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Determine expressions for the distributions of bending moment and shearing force for the horizontal beam of Figure. Hence, sketch the bending and shearing force diagrams.

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Firstly, it will be necessary to calculate the unknown reactions RA and RB.

Taking moments about B gives:

RA × 5 m + 10 kN × 1 m = 5 kN × 7 m + 6 kN × 3 m

i.e. 5RA + 10 = 35 + 18 5RA = 35 + 18 − 10 = 43

from which,RA = 53/5 = 8.6 kN

Resolving forces vertically gives:

RA + RB = 5 kN + 6 kN + 10 kN

i.e. 8.6 + RB = 21

from which, RB = 21 − 8.6 = 12.4 kN

For the range C to A, see Figure(b)

To calculate the bending moment distribution (M), only the resultant of the moments to the left of the section at x will be considered, as the resultant of the moments on the right of the section of x will be exactly equal and opposite.

Bending moment (BM)

From Figure (b), at any distance x,

M = −5 × x (hogging) = −5x  ......(6.7)

Equation (6.7) is a straight line between C and A.

At C, x = 0, therefore MC = −5 × 0 = 0 kN m

At A, x = 2 m, therefore MA = −5 × 2

= −10 kN m

Shearing force (SF)

To calculate the shearing force distribution (F) at any distance x, only the resultant of the vertical forces to the left of x will be considered, as the resultant of the vertical forces to the right of x will be exactly equal and opposite. From Figure (b), at any distance x,

F = −5 kN .....(6.8)

It is negative, because as the left of the section tends to slide downwards the right of the section tends to slide upwards. (Remember, right hand down is positive).

For the range A to D, see Figure(c)

Bending moment (BM)

At any distance x between A and D

Shearing force (SF)

At any distance x between A and D,

F = −5 kN + 8.6 kN = 3.6 kN (constant) ......(6.10)

For the range D to B, see Figure(d).

Bending moment (BM)

For the range B to E, see Figure.(e)

Bending moment (BM) 

In this case it will be convenient to consider only the resultant of the couples to the right of x (− remember that only one side need be considered, and in this case, there is only one load to the right of x).

At x, M = −10 × (8 − x) = −80 + 10x ....(6.13)

Equation (6.13) can be seen to be a straight line between B and E.

Equation (6.14) is positive because the shearing force is causing the right side to slide downwards. The bending moment and shearing force diagrams are plotted in Figure(f) with the aid of equations (6.7) to (6.14) and the associated calculations at C, A, D, B and E. 

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