Chapter 5, Problem 5.5.26P

Consider the compound beam with segments AB and BCD joined by a pin connection (moment release) just right of B (see figure part a). The beam cross section is a double-T made up from three 50 mm × 150 mm wood members (actual dimensions, see figure part b),

(a) Find the cent raid C of the double-T cross section (c₁:c₂): then compute the moment of inertia, [I₂ (mm⁴ )].

(b) Find the maximum tensile normal stress i_fand maximum compressive normal stress t_t. (kPa) for the loading shown. (Ignore the weight of the beam.)

  

To determine

The centroid $C$ of the double T cross section.

The moment of inertia of the section.

Answer to Problem 5.5.26P

The centroid $C$ of the double T cross section.

The moment of inertia of the section.

Explanation of Solution

Given information:

Width of the beam is $150\text{mm}$ , height of the beam is $50\text{mm}$ .

The following figure shows the distribution of forces on the beam

  

Figure-(1)

  

Figure-(2)

Write the expression for the distance of the centroid from the bottom.

  $c_2=\frac{2A_x\times \left(\frac{h}{2}\right)+A_x\left(h+\frac{b}{2}\right)}{3A_x}$.....(I)

Here, the for the distance of the centroid from the bottom is $c_2$ , the area of the section is $A_x$ , the height of the section is $h$ and the width of the section is $b$ .

Write the expression for the sum of centroids.

  $c_1+c_2=b+h$.....(II)

Here, the for the distance of the centroid from the top is $c_1$

Write the expression for moment of inertia.

  $I_z=2\left(\frac{1}{12}\right)b{h}^3+2A{\left(c_2-\frac{h^2}{2}\right)}^2+\frac{h{b}^3}{12}+A{\left(c_1-\frac{b^2}{2}\right)}^2$ .....(III)

Here, the moment of inertia is $I_z$ .

Calculation:

Substitute $50\text{mm}$ for $h$ , $150\text{mm}$ for $b$ , and $7500{\text{mm}}^2$ for $A_x$ in Equation (I).

  $\begin{array}{c}{c}_2=\frac{2\times \left(7500{\text{mm}}^2\right)\times \left(\frac{150\text{mm}}{2}\right)\text{+}\left(\text{7500}{\text{mm}}^2\right)\times \left(150\text{mm}+\frac{50\text{mm}}{2}\right)}{\left(3\times 7500{\text{mm}}^2\right)}\\ =\left(\frac{150+175}{3}\right)\text{mm}\\ \text{=108}\text{.33}\text{mm}\end{array}$

Substitute $150\text{mm}$ for $b$ , $50\text{mm}$ for $h$ and $108.33\text{mm}$ for $c_2$ in Equation (II).

  $\begin{array}{c}{C}_2=150\text{mm}+50\text{mm}-108.33\text{mm}\\ \text{=200}\text{mm}-\text{108}\text{.33}\text{mm}\\ \text{=91}\text{.7}\text{mm}\end{array}$

Substitute $150\text{mm}$ for $b$ , $50\text{mm}$ for $h$ and $108.33\text{mm}$ for $c_2$ , $91.7\text{mm}$ for $c_1$ in Equation (III).

  $\begin{array}{c}{I}_z=\left[\begin{array}{l}\frac{\left(50\text{mm}\times {\left(150\text{mm}\right)}^3\right)}{6}+\left(\left(2\times 7.5\times {10}^3{\text{mm}}^2\right)\times {\left(108.33-\left(\frac{\left(50\text{mm}\right)}{2}\right)\right)}^2\right)\\ +\left(\frac{\left(150\text{mm}\times {\left(50\text{mm}\right)}^3\right)}{12}\right)+\left(\left(7.5\times {10}^3{\text{mm}}^2\right)\times {\left(91.7\text{mm}-\left(\frac{\left(25\text{mm}\right)}{2}\right)\right)}^2\right)\end{array}\right]\\ =28125000{\text{mm}}^4+104158333{\text{mm}}^4+1562500{\text{mm}}^4+47044800{\text{mm}}^4\\ =18.08\times {10}^7{\text{mm}}^4\end{array}$

Conclusion:

The centroid $C$ of the double T cross section is $200\text{mm}$ .

The moment of inertia of the section $18.08\times {10}^7{\text{mm}}^4$ .

To determine

The maximum tensile normal stress.

The maximum compressive normal stress

Answer to Problem 5.5.26P

The maximum tensile normal stress is ${\sigma }_t=5506\text{kPa}$ .

The maximum compressive normal stress is ${\sigma }_c=4659\text{kPa}$ .

Explanation of Solution

The following figure shows the distribution of forces on the beam.

  

Figure-(3)

  

Figure-(4)

Write the expression for the moment about point $B$ .

  $A_y\times AB+M_a=0$ .....(IV)

Here, the length of the member is $AB$ , the vertical reaction at point $A$ is $A_y$ , and the moment about point $A$ is $M_a$ .

