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# calculate deflection of a supported beam. Engineering calculation?

Calculate the deflection at mid-span of a simply-supported beam of span 5m and flexural stiffness EI = 2x106 Nm2 carrying a central point load, P = 10kN. Use the method of direct integration. (Hint: consider the left hand half of the beam only.)

Thank you!

### 1 Answer

- civil_av8rLv 77 years agoFavorite Answer
Thanks for the problem. It's been a while since I've seen a beam problem. I'm even having to get out my old Mechanics of Materials book for this one... LOL

FBD

^--------V-------^

A..... P...... B

|- L/2 -|

|------- L -------|

1st find the reaction forces

Sum of the moments about A (CCW = +) = 0

0 = -P*L/2 + Fb*L

Fb = P/2

Sum of the forces in the vertical direction (up is positive) = 0

0 = Fa - P + Fb = Fa - P + P/2

Fa = P/2

deflection

y'' = M / (EI) ==> M = y'' EI

Looking at the LHS of the Beam

P/2 ======= P ======= P/2

| ------ x----- |

| ----- L/2 ------- |

| ---------------- L ----------------- |

Moment at X on the LHS = P/2*x + P*(L/2 - x)

EI y'' = M = 1/2*P*x - P*(x - 1/2*L)

Integrate

EI y' = 1/4*P*x^2 - 1/2*P*(x - 1/2*L)^2 + C1

Integrate

EI y = 1/12*P*x^3 - 1/6*P*(x-1/2*L)^3 + C1x + C2

Boundary conditions

y = 0 at x = 0 and y = 0 at x = L

1st boundary condition

0 = 1/12*P*0 - 1/6*P*(0)^3 + 0 + C2

C2 = 0

2nd boundary condition, y = 0 at x = L

0 = 1/12*P*L^3 - 1/48*P*L^3 + C1*L

0 = 1/12*P*L^2 - 1/48*P*L^2 + C1

C1 = PL^2*(1/48 - 1/12)

C1 = -3/48 PL^2

C1 = -1/16*P*L^2

EI*y = 1/12*P*x^3-1/6*P*(x-L/2)^3-1/16*P*L^2*x

Max deflection occurs at x = L/2

EI*y_max = 1/12*P*(L/2)^3 - 0 - 1/16*P*L^2*(L/2)

y_max = -P*L^3 / (48*EI)

Plug in numbers

y_max = -(10000 N * (5m)^3) / (48 * 2*10^6 N m^2)

y_max = 0.013 m

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