Problems on single over hanging beam

Q. A beam of length 5 m, simply supported over a span of 4 m and is having 1 m overhang on the right hand support. It carries two point loads 2kN and 4 kN at the distances of 3 m and 5 m from LHS and a udi of 2 kN/m spread over 3 m from LHS. Draw SF and BM diagrama state the position and magnitude of max +ve BM and the position of point of contraflexure.

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SOLUTION: Total load = (2 * 3) + 2 + 4 = 12kN To find reactions Taking moments about A R{B} * 4 = 2 * 3 + 4 * 5 + (2 * 3) * 3/2 R{B} = 35/4 = 8.75kN R{A} = 12 – 8.75 = 3.25kN SFD ( SHEAR FORCE DIAGRAM) F{A} … Read more

Q. Draw the sketch of a beam of 9 m length simply supported at the left end and at 6 m from the left end. It carries a udl 5kN / m over the supported length and a concentrated load of 10 kN at the extreme right end.

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Q. Draw the sketch of a beam of 9 m length simply supported at the left end and at 6 m from the left end. It carries a udl 5kN / m over the supported length and a concentrated load of 10 kN at the extreme right end. SOLUTION: Total load – 5 * 6 … Read more

Q. A beam 8 m long is simply supported at left end and at 6 m from left end. It carries an udl of 5 kNim over the over hanging portion and a print load of 10 kN at 3 m from left hand support. Draw SF and BM diagrams and locate the point of contra flexure.

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SOLUTION: Total load= 10 + 5 * 2 = 20kN To find reactions Taking moments about A R{B} * 6 = 10 * 3 + (5 * 2)(2/2 + 6) = 30 + 70 = 100 R{B} = 100/6 = 16.67kN R{A} = 20 – 16.67 = 3.33kN SFD ( SHEAR FORCE DIAGRAM) F{A} = … Read more

Q. Q.A beam of length 11 m is simply supported on a span of 8 m with an over hang of 3 m in the right hand portion. It carries a udl of 10 kN/m on a length of 3 m from LHS and a point load are 3kN,6kN and 5kN at the distance 5m, 7m and 8m from LHS. Draw SF and BM diagrams and state the position of point of contra flexure.

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SOLUTION: Total load= 10 * 3 + 3 + 6 + 5 = 44 kN To find the reactions Taking moments about A R{B} * 8 = 5 * 11 + 6 * 7 + 3 * 5 + (10 * 3) 3/2 = 55 + 42 + 15 + 45 = 157 R{B} = … Read more

Q.A beam of length 9 m is simply supported on a span of 5 m with an over hang of 4 m in the right hand portion. It carries a udl of 3 kN/m on a length of 5 m from LHS and a point load of 4 kN at 9 m from LHS. Draw SF and BM diagrams and state the position of point of contra flexure.

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SOLUTION: Total load 3 * 5 + 4 = 19 kN To find reactions Taking moments about A R{B} * 5 = 4 * 9 + (3 * 5) * 5/2 = 73.5 R{B} = 73.5/5 = 14.7kN RA =Total load – R{B} = 19 – 14.7 = 4.3kN SFD SHEAR FORCE DIAGRAM  F{A} = … Read more

Q. simply supported beam ABC with suports at A and B, 6 m apart with an over hang of BC = 2m long carries a udl of the whole length. Draw SFD & BMD

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SOLUTION: Total load = 20 * 8 = 160 kN To find the reactions Taking moments about A R{B} * 6 = (20 * 8) * 8/2 = 640 R{B} = 640/6 = 106.67kN R{A} = 160 – 106.67 = 53.33kN SFD (SHEAR FORCE DIAGRAM) F{A} = 53.33kN FB =just left ofB = – 66.67kN … Read more

Q. A beam of length 6 m is supported at 2 m and 6 m from the left end of the beam and carrying 10 kN and 20 kN are acting at 2 m and 6 m from the right end of the beam. Find the support reactions.

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Q. A beam of length 6 m is supported at 2 m and 6 m from the left end of the beam and carrying 10 kN and 20 kN are acting at 2 m and 6 m from the right end of the beam. Find the support reactions. SOLUTION: Total load= 20 + 10 = … Read more