STRENGTH OF MATERIAL

Q. A simply supported beam 6m span is carrying a udl of 4kN / m over its right half of its span and a point load of 10 kN at mid span. Draw SF and BM diagrams and find value of maximum hending moment.

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SOLUTION: Total load= 10 + 4 * 3 = 22kN To find the reactions Taking moments about A R{B} * 6 =10 * 3 + (4 * 3) (3/2+3) = 84 R_{B} = 84/6 = 14kN R{A} = 22 – 14 = 8kN SFD SHEAR FORCE DIAGRAMĀ  F{A} = 8kN (same in AC) Fc (just … Read more

Q. A simply supported beam AB of span 6m long carries a udl of 10 kN/m over a length of 2 m from the right support. In addition to that it carries two point loads of 12 kN and 15 kN at a distances of 2 m and 3 m respectively from left support. Draw SF and BM diagrams and indicating values at salient points.

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SOLUTION: Total load = 12 + 15 + (10 * 2) = 47kN To find the reactions Taking moments about B R{A} * 6 = (10 * 2) * 2/2 + 15 * 3 + 12 * 4 = 20 + 45 + 48 = 113 R{A} = 113/6 = 18.83kN R{B} = 47 – … Read more

Q. A simply supported beam of span 6 m carries udl of 10kN / m upto a distance of 2 m from left support and a point load of 15 kN is placed at a distance of 2 m from right support. Draw the SFD and BMD and indicate where max shear force and max B.M occurs.

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SOLUTION: Total load = 10 * 2 + 15 = 35 kN To find reactions Taking moment about A R_{B} * 6 = 15 * 4 + (10 * 2) * 2/2 = 80kN R_{B} = 80/6 = 13.33kN R_{A} = 35 – 13.33 = 21.67kN SFD (SHEAR FORCE DIAGRAM) F_{A} = 21.67kN F_{B} = … Read more

Q. A simply supported beam of span 12 m carries a udl of 5kN / m over a length of 6 m on middle portion of the beam. In addition to that it carries two point loads of 20 kN and 15 kN at 1.5 m and 10 m from left hand support. Draw the SF and BM diagrams and indicate their max values.

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SOLUTION: Total load = 20 + 15 + 5 * 6 = 65kN To find the reactions Taking moments about A R{B} * 12 = 20 * 1.5 + 15 * 10 + (5 * 6)(6/2 + 3) = 360 R{B} = 360/12 = 30kN R{A} = 65 – 30 = 35kN SFD ( SHEAR … 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

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.

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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.

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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.Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  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

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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 Ɨ 4) = 8 kN As the load on the beam is symmetrical Hence RA =RB = Total load / 2 … Read more