Physics, asked by samirkhan4413, 2 months ago

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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Answered by Anonymous
2

Answer:

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

Answered by amikkr
0

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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