Problem 10.9. Vibration of a timber plate

A rectangular timber slab with dimensions Lx = 8.0 m and Ly = 6.0 m has stiffness values Dx = D11 = 3000 kNm, Dy = D22 = 750 kNm, Dt = 2D66
= 0 kNm (neglected). The distributed mass is: m = 200 kg/m2. Determine the fundamental frequency of the slab if
a) all four edges are simply supported,
b) one edge parallel to the y axis is free and the other three edges are simply supported (in the Figure EI = 0).
c) What should be the bending stiffness (EI =?) of the edge beam if we want the fundamental frequency to be the average of the frequencies
of case a) and case b)?

Solve Problem

Solve

Problem a)

Natural frequency with hinged supports, fn [Hz]=

Problem b)

Natural frequency with one free edge, fn [Hz]=

Problem c)

Bending stiffness of the supporting beam, EI [kNm2]=

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Steps

Step by step

Problem a)

Step 1.  Give the natural frequency assuming hinged edges.

Show natural frequency

See Table 10.6 first row or Eq.(10-99)

f2=f2x+f2y+f2t=π2D114mL4x+π2D224mL4y+2π2D664mL2xL2y=π2D114mL4x+π2D224mL4y=π2×3000×1034×200×8.04+π2×750×1034×200×6.04f=4.02 

Note that the eigenfrequencies due to the three stiffnesses are fx2=9.04 Hz, fy2=7.14 Hz,  ft2=0 Hz . The above expression is equivalent to the application of Southwell’s expression.

Problem b)

Step 1.  Give the natural frequency assuming one free edge.

Show natural frequency

See Table 10.8 first row, EI =0

f2=1fx2+ft20.25f662+1fEI21+fy2+0.25f662since EI=0      fEI2=3π2EI4mLxLy4=0   and   f662=ft2=0  f= fy=π2D224mLy4=π2×750×1034×200×6.04=2.672 Hz 

Problem c)

Step 1.  Determine the necessary bending stiffness of the supporting beam to obtain the average of the above two frequencies.

Show bending stiffness

See Table 10.8 first row, EI =?

f=4.02+2.6722=3.34 Hzf2=1fx2+ft20.25f662+1fEI21+fy2+0.25f662=1fx2+1fEI21+fy2      fEI2=1f2 fy21fx21=13.342 7.1419.041=7.38 Hz fEI2=3π2EI4mLxLy4      EI=43fEI2mLxLy4π2=437.382200×103×8.0×6.04π2=2068 kNm2

Results 

Worked out solution

Problem a)

The natural frequency of the plate with four hinged edges is:

See Table 10.6 first row or Equation (10-99)

f2=fx2+fy2+ft2=π2D114mLx4+π2D224mLy4+2π2D664mLx2Ly2=π2D114mLx4+π2D224mLy4=π2×3000×1034×200×8.04+π2×750×1034×200×6.04f=4.02 Hz

f2=fx2+fy2+ft2=π2D114mLx4+π2D224mLy4+2π2D664mLx2Ly2=π2D114mLx4+π2D224mLy4=π2×3000×1034×200×8.04+π2×750×1034×200×6.04f=4.02 Hz

Note that the eigenfrequencies due to the three stiffnesses are fx2=9.04 Hz, fy2=7.14 Hz,  ft2=0 Hz . The above expression is equivalent to the application of Southwell’s expression.

Problem b)

The natural frequency assuming one free edge:

See Table 10.8 first row, EI =0

f2=1fx2+ft20.25f662+1fEI21+fy2+0.25f662since EI=0      fEI2=3π2EI4mLxLy4=0   and   f662=ft2=0  f= fy=π2D224mLy4=π2×750×1034×200×6.04=7.14 Hz 

Problem c)

The necessary bending stiffness of the supporting beam is calculated from its natural frequency, which results the average frequency of the plate of case a) and b).

See Table 10.8 first row, EI =?

f=4.02+7.142=3.34 Hzf2=1fx2+ft20.25f662+1fEI21+fy2+0.25f662=1fx2+1fEI21+fy2      fEI2=1f2 fy21fx21=13.342 7.1419.041=7.38 Hz fEI2=3π2EI4mLxLy4      EI=43fEI2mLxLy4π2=437.382200×103×8.0×6.04π2=2068 kNm2