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Fig. 495 shows a fixed bcam AB of span / carrying a uniformly distiibutcd load h> per unit run over ihe whole span

By symmetry, tlić end momenls M« and A/* are equal.

The free B.M. diagram is a parabola whose central ordinate

Equating tbe areas of the fixed and free B.M.diagrams, wchave,

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stunoth of HAnmiAU

Hcnce at any scction distant x from the left cod A the actual <• bending moment is given by

M — Free B M.—Fixcd B.M.

_u_ wl wjr wP " 2* 2 “12

For tbe points of contrafleiurc,

wl wx* wP _

2* 2    i2~°

Solving this quadratic we get.

/

*"■ 2 ±2^3

Hence two points of contraflexure occur. Thesc arc cquidistanf from thc centre of thc span Eacb point of contraflexure is at a

distaoce of from the centrc of the span.

Slop* and defltetion

Tbe bending moment at any scction is given by,

^{M El}óx'~ 2^X 2    12

Infegrating, we get.

eA -*¹?—»+Ci <Sfo(*

QX 4 O 12

At

x~0, -

dx

= 0

Ci* o

Integrating again, we get,

wlxⁱ wx* wPx*

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eguahon) At

12 “'”24    24~ ^+C: iDffi^fCllon

x~Q,y~0

Ct-0

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For tbe deflection at the centre, puttingx= j-in the deflec-tion equatioo, we get.

11:26 PM 6/12/2017