UNIFORMLY DISTRIBUTED LOADS The figure below is a

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 = 10 Kn/m This will give a CLOCKWISE MOMENT RL 11 M RR

UNIFORMLY DISTRIBUTED LOADS Udl = 10 Kn/m This will give a CLOCKWISE MOMENT RL 11 M RR

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 k. N § Therefore: § RL = (10 x 11) – 55 = 55 k. N

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

Worked example UDL Udl = 10 Kn/m 6 m RL 20 k. N 4 m 4 m RR

Reaction Calculations § Take Moments about RL RR x 14 = (10 x 6 x 3) + (20 x 10) = 180 + 200 RR = 380/14 = 27. 14 k. N Therefore RL = (20 + 60) – 27. 14 = 52. 86 k. N

SHEAR FORCE DIAGRAM 52. 86 -7. 14 -27. 14

Worked example UDL Locate points on UDL & find BM values 20 k. N Udl = 10 Kn/m a b c d f e 6 m RL = 52. 86 Kn 4 m 4 m RR = 27. 14 Kn

Worked example UDL Consider all loads to the left of the point The BM value at the ends of a simply supported beam is ZERO. you are considering. CWM are +ve ACWM are -ve Udl = 10 Kn/m a b c d 6 m f 4 m b RL = 52. 86 Kn 2 m 20 k. N e 4 m BMb = +(52. 86 x 2) – (10 x 2 x 1) = 105. 72 – 20 = 85. 72 k. Nm This is the BM at b.

Worked example UDL Consider all loads to the left of the point you are considering. Udl = 10 Kn/m a b 20 k. N c d 6 m f 4 m c RL = 52. 86 Kn 4 m e 4 m BMc = +(52. 86 x 4) – (10 x 4 x 2) = 211. 44 – 80 = 131. 44 k. Nm This is the BM at c.

Worked example UDL Consider all loads to the left of the point you are considering. 20 k. N Udl = 10 Kn/m a b c d 6 m f 4 m c RL = 52. 86 Kn 6 m e 4 m BMd = +(52. 86 x 6) – (10 x 6 x 3) = 317. 16 – 180 = 137. 16 k. Nm This is the BM at d.

Worked example UDL Nb We don’t consider the load acting at e. Consider all loads to the left of the point you are considering. Udl = 10 Kn/m 20 k. N a b c d 6 m 4 m RL = 52. 86 Kn 6 m f 4 m e e 4 m BMe = +(52. 86 x 10) – (10 x 6 x 7) = 528. 6 – 420 = 108. 60 k. Nm This is the BM at e.

SHEAR FORCE DIAGRAM 52. 86 Point of zero SF is the point of maximum BM SHEAR FORCE DIAGRAM -7. 14 -27. 14 BENDING MOMENT DIAGRAM b c d a e -27. 14 f Plot +ve values downward and –ve values upward

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