Basic Engineering Mathematics, Fifth Edition

258 Basic Engineering Mathematics

(i) Divide baseADinto any number of equal

intervals, each of widthd(the greater the

number of intervals,thegreater theaccuracy).

(ii) Erect ordinates in the middle of each interval

(shown by broken lines in Figure 28.2).

(iii) Accurately measure ordinatesy 1 ,y 2 ,y 3 ,etc.

(iv) AreaABC D

=d(y 1 +y 2 +y 3 +y 4 +y 5 +y 6 ).

In general, the mid-ordinate rule states

Area=(width of interval)(sum of

mid-ordinates)

(d) Simpson’s rule

To determine the areaPQRSof Figure 28.1,

(i) Divide basePSinto anevennumber of inter-

vals, each of widthd(the greater the number

of intervals, the greater the accuracy).

(ii) Accurately measure ordinatesy 1 ,y 2 ,y 3 ,etc.

(iii) AreaPQRS=

d

3

[(y 1 +y 7 )+ 4 (y 2 +y 4 +y 6 )

+ 2 (y 3 +y 5 )]

In general, Simpson’s rule states

Area=

1

3

(

width of

interval

)[(

first+last

ordinate

)

+ 4

(

sum of even

ordinates

)

+ 2

(

sum of remaining

odd ordinates

)]

Problem 1. A car starts from rest and its speed is

measured every second for 6s.

Timet(s) 01 2 3 4 5 6

Speedv(m/s) 0 2.5 5.5 8.75 12.517.5 24.0

Determine the distance travelled in 6 seconds (i.e.

the area under thev/tgraph), using (a) the

trapezoidal rule (b) the mid-ordinate rule

(c) Simpson’s rule

A graph of speed/time is shown in Figure 28.3.

(a) Trapezoidal rule(see (b) above)

The time base is divided into 6 strips, each of

width 1s, and the lengthof the ordinates measured.

Thus,

area=( 1 )

[(

0 + 24. 0

2

)

+ 2. 5 + 5. 5

+ 8. 75 + 12. 5 + 17. 5

]

=58.75m

Graph of speed/time

Speed (m/s)

30

25

20

15

10

5

2.54.05.57.08.7510.7512.515.017.520.2524.0

(^0123)
Time (seconds)
456
1.25
Figure 28.3
(b) Mid-ordinate rule(see (c) above)
The time base is divided into 6 strips each of
width 1s. Mid-ordinates are erected as shown in
Figure28.3 by the broken lines. The length of each
mid-ordinate is measured. Thus,
area=( 1 )[1. 25 + 4. 0 + 7. 0 + 10. 75 + 15. 0

• 20 .25]=58.25m
  (c) Simpson’s rule(see (d) above)
  The time base is divided into 6 strips each of width
  1s and the lengthof the ordinatesmeasured. Thus,
  area=
  1
  3
  ( 1 )[( 0 + 24. 0 )+ 4 ( 2. 5 + 8. 75


• 
  
  
  
  17. 5 )+ 2 ( 5. 5 + 12. 5 )]=58.33m
      Problem 2. A river is 15m wide. Soundings of
      the depth are made at equal intervals of 3m across
      the river and are as shown below.
      Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0
      Calculate the cross-sectional area of the flow of
      water at this point using Simpson’s rule
      From (d) above,
      Area=
      1
      3
      ( 3 )[( 0 + 0 )+ 4 ( 2. 2 + 4. 5 + 2 .4)+ 2 ( 3. 3 + 4. 2 )]
      =( 1 )[0+ 36. 4 +15]=51.4m^2