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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; Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

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 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

Point Loads Moment about a point (pivot) Moment = Force x Distance about a point Force Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing Distance

Point Loads Moment about a point This gives a CLOCKWISE MOMENT Pivot Force (F) MOMENT = F x D Distance (D) Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

Worked example 50KN 100KN 4M 4M 3M 11M RL RR Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing 11M RL RR

Worked example 50KN 100KN 4M 4M 3M 11M RL RR ACM = RR X 11 CM = 50 X 4 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing 11M RL RR

Worked example Calculate the left and right reactions Take the left-hand reaction as the PIVOT: Anti-clockwise moments = Clockwise moments Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

Worked example Anti-clockwise moments = Clockwise moments RR x 11 = (100 x 8) + (50 x 4) RR x 11 = 1000 RR = 1000/11 = 90.91 Kn Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

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 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

Worked example RR = 90.91Kn RL = 59.09kN Check ΣUPWARD FORCES = ΣDOWNWARD FORCES 59.09 + 90.91 = 150 (50+100) correct Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

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 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

UNIFORMLY DISTRIBUTED LOADS Udl = 10Kn/m This will give a CLOCKWISE MOMENT 11M RL RR Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

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 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

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 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

Worked example UDL Udl = 10Kn/m 20kN 6m 4m 4m RL RR Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing 6m 4m 4m RL 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 kN Therefore RL = (20 + 60) – 27.14 = 52.86 kN Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

SHEAR FORCE DIAGRAM 52.86 -7.14 -27.14 -27.14 Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing -27.14 -27.14

BENDING MOMENT DIAGRAM content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing

title content Names of who is presenting and positions What we are here to do today NB: you can also copy and use this slide as a template for further slides required with writing