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Simply supported beam with both side overhanging subjected to U.D.L

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Consider a simply supported beam with overhand on both sides. The beam is carrying a U.D.L., over the entire

length as shown in figure.

Consider a simply supported beam with overhang on both sides. The beam is carrying a U.D.L., over the entire

length as shown in figure.

To draw SFD and BMD we need RA and RB.

Since, the beam shown in figure is symmetrically loaded,

RA = RB = wL/2 …(1)

Taking section (X – X’) between D and B, at a distance x from point D, the SF at the section is given by

Fx = + w. x …(2)

Therefore, at x = 0 FD = 0

x = a; FB = + w

Now, taking section between B and A, at a distance x from D.

Fx = + wx = wL/2 …(3)

At x = a; FB = + w (a – L/2)

The position where Fx = 0 can be obtained from equation (3)

w.x w.L/2 = 0 x = L/2

The values of SF at remaining salient points can be obtained by symmetry.

The bending moment at section (X – X’) between D and B is given by:

Mx = –wx2/2 …(4)

Therefore, at

x = 0; MD = 0

x = a; MB = –wa2 / 2

The bending moment at point E is;

ME = –wL/2 × L/4 + wL/2 (L/2–a)

ME = –wL2/8 + wL2/4 – w.L.a/2

ME = w.L2/8 – wL.a/2 …(5)

Now, there may be three cases

A < L / 4

⇒ wL2/8 – w.L.a/2 > 0

⇒ ME > 0 i.e., ME is positive.

As shown in figure

a > L/4

wL2/8 – w.L/4 . L/4 = 0

⇒ ME = 0 i.e., ME is zero,

As shown in figure

a > L/4

wL2/8 – w.L.a/2 < 0

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⇒ ME < 0 i.e., ME is negative

As shown in figure

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