Thursday, 7 July 2011

DEFLECTION BEAM PROBLEM

Problem 403

Beam loaded as shown in Fig. P-403. See the instruction.

Concentrated loads in simple beam

Solution 403

From the load diagram:
$ \Sigma M_B = 0 $
$ 5R_D + 1(30) = 3(50) $
$ R_D = 24 \, \text{kN} $

$ \Sigma M_D = 0 $
$ 5R_B = 2(50) + 6(30) $
$ R_B = 56 \, \text{kN} $

Segment ABSegment AB:
$ V_{AB} = -30 \, \text{kN} $
$ M_{AB} = -30x \, \text{kN}\cdot\text{m} $



403-segment-bc.jpgSegment BC:
$ V_{BC} = -30 + 56 $
$ V_{BC} = 26 \, \text{kN} $
$ M_{BC} = -30x + 56(x - 1) $
$ M_{BC} = 26x - 56 \, \text{kN}\cdot\text{m} $



403-segment-cd.jpgSegment CD:
$ V_{CD} = -30 + 56 - 50 $
$ V_{CD} = -24 \, \text{kN} $
$ M_{CD} = -30x + 56(x - 1) - 50(x - 4) $
$ M_{CD} = -30x + 56x - 56 - 50x + 200 $
$ M_{CD} = -24x + 144 \, \text{kN}\cdot\text{m} $



Load, Shear, and Moment Diagrams

To draw the Shear Diagram:

  1. In segment AB, the shear is uniformly distributed over the segment at a magnitude of -30 kN.
  2. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.
  3. In segment CD, the shear is uniformly distributed at a magnitude of -24 kN.

To draw the Moment Diagram:

  1. The equation MAB = -30x is linear, at x = 0, MAB = 0 and at x = 1 m, MAB = -30 kN·m.
  2. MBC = 26x - 56 is also linear. At x = 1 m, MBC = -30 kN·m; at x = 4 m, MBC = 48 kN·m. When MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B.
  3. MCD = -24x + 144 is again linear. At x = 4 m, MCD = 48 kN·m; at x = 6 m, MCD = 0.


Problem 409

Cantilever beam loaded as shown in Fig. P-409. See the instruction.

 

Uniform load in a segment of cantilever beam

 

Solution 409

409-segment-ab.jpgSegment AB:
$ V_{AB} = -w_ox $
$ M_{AB} = -w_ox(x/2) $
$ MAB = -\frac{1}{2}w_ox^2 $

 

409-segment-bc.jpgSegment BC:
$ V_{BC} = -w_o(L/2) $
$ V_{BC} = -\frac{1}{2} w_oL $
$ M_{BC} = -w_o(L/2)(x - L/4) $
$ M_{BC} = -\frac{1}{2} w_oLx + \frac{1}{8} w_oL^2 $

 

409-load-shear-moment.jpg

To draw the Shear Diagram:

  1. VAB = -wox for segment AB is linear; at x = 0, VAB = 0; at x = L/2, VAB = -½woL.
  2. At BC, the shear is uniformly distributed by -½woL.

To draw the Moment Diagram:

  1. MAB = -½wox2 is a second degree curve; at x = 0, MAB = 0; at x = L/2, MAB = -1/8 woL2.
  2. MBC = -½woLx + 1/8 woL2 is a second degree; at x = L/2, MBC = -1/8 woL2; at x = L, MBC = -3/8 woL2.

Problem 605

Determine the maximum deflection δ in a simply supported beam of length L carrying a concentrated load P at midspan.

 

Solution 605

Concentrated Load at Midspan of Simple Beam

 

$ EI \, y'' = \frac{1}{2}Px - P\langle x - \frac{1}{2}L \rangle $
$ EI \, y' = \frac{1}{4}Px^2 - \frac{1}{2}P\langle x - \frac{1}{2}L \rangle^2 + C_1 $
$ EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 + C_1x + C_2 $

 

At x = 0, y = 0, therefore, C2 = 0

 

At x = L, y = 0
$ 0 = \frac{1}{12}PL^3 - \frac{1}{6}P\langle L - \frac{1}{2}L \rangle^3 + C_1L $
$ 0 = \frac{1}{12}PL^3 - \frac{1}{48}PL^3 + C_1L $
$ C_1 = -\frac{1}{16}PL^2 $

 

Thus,
$ EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 - \frac{1}{16}PL^2x $

 

Maximum deflection will occur at x = ½ L (midspan)
$ EI \, y_{max} = \frac{1}{12}P(\frac{1}{2}L)^3 - \frac{1}{6}P (\frac{1}{2}L - \frac{1}{2}L)^3 - \frac{1}{16}PL^2 (\frac{1}{2}L) $
$ EI \, y_{max} = \frac{1}{96}PL^3 - 0 - \frac{1}{32}PL^3 $
$ y_{max} = -\dfrac{PL^3}{48EI} $

 

The negative sign indicates that the deflection is below the undeformed neutral axis.

 

Therefore,
$ \delta_{max} = \dfrac{PL^3}{48EI}\,\, $            answer


Problem 607

Determine the maximum value of EIy for the cantilever beam loaded as shown in Fig. P-607. Take the origin at the wall.

 

Cantilever Beam with Point Load

 

Solution 607

$ EI \, y'' = -Pa + Px - P\langle x - a \rangle $
$ EI \, y' = -Pax + \frac{1}{2} Px^2 - \frac{1}{2} P\langle x - a \rangle^2 + C_1 $
$ EI \, y = -\frac{1}{2}Pax^2 + \frac{1}{6} Px^3 - \frac{1}{6} P\langle x - a \rangle^3 + C_1x + C_2 $

 

607-deflection-of-cantilever-beam.jpg

 

At x = 0, y' = 0, therefore C1 = 0
At x = 0, y = 0, therefore C2 = 0

 

Therefore,
$ EI \, y = -\frac{1}{2}Pax^2 + \frac{1}{6} Px^3 - \frac{1}{6} P\langle x - a \rangle^3 $

 

The maximum value of EI y is at x = L (free end)
$ EI \, y_{max} = -\frac{1}{2}PaL^2 + \frac{1}{6} PL^3 - \frac{1}{6} P(L - a)^3 $
$ EI \, y_{max} = -\frac{1}{2}PaL^2 + \frac{1}{6} PL^3 - \frac{1}{6} P(L^3 - 3L^2a + 3La^2 - a^3) $
$ EI \, y_{max} = -\frac{1}{2}PaL^2 + \frac{1}{6}PL^3 - \frac{1}{6}PL^3 + \frac{1}{2}PL^2a - \frac{1}{2}PLa^2 + \frac{1}{6}Pa^3 $
$ EI \, y_{max} = -\frac{1}{2}PLa^2 + \frac{1}{6}Pa^3 $
$ EI \, y_{max} = -\frac{1}{2}PLa^2 + \frac{1}{6}Pa^3 $
$ EI \, y_{max} = -\frac{1}{6}Pa^2(3L - a) \,\, $            answer






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