STRENGTH OF MATERIAL

Q. A simply supported beam of span 10 m carries a udl of 40kN / m over a length of 5 m from LHS and over a length of 2 m beginning from RHS. In addition it carries a point load of 100 kN at 2 m from the RHS. Draw SF and BM diagrams and state the position and magnitude of max. BM.

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SOLUTION: Total load = 40 * 5 + 100 + 40 * 2 = 380kN To find the reactions Taking moments about A R{B} * 10 = (40 * 5) * 5/2 + 100 * 8 + (40 * 2)(2/2 + 8) R{B} = 2020/10 = 202kN R{A} = 380 – 202 = 178kN SFD … 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.                                  R{A} = … Read more

Q. A simply supported beam of span 6 m. A udl of 2 kN/m is acting on the right half of the span. A point load of 10 kN is at 2 m from the LHS. Draw SFD and BMD.

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SOLUTION: Total load = 10 + 2 * 3 = 16 kN To find reactions Taking moments about B R{A} * 6 = 10 * 4 + (2 * 3) * 3/2 = 49 R{A} = 49/6 = 8.17kN R{B} = 16 – 8.17 = 7.83kN SFD( SHEAR FORCE DIAGRAM) F{A} = 8.17kN F{B} = … 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 × 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.

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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.

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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.

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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.

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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. simply supported beam of span 6 m carrying concentrated loads of 30KN, 20 kN, 30 kN and 20 kN at distances of 1 m, 2 m, 3m and 4 m from L.H.S. Find the supported reactions.

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Q. simply supported beam of span 6 m carrying concentrated loads of 30KN, 20 kN, 30 kN and 20 kN at distances of 1 m, 2 m, 3m and 4 m from L.H.S. Find the supported reactions. SOLUTION: To find reactions : Total load = 30 + 20 + 30 + 20 = 100 kN … Read more