Write the expression for the vertical equilibrium of forces.

  $A_y+B_y=0$.....(V)

Here, the vertical reaction at point $B$ is $B_y$ .

Write the expression for moment at point $D$ .

  $A_y\times BD+C_y\times CD=F_E\times ED+F_F\times DF+F_G\times GD+M_a$.....(VI)

Here, the distance between the point $B$ and $D$ is $BD$ , the distance between the points $D$ and $E$ is $ED$ , the distance between the points and $D$ is $CD$ , the distance between the points $D$ and $F$ is $DF$ , the distance between the points $G$ and $D$ is $GD$ , the force on point $E$ is $F_E$ , the force on the point $F$ is $F_F$ , the force on the point $G$ is $F_G$ ,and the vertical reaction at point $C$ is $C_y$ .

Write the expression for the vertical equilibrium of forces of free body $BCD$ .

  $C_y+D_y=F_G+F_F+F_E$ .....(VII)

Here, the vertical reaction at point $D$ is $D_y$ .

Write the expression for the moment about point E

  $M_E=D_y\times DE$ .....(VIII)

Here, the moment about point $E$ is $M_E$ .

Write the expression for the moment about point $C$ .

  $M_c=D_y\times CD-F_E\times DE$.....(IX)

Here, the moment about point $C$ is $M_C$ .

Write the expression for moment about point F

  $M_F=D_y\times CD-F_E\times DE-C_y\times CD$.....(X)

Here, the moment about point $F$ is $M_F$ .

Write the expression for moment about point $G$ .

  $M_G=D_y\times CD-F_E\times DE-C_y\times CD-F_F\times GF$ .....(XI)

Here, the moment about point $G$ is $M_G$ .

Write the expression for moment about point $G$ .

  $M=D_y\times CD-F_E\times DE-C_y\times CD-F_F\times GF-F_G\times BG+M_a$ .....(XII)

Here, the total moment about point $B$ is $M_B$ .

Write the expression for maximum tensile stress.

  ${\sigma }_t=\frac{M\times c_2}{I_z}$ .....(XIII)

Here, the maximum tensile stress is ${\sigma }_{ten}$ , the distance of centroid is $c_2$ .

Calculation:

Substitute $3\text{m}$ for $AB$ , $600\text{N}\cdot \text{m}$ for $M_a$ , and in Equation (IV).

  $\begin{array}{l}\left(A_y\times 3\text{m}\right)+\left(600\text{N}\cdot \text{m}\right)\text{=0}\\ \left(A_y\times 3\text{m}\right)=-\left(600\text{N}\cdot \text{m}\right)\\ A_y=-200\text{N}\end{array}$

Substitute $-200\text{N}$ for $A_y$ , in Equation (V).

  $\begin{array}{l}{B}_y=-A_y\\ B_y=200\text{N}\end{array}$

Substitute $-200\text{N}$ for $A_y$ , $6\text{m}$ for $BD$ , $3\text{m}$ for $CD$ , $1730\text{N}\text{}$ for $F_E$ , $\text{1380}\text{N}$ for $F_F$ , $\text{690}\text{N}$ for $F_G$ ,and $\text{600}\text{N}\cdot \text{m}$ for $M_a$ , $1.5\text{m}$ for $ED$ , $\text{4}.5\text{m}$ for $GF$

  $5\text{m}$ for $GD$ rin Equation (VI).

  $\begin{array}{l}\left(-200N\times 6\text{m}\right)+\left(C_y\times 3\text{m}\right)=\left[\left(\begin{array}{l}\left(1730\text{N}\times 1.5\text{m}\right)+\left(1380\text{N}\times 4.5\text{m}\right)\\ +\left(690\text{N}\times 5\text{m}\right)+600\text{N}\cdot \text{m}\end{array}\right)\right]\\ \left(-\text{1200}\text{N}\cdot \text{m}\right)\text{+}\left({\text{C}}_y\times 3\text{m}\right)=\left(2595\text{N}\cdot \text{m}+6210\text{N}\cdot \text{m}+3450\text{N}\cdot \text{m}+600\text{N}\cdot \text{m}\right)\\ \left(C_y\times 3\text{m}\right)=\left(1105\text{N}\cdot \text{m}+1200\text{N}\cdot \text{m}+600\text{N}\cdot \text{m}\right)\\ C_y=4285\text{N}\end{array}$

Substitute $4285\text{N}$ for $C_y$ , $\text{690}\text{N}$ for $F_G\text{}$

  $690\text{N}$ for $F_G$ , $1380\text{N}$ for $F_F$ , $1730\text{N}$ for $F_E$ in Equation (VII).

