Q. A simply supported beam of span 5 m is carrying a udl of 2kN / m is acting from 0 m to 3 m from LHS and a point load of 4 kN is a acting at 1 m from RHS. Find the reactions.

Q. A simply supported beam of span 5 m is carrying a udl of 2kN / m is acting from 0 m to 3 m from LHS and a point load of 4 kN is a acting at 1 m from RHS. Find the reactions. SOLUTION: To find reactions Total load= (2 * 3) + … Read more

Q. A simply supported beam of span 6 m, carries a udl of 4 kN/m over its right half of the span. A point load 6 kN is acting at 2 m from left hand support. Draw SFD and B.MD.

SOLUTION: Total load = 6 + 4 * 3 = 18kN To find reactions Taking moments about B R{A} * 6 = 6 * 4 + (4 * 3) * 3/2 = 24 + 18 = 42kN.                                  R{A} = … Read more

Q. A simply supported beam of span 6 m, a udl of 2 kN/m is of length 4 m acting at middle of the span. Draw S.F.D and B.M.D

Q. A simply supported beam of span 6 m, a udl of 2 kN/m is of length 4 m acting at middle of the span. Draw S.F.D and B.M.D SOLUTION: Total load = (2 × 4) = 8 kN As the load on the beam is symmetrical Hence RA =RB = Total load / 2 … Read more

Q. A simply supported beam of span 4 m carries a udl of 2 kN/m over its left half of the span. Draw SFD and BMD.

Q. A simply supported beam of span 4 m carries a udl of 2 kN/m over its left half of the span. Draw SFD and BMD. SOLUTION: Total load= 2×2=4 kN To find reactions Taking moments about A RB×4=(2×2)2/2=4 RB=4/4 = 1kN RA = Total load – RB = 4 – 1 = 3kN SFD … Read more

Q. Calculate the max bending moment for a simply supported beam with udl of 2 kN/m throughout its span. Length of beam is 4 m.

Q. Calculate the max bending moment for a simply supported beam with udl of 2 kN/m throughout its span. Length of beam is 4 m. SOLUTION: Given loading on the beam is Symmetrical hence RA = RB= 4 × 2 /2= 4kN BMD (BENDING MOMENT DIAGRAM) MA = MB = 0 MC = 4 × … Read more

Q.A Simply supported beam of span 4 m is carrying a udl of 10 kN/m on entire span. Draw SFD and BMD.

Q.A Simply supported beam of span 4 m is carrying a udl of 10 kN/m on entire span. Draw SFD and BMD. SOLUTION: The load on the beam is symmetrical RA = RB = (10 × 4)/2 = 20kN SFD (SHEAR FORCE DIAGRAM) F{A} = 20kN F{C} = 0kN F{B} = – 20 kN Max … Read more

Q. A simply supported beam of span 6 m subjected to a udl of 2 kN/m over the entire span. Find the support reactions.

Q. A simply supported beam of span 6 m subjected to a udl of 2 kN/m over the entire span. Find the support reactions. SOLUTION: Total load= 6 × 2 = 12kN As the given loading is symmetrical hence RA =RB = Total load / 2 = 12/2 = 6kN .       Q. … Read more

Q. A simply supported beam 8 m long carries point loads of 10 kN, 8 kN and 6 kN at distances of 2 m, 5 m and 6 m respectively from the left support. Draw SF and BM diagrams.

Q. A simply supported beam 8 m long carries point loads of 10 kN, 8 kN and 6 kN at distances of 2 m, 5 m and 6 m respectively from the left support. Draw SF and BM diagrams. SOLUTION: Total load = 10 + 8 + 6 = 24kN To find reactions Taking moments … Read more

Q. A simply supported beam of span 5 m three concentrated loads 10 kN, 15 kN and 10 kN are acting at 1 m, 3 m and 4 m respectively from left hand support. Find the reactions.

Q. A simply supported beam of span 5 m three concentrated loads 10 kN, 15 kN and 10 kN are acting at 1 m, 3 m and 4 m respectively from left hand support. Find the reactions. SOLUTION: Total load = 10 + 25 + 10 = 45 kN To find the reactions Taking moments … Read more

Q. A simply supported beam of span 6 m. Three concentrated load of 10 kN, 20 kN and 30 kN are acting at 2 m, 3 m and 5 m respectively from left hand support. Draw SFD and BMD.

Q. A simply supported beam of span 6 m. Three concentrated load of 10 kN, 20 kN and 30 kN are acting at 2 m, 3 m and 5 m respectively from left hand support. Draw SFD and BMD. SOLUTION: Total load = 10 + 20 + 30 = 60kN To find the reactions Taking … Read more