1. Welcome to Neath Port Talbot College!

2. Point loads
The figure below illustrates a simply supported beam with reactions at each end and two point loadings of 50kN and 100kN.
We shall calculate the value of the end reactions. Before undertaking this calculation, we need to recap some of the basic laws associated with this type of structure;
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3. Point Loads
The sum of the forces in one direction must equal the sum of the forces in the opposite direction ie the sum of downward forces equals the sum of the upward forces.
Anti-clockwise moments equal clockwise moments
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4. Point Loads Moment about a point (pivot)
Moment = Force x Distance about a point
Force
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Distance

5. Point Loads Moment about a point This gives a CLOCKWISE MOMENT Pivot
Force (F)
MOMENT = F x D
Distance (D)
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6. Worked example 50KN 100KN 4M 4M 3M 11M RL RR
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11M RL RR

7. Worked example 50KN 100KN 4M 4M 3M 11M RL RR ACM = RR X 11 CM = 50 X 4
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11M RL RR

8. Worked example Calculate the left and right reactions
Take the left-hand reaction as the PIVOT:
Anti-clockwise moments = Clockwise moments
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9. Worked example Anti-clockwise moments = Clockwise moments
RR x 11 = (100 x 8) + (50 x 4)
RR x 11 = 1000
RR = 1000/11
= Kn
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10. Worked example Similarly, if we start at the right hand reaction:
Anti-clockwise moments= clockwise moments
(100 x 3) + (50 x 7) = RL x 11
650 = RL x 11
650/11 = RL = 59.09Kn
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11. Worked example RR = 90.91Kn RL = 59.09kN Check
ΣUPWARD FORCES = ΣDOWNWARD FORCES
= 150 (50+100) correct
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12. UNIFORMLY DISTRIBUTED LOADS
The figure below is a simple uniformly distributed load; the total load is calculated on the beam and is considered to be acting at the centre of the UDL.
Udl = 10Kn/m
11M RL RR
This will give a CLOCKWISE MOMENT
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13. UNIFORMLY DISTRIBUTED LOADS
Udl = 10Kn/m
This will give a CLOCKWISE MOMENT
11M RL RR
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14. Anti-clockwise moments = clockwise moments RR x 11 = (10 x 11) x 5.5
UNIFORMLY DISTRIBUTED LOADS
Anti-clockwise moments = clockwise moments
RR x 11 = (10 x 11) x 5.5
RR x 11 = 110 x 5.5
RR=605/11 = 55 kN
Therefore:
RL = (10 x 11) – 55 = 55 kN
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15. In this case where the UDL is carried for the whole length of the beam
UNIFORMLY DISTRIBUTED LOADS
In this case where the UDL is carried for the whole length of the beam
We can divide the total weight by two, then this should equal the end reactions, which it does. Therefore RR=RL
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16. Worked example UDL Udl = 10Kn/m 20kN 6m 4m 4m RL RR
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17. Reaction Calculations
Take Moments about RL
RR x 14 = (10 x 6 x 3) + (20 x 10) =
RR = 380/14
= kN
Therefore
RL = ( ) – = kN
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18. SHEAR FORCE DIAGRAM 52.86 -7.14 -27.14 -27.14
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-27.14
-27.14

19. BENDING MOMENT DIAGRAM
content
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