Answer to Question #91440 in Mechanics | Relativity for ernest

Question

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.

Answer 1

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

$$\sigma=\frac {My}{I}$$

where:

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

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

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

$$y=D/2=0.04 \space m$$

M - the maximum bending moment:

$$M=\frac 1 2 (\frac L 2) (\frac {wL} 2)$$

weight of the mass 320 kg:

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

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

$$wL=W$$

w - pressure per unit length

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

So

$$\sigma=\frac{My}{I}=\frac{1}{2}(\frac{L}{2})(\frac{wL}{2})(\frac{D}{2})(\frac{12}{BD^3})=\frac{3mgL}{4BD^2}=3678750\space Pa$$

Answer: 3678750 Pa