STRUCTURAL MECHANICS

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;

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

Point Loads Moment about a point (pivot) Moment = Force x Distance about a point Force 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

Example 1 50KN 100KN 4M 4M 3M 11M

Example 1 Calculate the left and right reactions Take the left-hand reaction as the PIVOT: Anti-clockwise moments = Clockwise moments

Example 1 50KN 100KN 4M 4M 3M 11M RL RR ACM = RR X 11 CM = 50 X 4

Example 1 Anti-clockwise moments = Clockwise moments RR x 11 = (100 x 8) + (50 x 4) RR x 11 = 1000 RR = 1000/11 = 90.91 Kn

Example 1 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

Example 1 RR = 90.91Kn RL = 59.09kN Check ΣUPWARD FORCES = ΣDOWNWARD FORCES 59.09 + 90.91 = 150 (50+100) correct

Cantilever 10KN 2KN 6KN 1M 1M 1M 2M RR RL

Cantilever RR x 3 = (10 x 1) + (6 X 2) + (2 x 5) RR x 3 = 10 + 12 + 10 RR = 10.67kN 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

Cantilever RR = 10.67kN Total Downward Force = 18kN Therefore, RR + RL = 18 RL = 18 – RR RL = 18 – 10.67 RL = 7.33kN 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 10KN 2KN 6KN 1M 1M 1M 2M RR RL 7.33 7.33 2.00 2.00 2.67 2.67 8.67 8.67

BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E 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 Find the BM values at points A, B, C, D & E The BM value at the ends of a simply supported beam is always ZERO. In this case BMA & BME equal 0 Knm 16

BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E 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 Consider all loads to the left of the point where a BM value is required. Find the BM values at points B, C & D 17

BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E BM value at points B. Consider the following: 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 B BMB = +(7.33x 1) = +7.33 Knm 1m CLOCKWISE MOMENTS CONSIDERED TO BE POSITIVE. 7.33 18

BENDING MOMENT B C D E 10KN 6KN 1M 2M 2KN A 10Kn BM value at points C. Consider the following: 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 BMC = +(7.33x2) – (10x1) = +4.66Knm C 1m 1m ANTI CLOCKWISE MOMENTS CONSIDERED TO BE NEGATIVE. 7.33Kn 19

BENDING MOMENT 6KN 10KN 2KN 1M 2M A D B C E BM value at points D. Consider the following: 10Kn 6Kn 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 BMD = +(7.33x3) – (10x2) – (6x1) = -4.01Knm D 1m 1m 1m 7.33Kn 20

BENDING MOMENT DIAGRAM C D E 10KN 6KN 1M 2M 2KN A Note! Positive values plotted BELOW Negative values plotted ABOVE 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 -4.01 A B C D E 4.66 7.33 21

PROBLEM 15KN 10KN 3KN 2M 2M 1M 1M RR RL

15KN 10KN 3KN 2M 2M 1M 1M (15 x 2) + (10 x 4) + (3 x 6) = RR x 5 RR = 17.6 kN RR + RL = 28kN RL = 28 – 17.6 RL = 10.4 kN 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 23

SFD 15KN 10KN 3KN 2M 2M 1M 1M RR RL 3 15 3 10.4 3 -4.6 -4.6 17.6 10 -14.6 -14.6

BMD = +(10.4 x 5) –(15 x 3) –(10 x 1) = -3.00 kNm RR C D A E RL BMA & E = 0 BMB = +(10.4 x 2) = +20.80 kNm BMC = +(10.4 x 4) – (15 x 2) = +11.60 kNm BMD = +(10.4 x 5) –(15 x 3) –(10 x 1) = -3.00 kNm 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 25

BMD 15KN 10KN 3KN 2M 2M 1M 1M 3.0 A B C E D 11.6 20.8