ENGR 1990 Engineering Mathematics
Applications of Integration in ME 2560, ME 2570
Example 1: Resultant of a distributed load
R
Y
The diagram shows a cantilevered
beam with a linearly varying load
intensity. The maximum load intensity is
w (lb/ft) at the right end of the beam. To A
calculate the support forces at A, the
distributed load can be replaced by a
single resultant load R acting at a
distance  from the left end.
w (lb/ft)

X
L
The load R and distance  are found by solving the following equations
L
L
L
R   w x dx and R      w x  x dx   w x 2 dx
0
0
(1)
0
The first of Eq. (1) equates the resultant force R with the summation of the load intensity,
and the second one equates the moment of R about A with the sum of the moments of the
of the load intensity.
Given: L  10 (ft) , w  100 (lb/ft)
Find: (a) the resultant force R; and (b) the distance  that it acts from the support.
Solution:
(a) The resultant load is
10
R   10 x dx   5 x
0
2

10
0
10
 5  100  500 (lb) or R   10 x dx  12  10  100  500 (lb)
0
(b) The distance  is
10
3
500    10 x 2 dx   10
3 x 
0
10
0
 
10  1000 2
 10   6.6 (ft)
3  5000 3
So, for a linearly varying load (starting at zero), the resultant acts
2
3
of the way along the
distributed load. If the load started at w (lb/ft) at the wall, then the resultant would be
located
1
3
of the way along the load.
1/5
Example 2:
R
w (lb/ft)
Y
Given: L  10 (ft) , w  100 (lb/ft)
Find: (a) the resultant force R; and (b) the
distance  that it acts from the support.
A
Solution:
(a) The resultant load is

X
L
10
R   100 dx  10  100  1000 (lb)
0
(b) The distance  is
10
1000    100 x dx  12  10  1000   
0
5000 1
 10   5 (ft)
1000 2
So, for a uniformly distributed load, the resultant acts
Example 3:
The diagram of the internal
shearing force for the simply
supported beam with a concentrated
load is shown. The internal bending
moment is related to the shearing
force by the equation
M ( x)   V ( x)dx
Given: P  100 (lbs) , L  5 (ft) ,
a  3.5 (ft) , b  1.5 (ft) , and
M (0)  M ( L)  0
Find: (a) the moment diagram for the
beam; and (b) M max the maximum
bending moment in the beam.
1
2
of the way along the load.
Y
P
a
b
X
A
B
L
FA 
V
x
bP
FA 
L
x
bP
L
P
V
V
bP
L
L
aP
L
a
X
b
2/5
Solution:
(a) We can construct the moment diagram from the shear diagram. Where the shear is
constant, the moment varies linearly with x.
Va  bP L  1.5  100 5  30 (lb) and Vb   aP L  3.5  100 5  70 (lb)
M ( x)   30dx  30 x  D  30 x for 0  x  3.5 (recall that M (0)  0 )
M ( x)   70dx  70 x  D  70 x  350 for 3.5  x  5
100 (lb)
Y
(b) The maximum bending moment occurs at
the concentrated load. It is the area under the
shear diagram from 0 to 3.5 feet.
M max  30  3.5  105 (ft-lb)
 M (3.5)  105 (ft-lb) 
3.5 (ft)
1.5 (ft)
X
A
L
B
V (lb)
30
X
70
M (ft-lb)
105
X
3/5
Example 4:
w (lb/ft)
Y
Given: L  10 (ft) , w  100 (lb/ft) ,
M (0)  M ( L)  0 , and
V ( x )  500  100 x (lb)
X
A
0  x  L
Find: (a) the bending moment diagram;
and (b) the maximum bending moment.
Solution:
(a) Again, we can construct the moment
diagram from the shear diagram. Where
the shear is varies linearly, the moment
varies quadratically with x.
M ( x)    500  100 x  dx
 500 x  50 x 2  D
B
L
wx
x
2
x
2
x
wL
2
V
V (lb)
500
X
L
2
-500
 M (0)  0 
 500 x  50 x 2 (ft-lb)
w (lb/ft)
Y
(b) The maximum bending moment occurs at
the midpoint of the beam. It is the area under
the shear diagram from 0 to L 2 .
A
X
L
M max  12  5  500  1250 (ft-lb)
B
V (lb)
500
X
L
2
-500
M (ft-lb)
1250
X
4/5
Example 5:
w (lb/ft)
Y
Given: L  10 (ft) , w  100 (lb/ft) ,
M ( L)  0 , and
V ( x )  1000  100 x (lb)
0  x  L
A
x
Find: (a) the bending moment diagram;
and (b) the maximum bending moment.
Solution:
wx
x
2
MA
x
2
V
x
(a) In this case, the shear varies linearly
with x, so the moment will vary
quadratically with x. Given M (10)  0 ,
we find
X
L
wL
V (lb)
1000
M ( x)   1000  100 x  dx
 1000 x  50 x 2  D
X
 5000  1000 x  50 x (ft-lb)
2
(b) The maximum moment occurs at the
left end of the beam and is equal to the
area under the shear diagram from 0 to L.
Why?
L
w (lb/ft)
Y
A
X
L
M max  12  10  1000  5000 (ft-lb)
V (lb)
1000
X
M (ft-lb)
10
-5000
5/5