Strength of Materials
by R.S. Khurmi
468
19.7. Simply Supported Beam with an Eccentric Point Load
Fig. 19.4. Section X in CB.
Consider a simply supported beam AB of length l and carrying an eccentric point load W at C as
shown in Fig. 19.4. From the geometry of the figure, we find that the reaction at A,
RA =
Wb
l
and RB =
Wa
l
Now consider a section X in CB at a distance x from B, such that x is less than b (i.e., x < b). We
know that the bending moment at this section,
MX = RB · x =
∴
Integrating the above equation,
Wax
l
...(Plus sign due to saging)
d2y
Wax
=
l
dx 2
E l.
E l.
Wax 2
dy
+ C1
=
2l
dx
...(i)
where C1 is the first constant of integration. We know that at C, x = b and
∴
E liC =
or
dy
=i .
dx C
Wab 2
+ C1
2l
C1 = (El · iC) –
Wab 2
2l
Substituting this value of C1 in equation (i),
dy = Wax 2
Wab 2
+ ( El · iC ) −
2l
2l
dx
Integrating the above equation once again,
El
El . y =
Wax3
Wab2 x
+ (El · iC · x) −
+ C2
6l
2l
...(ii)
...(iii)
where C2is the second constant of integration. We know that when x = 0, then y = 0. Substituting these
values in equation (iii), we get C2= 0.
3
2
Wax
Wab x
+ ( El · iC x) −
...(iv)
6l
2l
The equations (ii) and (iv) are the required equations for slope and deflection at any point in the
section AC. A little consideration will show that these equations are useful, only if the value of iC is known.
∴
El . y =
Fig. 19.5. Section X in AC.
Now consider a section X in AC, at a distance x from B such that x is greater than b (i.e., x > b) as
shown in Fig. 19.5. We know that bending moment at this section,
MX =
Wax
l − W ( x − b)
d2y
Wax
− W ( x − b)
=
2
l
dx
Integrating the above equation,
∴
El
El ·
dy
dx
=
2
Wax 2 W ( x − b)
−
+ C3
2l
2
...(v)
...(vi)
where C3 is the third constant of integration. We know that at C, x = b and
2
∴
EI · iC =
or
Wab
+ C3
2l
C3 = (EI · iC) –
Substituting this value of C3 in equation (vi),
dy
=i .
dx C
Wab 2
2l
2
2
2
W ( x − b)
dy
Wab
Wax
+ ( EI . iC ) −
−
=
2
2l
2l
dx
Integrating the above equation once again,
EI ·
EI · y =
...(vii)
3
Wax3 W ( x − b)
Wab 2
−
+ ( EI · iC ) x −
x + C4
6l
6
2l
...(viii)
where C4 is the fourth constant of integration. We know that when x = l, then y = 0. Substituting these
values in the above equation,
2
3
2
0 = Wal − Wa − Wab + ( EI · iC · l ) + C4
6
6
2
2
3
2
C4 =
Wab
Wa Wal
+
−
− (EI · iC ) · l
2
6
6
=
Wab
Wa 2 2
+
(a − l ) − (EI · iC · l )
2
6
=
Wab 2 Wa
+
[(l + a) (l − a)] − ( EI · iC · l )
2
6
=
Wab 2 Wab
−
(l + a) − ( EI · iC · l )
2
6
... [∵ (x – b) = a]
2
∴
2
2
... [∵ l – a = (l + a) (l – ay)]
[∵ (l – a) = b]
Wab
[3b − (l + a)] − (EI · iC · l)
6
Wab
[3b − (a + b + a)] − (EI · iC l)
=
6
=
Substituting this value of C4 in equation (viii),
=
Wab
(2b − 2a) − (EI · iC · l )
6
=
Wab
(b − a) − (EI · iC · l )
3
EI . y =
(∵ l = a + b)
3
Wab 2 x + Wab (b − a) − (EI · i · l )
Wax3 W ( x − b)
−
+ (EI · iC · x) −
C
3
2l
6l
6
...(ix)
The equations (vii) and (ix) are the required equations for the slope and deflection at any point in
the section AC. A little consideration will show that these equations are useful, only if the value of iC
is known. Now to obtain the value of iC, let us first find out the deflection at C from the equations for
sections AC and CB.
