A simply supported steel shaft of length L = 37 in is loaded with a uniform distributed load w = 6.5 lbf/in and a point load PA = 350 lbf located at a distance a = 11 in from the left end. Using singularity functions, find the deflections at A and at midspan. Note the percent difference between these two values.

The deflection at  A is 1.93×10⁻⁵ in and

The deflection at the midspan is 2.24×10⁻⁵ in

The percentage difference is 15.11 %

Explanation:

Here we have

Taking moments about the right end

350 × 26 + 6.5 × 37²/2 = 13549.25 = R₁×37

R₁ = 13549.25 /37 = 366.2 lbf

Weight of steel shaft=  6.5 × 37 = 240.5 lbf

Total downward force = 350 + 240.5 = 590.5 lbf

∴ R₂ = 590.5 lbf - 366.2 lbf = 224.3 lbf

The bending moment equations are

By integrating once we have

..................(1)

Second integration gives

......................(2)

The boundary conditions are, at x = 0, y = 0 and at x = 37. y = 0

From equation 2 we have at x = 0

Which gives  0 - 0 - 0 + 0 + B Since we ignore the bracket  with negative value

When x = 37, equation 2 becomes

37·C = -1558669.5

C = -42126.2

Therefore we have

When x = A = 11 in, we have

= -386118.133    

y = -386118.133 /EI  = -386118.133 /(200×10⁸ lbf·in²) = -1.93×10⁻⁵ in or 1.93×10⁻⁵ in

At the midspan, we have x = 37/2 in = 18.5

EIy = -449228.023 lbf·in⁴

=  -449228.023 lbf·in⁴/(200×10⁸ lbf·in²) = -2.24×10⁻⁵ in or

2.24×10⁻⁵ in

The percentage difference is

= 0.1511×100 = 15.11 %.

answer: 500 wph.

explanation:

answer:

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