# Engineering Mechanics

(Joyce) #1

Chapter 12 : Support Reactions  225

``````Since the reactions RC and RD are equal, therefore reaction at support
11
5.5 kN
2``````

==
We know that anticlockwise moment due to RC and RD about A
= 5.5 × x + 5.5 (x + 3) = 5.5 x + 5.5 x + 16.5 kN-m
= 11x + 16.5 kN-m ...(i)
= (0.6 × 5) 2.5 + 5 × 5 = 32.5 kN-m ...(ii)
Now equating anticlockwise and clockwise moments given in (i) and (ii)
11 x + 16.5 = 32.5 or 11 x = 16

``∴``

``````16
1.45 m
11``````

``x==``

It is thus obvious that the first support will be located at distance of 1.45m from A and second
support at a distance of 1.45 + 3 = 4.45 m from A. Ans.

Alternative Method

``````We know that sum of anticlockwise moments due to RD and point load at A about the support C
= (5.5 × 3) + (3 × x) = 16.5 + 3x``````

``````= 5 (5 – x) + 0.6 × 5(2.5 – x) = 25 – 5x + 7.5 – 3x = 32.5 – 8x
Equating anticlockwise and clockwise moments given in (iii) and (iv)
16.5 + 3x = 32.5 – 8x or 11 x = 16``````

or

``````16
1.45 m
11``````

``x== Ans.``

Note: In the second method, the uniformly distributed load between A and C will cause anticlockwise
moment about C, while the load between C and B will cause clockwise moment. But for the sake of simplicity, we
have taken the entire load from A to B (equal to 0.6 × 5) acting at its centre (i.e. 2.5 m from A or 2.5 - x) from C.

EXERCISE 12.1

1. A simply supported beam AB of span 4 m is carrying a point loads of 5, 2 and 3 kN at 1, 2
and 3 m respectively from the support A. Calculate the reactions at the supports A and B.
[Ans. 5.5 kN and 4.5 kN]

2. A simply supported beam of span 6 m is carrying a uniformly distributed load of 2 kN/m
over a length of 3 m from the right end B. Calculate the support reactions.
[Ans. RA = 1.5 kN, RB = 4.5 kN]

3. A simply supported beam AB of span 6 m is loaded as shown in Fig. 12.14.

``````Fig. 12.14.
Determine the reactions at A and B.[Ans. 6.875 kN, 9.125 kN]``````