Now substituting x = b in equation (iv) and equating the same with equation (ix),
3
Wab3
Wab3
Wab3 W (b − b)
Wab3 + Wab (b − a) − ( EI · i · l )
+ ( EI · iC · b) −
=
−
+ (EI · iC · b) −
C
6l
2l
3
6l
6
2l
Wab
(b − a)
3l
Substituting the value of EI . iC in equation (ii)
∴
EI . iC =
EI
dy
dx
2
=
Wax
Wab
Wab
+
(b − a) −
2l
3l
2l
=
Wa
[3x 2 + 2b (b − a) − 3b 2 ]
6l
2
Wa
(3x 2 + b2 − 2ab)
6l
This is required equation for slope at any section in BC. We know that the slope is maximum at
B. Thus for maximum slope, substituting x = 0 in equation (x),
=
EI . iB =
or
Wa
(− b 2 − 2ab)
6l
= −
Wa 2
(b + 2ab)
6l
= −
Wab
(b + 2a)
6l
= −
Wa
(l − a ) (l + a )
6l
= −
Wa 2
(l − a 2 )
6l
iB = –
=
Wa 2
(l − a 2 )
6EIl
Wa 2
(l − a 2 )
6EIl
...(Taking minus sign outside)
...(∵ a = l – b and a + b = l)
[Minus sign means that the tangent
at B, makes an angle with AB in the
negative or anticlockwise direction.]
Wb 2
2
(l − b )
...(Substituting b for a)
6EIl
For deflection at any point in AC, substituting the value of EI . iC in equation (iv),
Similarly,
iA =
EI · y =
=
Wax 2
2
[x + 2b (b − a) − 3b ]
6l
=
Wax 2
(x + 2b 2 − 2ab − 3b2 )
6l
=
or
Wax3 Wab
Wab 2 x
(b − a) x −
+
6l
3l
2l
Wax 2
( x − b 2 − 2ab)
6l
= −
Wax
[b (b + 2a) − x 2 ]
6l
= −
Wax
[(l − a) − x 2 ]
6l
= −
Wax 2
[l − a 2 − x 2 ]
6l
y = −
Wax 2
[l − a 2 − x 2 ]
6lEI
= −
Wax 2
2
2
(l − a − x )
6EIl
...(∵ b = l – a and a + b = l)
... (Minus sign means that the
deflection is downwards)
...(xi)
For deflection at C (i.e., under the load) substituting x = b in the above equation,
Wab 2
(l − a 2 − b 2 )
...(xii)
6EIl
We know that maximum deflection will occur in CB since b > a. Now for maximum deflection,
dy
= 0 . Therefore equating the equation (x) to zero,
let us substitute
dx
Wa
(3x 2 − b 2 − 2ab) = 0
6l
2
or
3x – b (b + 2a) = 0
2
3x – (l – a) (l + a) = 0
...(∵ b = l – a and a + b = l)
2
2
2
3x – (l – a ) = 0
3x2 = l2 – a2
yC = −
l 2 − a2
∴
x =
3
For maximum deflection, substituting this value of x in equation (xi),
⎛ l 2 − a 2 ⎞ ⎤ Wa
l 2 − a2 ⎡ 2
2
× ⎢l − a − ⎜
⎜ 3 ⎟⎟ ⎥ = 6EIl
3
⎝
⎠ ⎦⎥
⎣⎢
ymax =
Wa
6EIl
ymax =
Wa
(l 2 − a 2 )3/ 2
9 3 · EIl
l 2 − a2 ⎡ 2 2
× ⎢ (l − a 2 ) ⎤⎥
3
⎣3
⎦