SHEAR FORCE DIAGRAM and BENDING MOMENT DIAGRAM

Q.Draw S.FD and B.M.D for the following beam and also calculate the position of the point of contra flexure.

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Q.Draw S.FD and B.M.D for the following beam and also calculate the position of the point of contra flexure. SOLUTION: Total load = 30×11+80= 330 + 80 = 410 To find reactions: TakingĀ  moments about A: RB * 9 = (30 * 11) * 11/2 + 80 * 11 = 1815 + 880 = 2695 … Read more

Q. A beam ABCD is supported at A and D such that AB = 1.5 BC = 2.5 m and AD = 5m . It carries point loads of 30 kN and 20 kN at B and C respectively. It also carries a udl of 10 kN/m including self weight on entire span. Draw SFD and BMD.

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SOLUTION: Total load= 10 * 5 + 30 + 20 = 100kN To find reactions: R{D} * 5 = (10 * 5) * 5/2 + 30 * 1.5 + 20 * 4 = 125 + 45 + 80 = 250 R{D} = 250/5 = 50kN R{A} = 100 – 50 = 50kN SFD (SHEAR FORCE … Read more

Q.A simply supported beam of span 10 m is subjected to 20 kN, 30 kN, 50 kN and 20 kN loads respectively at 1 m, 3m, 5 m and 8 m from left hand support. Draw the S.F.D and B.M.D indicating the values at salient points.

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SOLUTION: Total load = 20 + 30 + 50 + 20 = 120kN To find reactions: Taking moments about A R{B} * 10 = 20 * 1 + 30 * 3 + 50 * 5 + 20 * 8 . .= 20 + 90 + 250 + 160 = 520kN R{B} = 520/10 = 52kN … Read more

Q. A cantilever beam of span 2 m carries three point loads of 2 kN, 1 kN and 1 kN at a distances of 0.3 m. 1.2 m and 2 m respectively from fixed end. Draw S.F and B.M diagrams and find maz, bending moment and indicate where it occurs.

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SOLUTION: Total load on the beam = 2 + 1 + 1 + 4 KN S.F.D: (SHEAR FORCE DIAGRAM) S.F at A = 4kN SFat C = 4 – 2 = 2kN(just right of C) SF at C = 4kN (just left D) SF at D = 2 – 1 = 1kN(just right of D) … Read more

Q. A beam ABCDE of 8 m length with double over hang is loaded as shown in figure. Draw SFD and BMD.

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Q. A beam ABCDE of 8 m length with double over hang is loaded as shown in figure. Draw SFD and BMD. SOLUTION: Total load= 10 + 40 + 20 + (6 * 6) = 106 kN To find reactions Taking moments about B (Anticlockwise moment = clockwise moment) R{D} * 5 + 10 * … Read more

Q. A beam of length 6 m is loaded as shown in figure. Draw SF and BM diagrams by indicating values at silent point.

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Q. A beam of length 6 m is loaded as shown in figure. Draw SF and BM diagrams by indicating values at silent point. SOLUTION: Total load = 15 + 10 + 10 * 1 = 35kN Taking moments about A R{B} * 4 + (10 * 1) * 1/2 = 15 * 3 + … Read more

Q. A beam of length 4 m. It is supported at 1 m and 3 m from left end of the beam. A udl 2 kN/m is acting on complete length Find the reactions.

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Q. A beam of length 4 m. It is supported at 1 m and 3 m from left end of the beam. A udl 2 kN/m is acting on complete length Find the reactions. SOLUTION: Total load = 2 * 4 = 8kN As beam and loading is symmetrical Hence R A =R B = … Read more

Q. A simply supported beam of span 8 m, carries a udl of 20kN / m upto a distance of 4 m from left support and a concentrated load of 40 kN at a distance of 2 m from right support. Draw SFD and BMD.

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SOLUTION: Total load 20 * 4 + 40 = 120kN To find the reactions Taking moments about A R{B} * 8 = 40 * 6 + (20 * 4) * 4/2 = 240 + 160 = 400 R{B} = 400/8 = 50kN R A =Total load – R{B} = 120 – 50 = 70kN SFD … Read more

Q. A simply supported beam of span 5 m carries a udl of 3 kN/m over its left half of the span and also a point load of 10 kN, 1 m from right end. Draw SFD.

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Q. A simply supported beam of span 5 m carries a udl of 3 kN/m over its left half of the span and also a point load of 10 kN, 1 m from right end. Draw SFD. SOLUTION: Total load = 3 * 2.5 + 10 = 17.5kN Taking moments about A R{B} * 5 … Read more

Q. A simply supported beam of span 8 m carries point loads of 10 kN, 30 kN and 20 kN at 2 m, 5 m and 7 m respectively from its LHS in addition to a udl of 7.5 kN/m over its entire length. Construct SF and BM diagrams and state the position and Magnitude of max. Š’.Šœ.

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SOLUTION: Load load = 10 + 30 + 20 + 7.5 * 8 = 120kN To find reactions R{B} * 8 = (7.5 * 8) 8/2 + 10 * 2 + 30 * 5 + 20 * 7 R{B} = 550/8 = 68.75kN R{A} = 120 – 68.75 = 51.25kN SFD ( SHEAR FORCE DIAGRAM) … Read more