A simply supported beam AB of span 4m is carrying point loads l0N, 6N & 4N at lm, 2m & 3m respectively from support A. Calculate a) reactions at supports A and B; b) magnitude and direction of the resultant
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A simply supported beam AB of span 4m is carrying point loads l0N, 6N & 4N at lm, 2m & 3m respectively from support A. Calculate a) reactions at supports A and B; b) magnitude and direction of the resultantA simply supported beam AB of span 4m is carrying point loads l0N, 6N & 4N at lm, 2m & 3m respectively from support A. Calculate a) reactions at supports A and B; b) magnitude and direction of the resultant
B.M at A and B= 0 Since support A and B are simple, B.M at C = 5.6 × 1.5=8.4 kN-m, B.M at D = 6.4× 1.5=9.6 kN-m.
- The reaction is nothing, and the process is reacting with one or more reactants or elements, giving a new product, which may be one or two products.
- The reaction is giving the combination or combined to provide the new reactants or products.
- There are many types of reactions present.
- Each reaction has to attain their work to give that product.
- Velocity is the state of the direction of the object or particle.
- Velocity is the scalar portion.
- It is the speed of transition of the distance.
- The standard unit is m/s.
- Velocity has both volume and movement.
- Velocity is the motion of an object or particle.
- Velocity is also called the rate of displacement.
Support Reactions:
∑MA=0
5×1.5+7×3.5−RB×5=0
5×RB=32
RB=6.4kN
∑Fy=0
RA+RB−5+7=0
RA+RB=12
RA=5.6kN
SF calculations
SFatA=+5.6kN
CL=+5.6kN
CR=5.6−5=0.6kN
DL=+0.6kN
DR=+0.6−7=−6.4kN
BL=−6.4kN
B=+6.4−6.4=0kN(∴ ok
B.M. calculation:-
B.M at A and B= 0 Since support A and B are simple.
B.M at C = 5.6 × 1.5=8.4 kN-m
B.M at D = 6.4× 1.5=9.6 kN-m
Therefore, B.M at A and B= 0 Since support A and B are simple, B.M at C = 5.6 × 1.5=8.4 kN-m, B.M at D = 6.4× 1.5=9.6 kN-m.
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