Question

Draw the shear and moment diagrams for the beam and loading shown. Determination of the shear and moment as functions of position is not required. Take P = 3.75 lb and Q = 8.5 lb.

y 19 P 4 lb с A EX E AB CD 5 in. 5 in. 5 in. 5 in.

Please explain step by step your solution

Thank you

Answer 1

The free Body diagram of the beams including reaction forces at B and BIS shown below, 4 lb р C 5in AB sin RB Sin sin RO Net We can find the value of RB and Roby taking moment about o and B. Moment should be zero at any point. Caefestexos al moment) Z MB=0 (moment force x distance) - 3.75x5 – 8.5x5 + Roxo + Rox10 - 4x15 10 RD- 5x8.5 +4x15 3.75x5 Roz 8-375 Note. Somse sign has to be given to moment in some divec E Mo- o tion) Roxio – 4 x 5 15x3.75 + 8.5x5 10 RB - 3.75*15 + 8.5x5 – 485 78.75 10 RB 78.75 7.875 lb. RB 10 Efy - internal a cat and Note the bet force in y direction is zero 3.75 + 8.5+4. 8.375-7-8750 Now, we hove to calculate the moment at every pants. This can be found by taking aby side i) AL A. Internal moment is zero at A. Since there is calculating moment from

11) ALB 3.75lb MB = – 5x 3.75 + Roxo B А 5 18.75 lbin S. taken The moment that causes sagging as & -ve and moment that is taken as positive by conventors caceses hogging Sagging hogging 3.75lb JQ Mci=-3.756 10 +7.875x + QxO A 5 3 5 C 1.875 lb in 875 lb (v) D 3.75 lb 8.5lb 8.58 Mo, = -3.75 x 15 +3.875x10 t s А S AB 5 5 Moi=-20 lb in. 7.875 lb Ro We got the shear forces at the points and Bending moment at given points cohere force ading. By convention the force downward is ove and force upward is positive.

4th 2 А 1 А. 121 SED (1) 3.15 - 4.37 - 87s BOp t, in) - 18.1S - 20