  $\begin{array}{l}4285\text{N}+D_y=\left(690\text{N}+1380\text{N}+1730\text{N}\right)\\ D_y=3800\text{N}-4285\text{N}\\ D_y=-485\text{N}\end{array}$

Substitute $-485\text{N}$ for $D_y$ and $1.5\text{m}$ for $DE$ in Equation (VIII).

  $\begin{array}{c}{M}_E=\left(-485\text{N}\times \text{1}\text{.5}\text{m}\right)\\ =-\text{727}\text{.5}\text{N}\cdot \text{m}\end{array}$

Substitute $-485\text{N}$ for $D_y$ and $1.5\text{m}$ for $DE$ , and $1730N$ for $F_E$ and $1.5\text{m}$ for $DE$ in Equation (IX).

  $\begin{array}{c}{M}_c=\left(-485\text{N}\times \text{3}\text{m}\right)-\left(\text{1730}\text{N}\times \text{1}\text{.5}\text{m}\right)\\ =\left(-1455\text{N}\cdot \text{m}\right)-\left(2595\text{N}\cdot \text{m}\right)\\ =-4050\text{N}\cdot \text{m}\end{array}$

Substitute $-485\text{N}$ for $D_y$ and $1.5\text{m}$ for $DE$ , and $1730N$ for $F_E$ and $1.5\text{m}$ for $DE$ , $4285\text{N}$ for $C_y$ and $1.5$ for $DY$ in Equation (X).

  $\begin{array}{c}{M}_F=\left(-485\text{N}\times 4.5\text{m}\right)-\left(1730\text{N}\times 3\text{m}\right)+\left(4285\text{N}\times 1.5\text{m}\right)\\ =\left(\left(-2182.5\text{N}\cdot \text{m}\right)-\left(1730\text{N}\times 3\text{m}\right)+\left(4285\text{N}\times 1.5\text{m}\right)\right)\\ =-945\text{N}\cdot \text{m}\end{array}$

Substitute $-485\text{N}$ for $D_y$ and $1.5\text{m}$ for $DE$ , and $1730N$ for $F_E$ and $1.5\text{m}$ for $DE$ , $4285\text{N}$ for $C_y$ and $1.5$ for $DY$ in Equation (XI).

  $\begin{array}{c}{M}_F=\left(-485\text{N}\times 5\text{m}\right)-\left(1730\text{N}\times 3.5\text{m}\right)+\left(4285\text{N}\times 2\text{m}\right)-\left(\text{1380}\text{N}\times \text{0}\text{.5}\right)\\ =\left(-3630\text{N}\cdot \text{m}\right)+\left(4285\text{N}\times 2\text{m}\right)-\left(\text{1380}\text{N}\times \text{0}\text{.5}\right)\\ =-600\text{N}\cdot \text{m}\end{array}$

  

Figure-(5)

Substitute $-4050\text{N}\cdot \text{m}$ for $M_c$ , $18.08\times {10}^7{\text{mm}}^4$ for $I_z$ and $108.3\text{mm}$ for $c_2$ in Equation (XII).

  $\begin{array}{c}{\sigma }_t=\frac{\left(4050\text{N}\cdot \text{m}\times 108.3\text{mm}\right)}{\left(7.969\times {10}^7{\text{mm}}^4\right)}\\ =\frac{\left(4050\text{N}\cdot \text{m}\times \left(108.3\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{\left(18.08\times {10}^7{\text{mm}}^4\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^4}\\ =\left(2568\times {10}^3\text{N}/{\text{m}}^2\right)\left(\frac{1\text{Pa}}{1\text{N}/{\text{m}}^2}\right)\\ =2568\times {10}^3\text{Pa}\end{array}$

  $\begin{array}{c}{\sigma }_t=\left(2568\times {10}^3\text{Pa}\right)\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =2568\text{kPa}\end{array}$

Substitute $-4050\text{N}\cdot \text{m}$ for $M_c$ , $18.08\times {10}^7{\text{mm}}^4$ for $I_z$ and $91.7\text{mm}$ for $c_1$ in Equation (XII).

  $\begin{array}{c}{\sigma }_c=\frac{\left(4050\text{N}\cdot \text{m}\times 91.7\text{mm}\right)}{\left(18.08\times {10}^7{\text{mm}}^4\right)}\\ =\frac{\left(4050\text{N}\cdot \text{m}\times \left(91.7\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{\left(18.08\times {10}^7{\text{mm}}^4\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^4}\\ =\left(20.541\times {10}^3\text{N}/{\text{m}}^2\right)\left(\frac{1\text{Pa}}{1\text{N}/{\text{m}}^2}\right)\\ =20.541\times {10}^3\text{Pa}\end{array}$

  $\begin{array}{c}{\sigma }_c=\left(20.541\times {10}^3\text{Pa}\right)\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =20.54\text{kPa}\end{array}$

Conclusion:

The maximum tensile normal stress is $2568\text{kPa}$ .

The maximum compressive normal stress is $20.541\text{kPa}$ .