Q. A horizontal beam of 12 m long simply supported at its ends, is subjected to vertical loads of 10 kN, 20 kN and 25 kN at 3 m, 7 m and 10 mm from left support respectively along with a udl of 10kN / m over a length of 3 m from left support and 6 kN/m over a length of 2 m from right support. Draw the SF and BM diagrams. Also calculate the max. B.M and its location.

by

SOLUTION:

Total load = 10 + 20 + 25 + 10 * 3 + 6 * 2 = 97 kN

To find the reactions

Taking moments about A

R_{B} * 12 = 10 * 3 + 20 * 7 + 25 * 10 + (10 * 3) * 3/2 (6 * 2)(2/2 + 10)

=30 + 140 + 250 + 45 + 132 = 597

R_{B} = 597/12 = 49.75kN

R_{A} = 97 – 49.75 = 47.25kN

(OR) Check: Taking moments about B

RA * 12 = 25 * 2 + 20 * 5 + 10 * 9 + (6 * 2)(2/2) + (10 * 3)(3/2 + 9)

= 50 + 100 + 90 + 12 + 315 = 567

R_{A} = 567/12 = 47.25kN

SFD ( SHEAR FORCE DIAGRAM)

F A =F max =47.25 kN

F_{c} = 17.25kN

F_{D} = 7.25kN (same in CD)

FD =12.75 (just right of D) (same in DE)

F_{F} = – 37.75kN

F_{B} = – 49.75kN

.. Max + ve SF occurs at A = 47.25kN and

max -veSF occurs at B = 49.75kN

BMD( BENDING MOMENT DIAGRAM)

M_{A} = M_{B} = 0

M_{C} = 47.25 * 3 – 10 * 3 * 3/2 = 96.75kNm

M_{D} = 47.25 * 7 – 10 * 4 – (10 * 3)(3/2 + 4)

= 330.75 – 40 – 165 = 125.75kNm

M_{E} = 49.75 * 2 – (6 * 2) * 2/2

= 99.5 – 12 = 87.5kNm

Max BM occurs at D (where SF changes it sign) and its value = 125.75 kNm (i.e., 7m from support A or 5m from support B).

Leave a Comment