Engineering Studies: Understanding Forces and Reactions in Beams
Explore how to calculate reaction forces in simply supported beams and apply Newton's Third Law and Conditions of Equilibrium. Understand non-concurrent forces and moments, as well as different support types in civil structures.
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Presentation Transcript
Simply Supported Beams Calculating Reaction Forces Year 12 Engineering Studies - Civil Structures
Newtons Third Law and Conditions of Equilibrium Applied Load F = mg Fv = 0 R = F Reaction Force
Newtons Third Law and Conditions of Equilibrium Applied Load mg Fv = 0 0 = mg-RL-RR RL = mg/2 RR = mg/2 Load is equally supported at either end
Newtons Third Law and Conditions of Equilibrium Applied Load mg Fv = 0 0 = mg-RL-RR RR = ? RL = ? More load on left hand side, less on right hand side
Non Concurrent Forces (Moments) Applied Load Applied Load M = Fd
Non Concurrent Forces (Moments) 850N Y X RL RR 1.5m 3.5m
Non Concurrent Forces (Moments) M@X = 0 0 = (RL x 0) + (850 x 1.5) (RR x 5) = 0 + 1275 5 RR RR = 1275/5 RR = 255N M@Y = 0 0 = (RR x 0) - (850 x 3.5) + (RL x 5) = 0 - 2975 + 5 RL RL = 2957/5 RL = 595N FV = 0 0 = -595 + 850 255 = 0 850N Y X RL RR 1.5m 3.5m
Non Concurrent Forces (Moments) 450N 850N Y X RL RR 1.0m 1.5m 2.5m
Non Concurrent Forces (Moments) M@X = 0 0 = (850x1.5) + (450x4) (RRx5) = 1275 + 1800 5 RR RR = 3075/5 RR = 615N M@Y = 0 0 = - (450x1) - (850x3.5) + (RLx5) = - 450 2975 +5RL RL = 3425/5 RL = 685N FV = 0 0 = - 685 + 850 + 450 - 615 = 0 450N 850N Y X RL RR 1.0m 1.5m 2.5m
Support Types Fixed Support can provide horizontal and vertical resistance Roller Support can provide vertical resistance only Fixed Roller
Support Types Accommodating movement in the structures caused by expansion, contraction and bending.
Treating Forces Applied at an Angle 850N Note: X is a roller support so is only capable of providing vertical reaction, therefore all horizontal loads must be resisted at the fixed support Y Y 60o X 1.5m 3.5m
Treating Forces Applied at an Angle 850N 850sin60 = 736.1N Method 1 Resolving Forces draw in reactions (assume direction for RR resolve forces into vertical and horizontal components and calculate values Y 60o X 850cos60 = 245N RRH 1.5m 3.5m RR RL RRV
Treating Forces Applied at an Angle 850N 850sin60 = 736.1N Take moments @ fixed support to calculate vertical reaction at roller support Y 60o X 850cos60 = 245N M@Y = 0 0 = - (736.1 x 3.5) + (RL x 5) RL = 2 575.4 / 5 RL = 515.3 N RRH 1.5m 3.5m RR RL RRV
Treating Forces Applied at an Angle Use F = 0 to calculate RRV and RRH 850N FV +ve = 0 850sin60 = 736.1N 0 = 736.1 515.3 RRV Y 60o X 850cos60 = 245N RRV = 220.8 RRH FH +ve = 0 1.5m 3.5m 0 = RRH 245 RR 515.3N RRH = 245N RRV
Treating Forces Applied at an Angle Calculate RR 850N 245N 850sin60 = 736.1N 220.8N RR Y 60o X 850cos60 = 245N 245N 1.5m 3.5m RR2 = 2452 +220.82 RR = 329.8N RR 515.3N tan = 220.8/245 = 420 220.8N RR = 329.8N 420
Treating Forces Applied at an Angle 850N Method 2 Perpendicular Distance d Calculate perpendicular distance from Y (fixed support) to line of action of force Y 60o X Sin 60 = d / 3.5 d = 3.5 x sin60 d = 3.03 1.5m 3.5m
Treating Forces Applied at an Angle 850N Take moments about fixed support to calculate vertical reaction @ X 3.03 MY = 0 Y 60o X 0 = - (850 x 3.03) + Rx x 5 Rx = 2575.5 / 5 Rx = 515.1 N 1.5m 3.5m
Treating Forces Applied at an Angle Resolve 850N force and then use F = 0 to calculate RRV and RRH 850N FV +ve = 0 850sin60 = 736.1N 0 = 736.1 515.1 RRV Y 60o X 850cos60 = 245N RRH RRV = 221 1.5m 3.5m FH +ve = 0 RR 0 = RRH 245 515.1N RRV RRH = 245N
Treating Forces Applied at an Angle Calculate RR 850N 245N 850sin60 = 736.1N 221N RR Y 60o X 850cos60 = 245N 245N 1.5m 3.5m RR2 = 2452 +2212 RR = 330 N RR 515.1N tan = 221/245 = 420 220.8N RR = 330 N 420
