points of contraflexureassume m1 m2 and m6 be


Points of Contraflexure:

Assume M1, M2, . . . and M6 be the points of contraflexure, where bending moment changes sign. To determine the position of M2, let a section XX at a distance x from the end C.

M x  = - 15.2 x + 24

- 15.2x + 24 = 0

∴          x2  = 1.579 m

To determine the position of M3 and M4, consider a section XX at a distance x from the end E.

M x   = 15.2x + 24 + 30 ( x - 3) - (( ½) ( x - 3) × 0.6 × ( x - 3) × (( x - 3)/3)

M x  = - 15.2 x + 24 + 30 x - 90 - 0.1 ( x - 3)3

= 14.8 x - 66 - 0.1 ( x - 3)3

= 14.8 x - 66 - 0.1 [ x3  - (3 × π2 × 3) + (3x × 9) - 33 ]

= 14.8x - 66 - 0.1 ( x3  - 9 x2  + 27 x - 27)

=- 0.1x3  + 0.9 x2  + 12.1x - 63.3

On changing the sign and equating it to zero, we obtain

0.1x3  - 0.9 x2  - 12.1 x + 63.3 = 0

Solving out by trial and error, we obatin

x1  = 4.4814 m

x2  = 14.357 m

 and    x3  = - 9.84 m

As, the value of x3 is negative, it must be ignored.

To determine the position of M3, assume a section XX at a distance x from the end E.

M x=  7.2 × x × (x /2) + 15.6

On equating it to zero, we obtain

- 3.6 x2 + 15.6 = 0

x = 2.082 m

The points of contraflexure are at distance of 1 m, 1.579 m, 4.4814 m & 14.357 m from the left end X & at distances of 1 m & 2.082 m from the right end F.

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Mechanical Engineering: points of contraflexureassume m1 m2 and m6 be
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