Question #91440
A rectangular plank of wood of length 2.8m is floating horizontally in still water, as shown in figure. If the cross-section of the wooden plank is of rectangular form, of width B= 280 mm and of thickness, D=80 mm, determine the maximum bending stress in the plank, assuming that concentrated mass of 320 kg is placed at its mid span. Let g=9.81m/s^2. Neglect the self-weight of the plank.
1
Expert's answer
2019-07-08T10:28:12-0400

bending stress (the value of the bending stress will vary linearly with distance from the neutral axis (below the yield strength of the material)):


σ=MyI\sigma=\frac {My}{I}

where:

I - second moment of area of the rectangular cross-section about its neutral axis:


I=BD312=0.000012 m4I=\frac{BD^3}{12}=0.000012\space m^4

y - the distance in the beam of rectangular cross section from the neutral axis:


y=D/2=0.04 my=D/2=0.04 \space m

M - the maximum bending moment:


M=12(L2)(wL2)M=\frac 1 2 (\frac L 2) (\frac {wL} 2)

weight of the mass 320 kg:


W=mg=320kg9.81ms2=3139.2 NW=mg=320kg\cdot9.81\frac{m}{s^2}=3139.2\space N

water pressure on the beam as a result of Newton's third law:


wL=WwL=W


w - pressure per unit length


w=mgL=1121.1 Nmw=\frac{mg}{L}=1121.1\space\frac N m

So


σ=MyI=12(L2)(wL2)(D2)(12BD3)=3mgL4BD2=3678750 Pa\sigma=\frac{My}{I}=\frac{1}{2}(\frac{L}{2})(\frac{wL}{2})(\frac{D}{2})(\frac{12}{BD^3})=\frac{3mgL}{4BD^2}=3678750\space Pa

Answer: 3678750 Pa


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