A cantilever beam of span 5m. A UDL loads of 10KN is acting between 2m to 5m from Fixed End. Find the reactions.

by

  SOLUTIONS: In this cantilever beam we are taken direct loads are Reactions RA = 10×3 = 30kN Reacting moment at A MA = -( 10×3) ( 3/2 +2 ) = 105kN.(anticlockwise)

A cantilever beam of span 1m and is fixed at A and free at B. It carries UDL loads of 10KN on whole span. Find the reactions.

by

  SOLUTIONS: In this cantilever beam we are taken direct loads are Reactions RA = 10×4 = 40kN Reacting moment at A MA = (10×4) 4/2 =( 40 ) 4/2 = 80kN/m (anticlockwise)

A cantilever beam of span 3m and is fixed at A and free at B. It carries three point loads of 10KN,15kN and 20kN at a distance of 1m, 2m, and 3m respectively From fixed End . Find the reactions.

by

  SOLUTIONS: In this cantilever beam we are taken direct loads are Reactions RA = 10+ 15+20 = 45kN Reacting moment at A MA= -(20×3+15×2 + 10×1) = -(60+30+10) = 100 kN/m

A simply supported beam of 6m span carries a udl of 10 kN/m over the left support 2 m length and also carries point loads of 15 kN acts at a distances of 4 m respectively from left support. Find the support reactions.

by

  SOLUTIONS: Total load= 10×2+15= 35kN Taking moments about A RB×6 = 15×4+ (10×2) 2/2 RB×6 = 80 RB = 80/6 = 13.33kN RA = Total load – RB RA= 35 -13.33 = 21.67kN

Q.A Simply Supported beam of length 4 m carries two point loads of 50 kN and 100 kN are placed at a distance of 1 m and 3 m from the left hand support. Find the reactions.

by

  SOLUTION : Total load  =50+100= 150kN Taking moments about A RB x 4 = 50 x 1 + 100 x 3 RB x 4 = 350 RB= 350/4 = 87.5kN RA= Total load – RB RA= 150 – 87.5 = 62.5kN ( or) Taking moments about B RA x 4 = 100×1 +50×3 RA … Read more

A simply supported beam of 6m span carries a udl of 4 kN/m over the middle 2 m length and also carries point loads of 2 kN and 4 kN acts at a distances of 1 m and 5 m respectively from left support. Find the support reactions.

by

SOLUTION : Total load = 2+4+4×2= 14kN Taking moments about. A RB x 6 = 2 x 1+4×5+(4×2) ( 2 / 2+2) RB x 6= 46 RB= 46 / 6 = 7.67kN RA = Total load – RB  RA= 14 – 7.67= 6.33kN ( or ) Taking moments about B RA x 6 = 4×1+2×5+(4×2) … Read more

A simply supported beam of span 4 m carries a udl of 10 kN/m over the entire span and a point load of 10 kN acting at distance of 1m from let end. Find the reactions at supports.

by

SOLUTION : Total load = 10 x 4 + 10 = 50 Taking moments about A RB x 4 = 10 x 1 + ( 10×4 ) 4/2 RB x 4 = 90kN RB = 90 / 4 = 22.5kN RA= Total load – RB RA= 50 – 22.5 = 27.5kN ( or ) Taking … Read more

A beam of length 8 m carries two point loads of 60 kN and 90 kN are placed at a distance of 3 m and 5 m from the left hand support. Find the reactions.

by

SOLUTION : Total Load = 60+90= 150kN Taking moments about A RB×8 = 60×3 + 90 x 5 RB x 8 = 630 RB = 630/8 = 78.75kN RA = Total load – RB RA= 150 – 78.75  = 71.25kN ( or ) Taking moments about B RA x 8 = 90×3 + 60×5 RA … Read more

A simply supported beam of span 4 m carries a point load 10 kN at a distance of 1 m from left support. Find the reactions at the supports

by

SOLUTION: Total load = 10kN Taking moments about A RB ×4=10×1 RB= 10 / 4 = 2.5kN RA= 10 – 2.5 = 7.5kN Taking moments about B’ RA x 4= 10 x 3 RA = 30/4 = 7.5 kN

A simply supported beam of 4 m span carries a point load of 10kN at its centre. Find the reactions at the supports.

by

A simply supported beam of 4 m span carries a point load of 10kN at its centre. Find the reactions at the supports. Solution: The load applied symmetrical. Hence  RA = RB = Total load/2                 = 10 / 2                 … Read more