Shear Force & Bending Moment | Mechanical Engineering SSC JE (Technical) PDF Download

5. Shear force and bending moment diagrams:
Sign Convention:
(i) Shear force
Positive Negative
SHEAR FORCE AND
5CHAPTER BENDING MOMENT
(ii) Bending moment STUDENT CORNER
Positive
(sagging
Nagative
(Hogging)
SFD and BMD for cantilever beams:
(i) Cantilever of length l carrying a concentrated load W at the free end
S x = + W
M x = – Wx
Mmax = – WL
(ii) Cantilever of length l carrying a uniformly distributed load of 'w' per unit run
over the whole length
Sx = + wx
Mx = –
wx2
2
Mmax = –
w 2
2
l
Smax = + wl
VA= Wl
Wl
w 2
2
l
BMD
(iii) Cantilever of length l carrying a uniformly distributed load of 'w' per unit run STUDENT CORNER
over the whole length and a concentrated load W at the free end
Sx = wx + W
Mx = –
wx2
wx
2
æ ö
ç + ÷
è ø
Smax = wl + W
Mmax = –
w 2
w
2
æ ö
ç + ÷
è ø
l
l
A
w per unit run
(iv) Cantilever of length l carrying a uniformly distributed load of 'w' per unit run for
a distance 'a' form free end
form D to B,
Sx = + wx
Mx = –
wx2
2
form A
Sx = + wa
Mx = –wa
x – a
2
æ ö
çè ø÷
W per unti run
M A D
l
B
e
v = wa
wa wa
B
w
D
D
A
A
Parabolic
Straight wa 1 a
2
æ ö
ç - ÷
è ø
(v) Cantilever of length 'l' carrying a load whose intensity varies uniformly from zero STUDENT CORNER
at free end to 'w' per unit run at the fixed end
Sx = +
wx2
2l
= area of load diagram between X and B, Smax = +
wl
2
Mx = Moment of load acting on XB about X
= area of the load diagram between X and
B × distance of centroid of this diagram
form X
= –
wx2 x
2 3
´
l
= –
wx3
6l
= Mmax =
w 2
6
l
A
A
V wl
2
=
wl
2
(vi) Cantilever carrying a load whose intensity varies uniformly form zero at the
fixed end to w per unit run at the free end
Sx =
w wx2 –
2 2
l
l
Mx =
w x wx2 w 2 – –
2 6 3
l l
l
STUDENT CORNER
3
2
6
wx
2
w 3 l 2
l
l - -
cubic
BMD
SFD and BMD for simply supported beams:
(i) Simply supported beam of span l carrying a concentrated load at mid span
Sx = +
W
2 (between AC)
Sx = –
W
2 (between CB)
Mx = +
W
2 x (between CB)
Mx = +
W
2 x (form A to C) (at a distance 'X' form A)
Mmax = Mc =
W
4
l
A
V w
2
= B
V w
2
=
(ii) Simply supported beam carrying a concentrated load placed eccentrically on STUDENT CORNER
the span
Sx = +
Wb
l
(form A to D)
= –
Wb
l
(form D to B)
Mx = +
Wb
l
x (form A to D)
at a distance 'x' form A
Mmax = MD =
Wab
l
A a
W
b B
D
B
SFD
D
BMD D B
A
l
A
V wb =
l B
V = wa
l
wab
l
w.a
l
w.b
l
NOTE: Maximum B.M. occurs where S.F. changes its sign.
(iii) Simply supported beam carrying a uniformly distributed load of w per unit run
over the whole span
Sx = +
W
2
l
-wx Mx =
W
2
l
x –
Wx2
2
, max
S WL
2
=
Mmax = Mc =
W 2
8
l
Va
2
l
l/2 l/2 wl2
8
(iv) Simply supported beam carrying a load whose intensity varies uniformly from STUDENT CORNER
zero at each end to 'w' per unit run at the mid span
2
x x w
4
S w
l
l = + -
Smax = +
W
4
l
Mx = 3 W W
x– x
4 3
l
l
Mmax =
M
c
=
W 2
12
l
+
12
wl2
2w x
l
w
4
V w b
l =
4
V w a
l
=
A
(v) Simply supported beam carrying a load whose intensity varies uniformly from
zero at one end to 'w' per unit run at the other end
Sx =
w wx2 –
6 2
l
l
Mx =
w wx3 x –
6 6
l
l
A
V wl
6
= b
V wl
3
=
wl
6
A B
STUDENT CORNER
Mmax B.M. occurs at x = 3
l
form end A
Mmax =
w 2
9 3
l
SFD and BMD for simply supported beams with overhang:
Simply supported beam with equal overhangs and carrying a uniformly
distributed load of 'w' per unit run over the whole length
S.F. at any section in EA at a distance x form E,
Sx = –wx
at any section in A.B,
Sx = ( ) w
2a – wx
2
l+
B.M. at any section in EA,
Mx = –
wx2
2
at any section in AB.
Mx = ( )( )
w wx2 l 2a x–a –
2 2
+
at x = 'a' and 'a + l' i.e., at A & B,
Mx = –
wa2
2
at x = a + i.e., atC
2
l
Mc = ( 2 2 ) w
– 4a
2
l
E
w unit per run
Cl
a a
A B D
C F
wa
D
a
A
o o o
D
E
E
n
V w(1 2a)
2
+
= n
V w(1 2a)
2
+
=
wl
2
wl
2
l2 4a2
2
-
wa2
2
wa2
2
wa
B
BMD
Case (a) : l2 > 4a2 Þ l > 2a STUDENT CORNER
l2 – 4a2 > 0 Þ Mc > 0 (Positive)
B.M.D. will be as shown in figure above of contraflexure O1 & O2 are at a
distance
2 4a2
2
1 = l - form centre.
Thus distance between point of contraflexure
O1 O2 = l2 - 4a2
Case (b) : l2 = 4a2 Þ p = 2a
B.M at C = Mc = 0
The beam will be subject to only hogging moments.
Points of contraflexure O1 & O2 will coincide with C.
B.M.D will be as shown in figure (a)
w per unit run
E
2
a wl
= l 2
a wl
=
V = w a l V = w b l
2
wl
2
wl
A
2
wl
2
wl
8
wl 2
8
wl 2
E
Case (c) l2 – 4a2 < 0 Þ l > 2a
E
wl
2
wl
E A
2
wl
2
wa2
2
wa2
BMD
(b)
(4a )
8
w 2 l2 -
C D
D
wa SFD
A C B
M STUDENT CORNER c is negative ,since l 2 < 4a2
Mc = – ( 2 2 ) w
4a –
8
l
B. M. will be zero only at ends A and D and at all other sections B.M. will be of
hogging type
B.M. and S.F due to a couple
Case (a): Cantilever
There will be no shear force
Case (b): Simple supported
Shear force is constant
Sx = –
M
l
B.M., Mx = –
Ma
l
(left of C)
= +
Mb
l
(right of C)
C
b
SFD
BMD
l
V M a = l
V M b =
A
l
Ma
l
Ma
l
Mb
l
Mb
A B