Question #6618

a square prism of wood 50*50 mm in cross section and 300 mm long is subjected to a tensile stress of 4 t/cm2 along it's longitudinal axis and lateral compressive to a tensile stress of 2 t/cm2 along one pair of sides and a lateral tensile stress of 1-0 t/cm2 acting along the other pair of sides.If the value of E of the material is 1.5*10 to the power 5 kg/cm2,calculate it's changed dimensions.poisson's ratio for wood=.4

Expert's answer

A square prism of wood 5050mm50*50 \, \text{mm} in cross section and 300mm300 \, \text{mm} long is subjected to a tensile stress of 4 t/cm² along its longitudinal axis and lateral compressive to a tensile stress of 2 t/cm² along one pair of sides and a lateral tensile stress of 1-0 t/cm² acting along the other pair of sides. If the value of ε\varepsilon of the material is 1.5101.5*10 to the power 5kg/cm25 \, \text{kg/cm}^2, calculate its changed dimensions. Poisson's ratio for wood = .4

Let


{lx=ly=50mm,lz=300mmσx=2000kgcm2,σy=1000kgcm2,σz=4000kgcm2E=1.5105kgcm2\left\{ \begin{array}{c} l _ {x} = l _ {y} = 5 0 m m, l _ {z} = 3 0 0 m m \\ \sigma_ {x} = 2 0 0 0 \frac {k g}{c m ^ {2}}, \sigma_ {y} = 1 0 0 0 \frac {k g}{c m ^ {2}}, \sigma_ {z} = 4 0 0 0 \frac {k g}{c m ^ {2}} \\ E = 1. 5 \cdot 1 0 ^ {5} \frac {k g}{c m ^ {2}} \end{array} \right.σ=EΔllΔl=σlE\sigma = E \frac {\Delta l}{l} \rightarrow \Delta l = \frac {\sigma l}{E}{Δlx=σxlE=20000.51.5105=666105cmΔly=σylE=10000.51.5105=333105cmΔlz=σzlE=4000301.5105=666105=0.8cm\left\{ \begin{array}{r l} & {\rightarrow \Delta l _ {x} = \frac {\sigma_ {x} l}{E} = 2 0 0 0 \cdot \frac {0 . 5}{1 . 5 \cdot 1 0 ^ {5}} = 6 6 6 \cdot 1 0 ^ {- 5} c m} \\ & {\rightarrow \Delta l _ {y} = \frac {\sigma_ {y} l}{E} = 1 0 0 0 \cdot \frac {0 . 5}{1 . 5 \cdot 1 0 ^ {5}} = 3 3 3 \cdot 1 0 ^ {- 5} c m} \\ & {\rightarrow \Delta l _ {z} = \frac {\sigma_ {z} l}{E} = 4 0 0 0 \cdot \frac {3 0}{1 . 5 \cdot 1 0 ^ {5}} = 6 6 6 \cdot 1 0 ^ {- 5} = 0. 8 c m} \end{array} \right.{Lx=lx+Δlx=0.5+666105cmLy=ly+Δly=0.5+333105cmLz=lz+Δlz=3.8cm\left\{ \begin{array}{l} \rightarrow L _ {x} = l _ {x} + \Delta l _ {x} = 0. 5 + 6 6 6 \cdot 1 0 ^ {- 5} c m \\ \rightarrow L _ {y} = l _ {y} + \Delta l _ {y} = 0. 5 + 3 3 3 \cdot 1 0 ^ {- 5} c m \\ \rightarrow L _ {z} = l _ {z} + \Delta l _ {z} = 3. 8 c m \end{array} \right.

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