A simply supported beam 100 feet in length Is pinned at its left end and support

Question

A simply supported beam 100 feet in length Is pinned at its left end and supported on a roller at its right end. There is a uniformly distributed load of 10 kips/ft along the entire length of the beam. There is also a concen- trated load of 20 kips downwards at 25 ft from the pinned end of the beam.

Using Table H-2 in the appendix to the text,

develop a mathematical expression

(equation) for the slope dv/dx

develop a mathematical expression

(equation) for the deflection v(x).

at what position (x) on the beam will

the deflection be greatest?

Explanation / Answer

Let Ra = reaction at pinned end and Rb = reaction at other end.

Balancing moments about pinned end:

Rb*100 - (10*100)*(100/2) - 20*25 = 0

Thus, Rb = 505 kips

Balancing vrtical direction forces, Ra + Rb = 10*100 + 20

Thus, Ra = 515 kips

At any section at distance x from pinned end, Moment M(x) = 515*<x-0> - (10/2)*<x-0>^2 - 20*<x-25> + 505*<x-100>

Integration of moment equation:

EIv'' = M(x) = 515*<x-0> - 5*<x-0>^2 - 20*<x-25> + 505*<x-100>

EIv' = (515/2)*<x-0>^2 - (5/3)*<x-0>^3 - (20/2)*<x-25>^2 + (505/2)*<x-100>^2 + C1

EIv = (515/6)*<x-0>^3 - (5/12)*<x-0>^4 - (20/6)*<x-25>^3 + (505/6)*<x-100>^3 + C1*x + C2

Boundary conditions: At x = 0 and at x = 100 ft, we have deflection v = 0.

Putting these, we get

C2 = 0 and

0 = (515/6)*<100-0>^3 - (5/12)*<100-0>^4 - (20/6)*<100-25>^3 + (505/6)*<100-100>^3 + C1*100 + 0

Solving this, C1 = -427604.167

a)

Thus, beam slope dv/dx = 1/(EI) * [(515/2)*<x-0>^2 - (5/3)*<x-0>^3 - (20/2)*<x-25>^2 + (505/2)*<x-100>^2 - 427604.167]

b)

Beam deflection = 1/(EI) *[(515/6)*<x-0>^3 - (5/12)*<x-0>^4 - (20/6)*<x-25>^3 + (505/6)*<x-100>^3 - 427604.167*x]

c)

Deflection will be max. when dv/dx = 0

0 = 1/(EI) * [(515/2)*<x-0>^2 - (5/3)*<x-0>^3 - (20/2)*<x-25>^2 + (505/2)*<x-100>^2 - 427604.167]

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