## Chapter 7

## Q. 7.4

A steel T-beam is used in an inverted position to span 400 mm. If, due to the application of the three forces as shown in the figure, the longitudinal strain gage at A (3 mm down from top of beam and at the x location shown) registers a compressive strain of −50 × 10^{−5}, how large are the applied forces?

Given: Dimensions of and strain in T beam.

Find: Magnitude of applied force P.

Assume: Hooke’s law applies.

## Step-by-Step

## Verified Solution

The gage at A, in the upper portion of the cross section, registers a negative strain. This tells us that the bending moment in the beam at A is negative. Using Hooke’s law, we will be able to relate this measured strain to a normal stress in the beam at this point, which we can then relate to the local bending moment. To do these calculations, we will need to know the location of the centroid and the second moment of area of the inverted T cross section.

As in Example 7.3, the centroid is clearly on the vertical line of symmetry. We need z_c , which we will calculate relative to the top of the section:

Hence, z_{\mathrm{C}}=\sum z A / \sum A=10 mm from the top, or 6 mm from bottom. Next comes the second moment of area I:

\begin{array}{cccc} \hline & \boldsymbol{b h}^{\mathbf{3 / 1 2}}\left(\mathrm{mm}^{\mathbf{4}}\right) & \boldsymbol{d}(\mathrm{mm}) & A \boldsymbol{d}^{\mathbf{2}} \left(\mathrm{mm}^{\mathbf{4}}\right) \\ \hline 1 & 576 & 4 & 768 \\ 2 & 64 & 4 & 768 \\ \sum & 640 & & 1536 \\ \hline \end{array}

I =\sum\left(\frac{b h^3}{12}+A d^2\right)=2176 \mathrm{~mm}^4

We are now ready to apply Hooke’s law and find the stress corresponding to the strain measured at A. The beam is steel, so its Young’s modulus is E = 200 GPa. In addition,

\sigma_{x x_A}=E \varepsilon_{x x_A}=\left(200 \times 10^9 \mathrm{~Pa}\right)\left(-50 \times 10^{-5}\right)=-10^8 \mathrm{~Pa}This is measured at z_A = 3 mm from the top of the T, or 7 mm from the neutral axis.

Next, we relate this stress to the internal bending moment at A:

\begin{aligned} \sigma_{x x_A} & =\frac{M_A z_A}{I}, \\ -10^8 \mathrm{~Pa} & =\frac{M_A(0.007 \mathrm{~m})}{2.176 \times 10^{-9} \mathrm{~m}^4}, \end{aligned}

so M_A=\frac{\left(-10^8 \mathrm{~Pa}\right)\left(2.176 \times 10^{-9} \mathrm{~m}^4\right)}{0.007 \mathrm{~m}}=-31.1 \mathrm{Nm} .

We then consider the loading on the beam to relate this local bending moment to the applied loads P. To do this, we must construct an FBD:

Equilibrium requires that \sum F_z=0, or R_1+R_2=5 P, and if we also impose \sum M_1=0 we will have 0.1P + 0.2P + 0.3(3P) − 0.4R_2 = 0, and solving these two equations we have R_1 = 2P and R_2 = 3P. We can then use the method of sections, cutting the beam and ensuring equilibrium of the right half, to find the bending moment at A:

\curvearrowleft \sum M_A=0=M_A-3 P \cdot(0.05 \mathrm{~m})+3 P \cdot(0.150 \mathrm{~m}),-

M_{\mathrm{about} A}=-0.450 P+0.150 P=-0.3 P .

So, knowing that M_A = −31.1 N m and that M_A = −0.3P, we find that

P=\frac{-31.1 \mathrm{Nm}}{-0.3 \mathrm{~m}}=104 \mathrm{~N}