Question #165331

A sagging floor is jacked up and a steel girder 10.0 m long whose cross-sectional area is 6.0 inches2 is put in place underneath. when a jack is removed, a sensitive strain gauge that shows that the girder has been compressed by 0.0080 cm. Find the load the girder is supporting. (use metric system units)


Expert's answer

Explanations & Calculations


  • The cross-sectional area of the girder in metric units is

A=6 inch2=6×(0.0254 m)2=0.00387 m2\qquad\qquad \begin{aligned} \small A&= \small 6\,inch^2\\ &= \small 6\times (0.0254\,m)^2\\ &= \small 0.00387\,m^2 \end{aligned}

  • If the load being supported by the girder is F\small F, then simply using the equation FA=YeL\frac{F}{A}=Y\frac{e}{L}, the needed magnitude of the load can be calculated
  • What happens here is that the girder is compressed within its elastic limits due to the load it supports & the compression is measured to be 0.008cm
  • Therefore, using the above equation,

F=AYeL=0.00387×Y×0.00008 m10m=3.096×10−8Y\qquad\qquad \begin{aligned} \small F&= \small AY\frac{e}{L}\\ &= \small 0.00387\times Y\times \frac{0.00008\,m}{10m}\\ &= \small 3.096\times10^{-8}Y \end{aligned}

  • Young's modulus differs with respect to the class of Steel: which varies from 190GPA-215GPA @room temperature. Therefore, using an appropriate value according to the requirements, the load it supports can be calculated.
  • As widely used structural steels have their modulus around 210GPA. Therefore, using that value to the situation described here, the load which was supported can be calculated

F=3.096×10−8 m2×210×109 Nm−2=6501.6 N\qquad\qquad \begin{aligned} \small F&= \small 3.096\times 10^{-8}\,m^2\times 210\times 10^9\,Nm^{-2}\\ &= \small \bold{6501.6\,N} \end{aligned}


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