Unit 2 Engineering Science - Statics and Dynamics
Statics and Dynamics
1. The diagram below shows the loading on one of the horizontal floor beams and associated supporting vertical columns for a proposed building, sited on an incline. In order to consider the worst case, assume that the beam is simply supported and the column is pin jointed at either end.
1.1 Determine the beam reaction forces at each support and .
Taking moments around
(2 x 50) + (6 x 70) + (10 x 20) = (8 x
∴ = = 90 KN
So, = 50 + 70 + 20 = + 90
= 50 KN
Modifications for UDL:
Taking moments around:
[(12 x 10) x 5] = 8 x (additional )
∴ Additional = 75 KN
So, additional = 120 = + 75
= 45 KN
So = 95 KN and = 165 KN
1.2 Draw the beam’s Shear Force diagram.
1.3 Determine the position of the maximum bending moment, measured from .
Maximum bending moment is where the force diagram crosses 0, at 8m and between 2m and 6m.
Bending Moment (BM) at 8m:
(95 x 8) – (50 x 6) – (70 x 2) - [(12 x 8) x 4] = -64KNm
BM between 2m and 6m =
0 = 21 – (12 x
1. The diagram below shows the loading on one of the horizontal floor beams and associated supporting vertical columns for a proposed building, sited on an incline. In order to consider the worst case, assume that the beam is simply supported and the column is pin jointed at either end.
1.1 Determine the beam reaction forces at each support and .
Taking moments around
(2 x 50) + (6 x 70) + (10 x 20) = (8 x
∴ = = 90 KN
So, = 50 + 70 + 20 = + 90
= 50 KN
Modifications for UDL:
Taking moments around:
[(12 x 10) x 5] = 8 x (additional )
∴ Additional = 75 KN
So, additional = 120 = + 75
= 45 KN
So = 95 KN and = 165 KN
1.2 Draw the beam’s Shear Force diagram.
1.3 Determine the position of the maximum bending moment, measured from .
Maximum bending moment is where the force diagram crosses 0, at 8m and between 2m and 6m.
Bending Moment (BM) at 8m:
(95 x 8) – (50 x 6) – (70 x 2) - [(12 x 8) x 4] = -64KNm
BM between 2m and 6m =
0 = 21 – (12 x
Bibliography: – Wikipedia