   wO a
2
L
2
L
2
w a
VC  O
2L
Part CB
Shear Force
1 wO a 2
V2 
2 L
 VC  VB
Shear force will be
constant in betn C to B
Mechanics of Solids
Bending Moment
1
1
a
2
M b1  wO x  wO a(  2) x  0
2
2
L
MA  0
1
2 b
M C  wO a ( )
2
L
Bending Moment
1
1
2
M b2 
wO a x  wO a 2
2L
2
1
b
M C  wO a 2 ( )
2
L
MB  0
65
1
a
V1  wO x  wO a (  2)  0
2
L
x
M b max
Mechanics of Solids
a ( L  b)
2L
wO a 2 ( L  b) 2

8 L2
66
Forces and Moments Transmitted by
Slender Members
SINGULARITY FUNCTIONS
 This is special mathematical tool to
handle discontinuous load functions
 Figure shows family of singularity
functions specially designed for this
purpose
fn  x   x  a
n
}
= 0; if x < a
= (x – a)n; if x > a
 The function x – a0 is called unit step
starting at x = a.
 The function x – a1 is called unit ramp
starting at x = a.
02.09.2015
BITS Pilani, K K Birla Goa Campus
Forces and Moments Transmitted by
Slender Members
 The first two members of the family shown in figure are
exceptional. To emphasize this, the exponent is written
below the bracket instead of above.
 The function x – a–1 is called unit concentrated load or unit
impulse function acting at x = a.
 The function x – a–2 is called unit concentrated moment or
unit doublet function acting at x = a.
 These functions are zero everywhere except at x = a where
they are infinite.
 They are, however, infinite in such a way that integration of
unit concentrated moment is unit concentrated load and
integration of unit concentrated load is unit step function.
02.09.2015
BITS Pilani, K K Birla Goa Campus
Forces and Moments Transmitted by
Slender Members
 Integration law for unit step and unit ramp functions is
x

n
x  a dx 

xa
n1
for
n 1
n0
 Integration law for unit concentrated load and unit
concentrated moment is
x

n
x  a dx  x  a
n 1
for
n0

x
Unit concentrated moment 
xa

Unit concentrated load
dx  x  a
x


02.09.2015
2
xa
dx  x  a
1
1
0
BITS Pilani, K K Birla Goa Campus
Forces and Moments Transmitted by
Slender Members
Loading intensities represented by Singularity functions
02.09.2015
BITS Pilani, K K Birla Goa Campus