Chapter 2, Problem 2.6.11P

The plane truss in the figure is assembled From steel C 10 X 20 shapes (see Table 3(a) in Appendix F). Assume that L = 10 ft and b = 0 71 L.

(a) If load variable P = 49 kips, what is the maximum shear stress T_maxin each truss member?

(b) What is the maximum permissible value of load variable P if the allowable normal stress is 14 ksi and the allowable shear stress is 7.5 ksi?

  

To determine

The maximum shear stress in the member AC.

The maximum shear stress in the member AB.

The maximum shear stress in the member BC.

Answer to Problem 2.6.11P

The maximum shear stress in the member AC is = $1.86\text{ksi}$.

The maximum shear stress in the member AB is = $7.42\text{ksi}$.

The maximum shear stress in the member BC is = $9.40\text{ksi}$.

Explanation of Solution

The following figure shows the forces on the truss:

  

       Figure-(1)

Write the expression for the length $AC$.

  $b=0.71L$....... (I)

Write the expression for the length CD.

  $CD=b\sin {\theta }_A$....... (II)

Here, the length of $AC$is $b$, and the angle between member AB and AC is ${\theta }_A$.

Write the expression for the length AD.

  $AD=b\cos {\theta }_A$....... (III)

Write the expression for the angle ${\theta }_B$.

  $\sin {\theta }_B=\frac{CD}{L}$....... (IV)

Here, the length of the member CB is $L$.

Write the expression for the length DB.

  $DB=\sqrt{L^2-C{D}^2}$....... (V)

Write the expression for the length AB.

  $AB=AD+DB$....... (VI)

Write the expression for the moment at point A.

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left(-P\right)\left(AD\right)-\left(2P\right)\left(CD\right)+\left(R_{By}\right)\left(AB\right)=0\end{array}$....... (VII)

Here, the variable load is $P$.

Write the equilibrium equation for the horizontal forces.

  $\begin{array}{l}{\displaystyle \sum F_H}=0\\ R_{Ax}+2P=0\end{array}$....... (VIII)

Write the equilibrium equation for the vertical forces.

  $\begin{array}{l}{\displaystyle \sum F_V}=0\\ R_{Ay}+R_{By}-P=0\end{array}$....... (IX)

Write the expression for the forces at joint A.

  $F_{AC}\sin {\theta }_A=-R_{Ay}$....... (X)

Write the expression for the horizontal forces at joint A.

  $R_{Ax}+F_{AC}\cos {\theta }_A+F_{AB}=0$....... (XI)

Write the expression for the horizontal forces at joint B.

  $F_{AB}-F_{BC}\cos {\theta }_B=0$....... (XII)

Write the expression for the maximum shear stress in the member AC.

  ${\tau }_{AC}=\frac{F_{AC}}{2A}$....... (XIII)

Write the expression for the maximum shear stress in the member AB.

  ${\tau }_{AB}=\frac{F_{AB}}{2A}$....... (XIV)

Write the expression for the maximum shear stress in the member BC.

  ${\tau }_{BC}=\frac{F_{BC}}{2A}$....... (XV)

Calculation:

Substitute $10\text{ft}$for $L$in Equation (I).

  $\begin{array}{c}b=\left(0.71\right)\left(10\text{ft}\right)\\ =7.1\text{ft}\end{array}$

Substitute $7.1\text{ft}$for $b$, and $60°$for ${\theta }_A$in Equation (II).

  $\begin{array}{c}CD=\left(7.1\text{ft}\right)\sin \left(60°\right)\\ =\left(7.1\text{ft}\right)\left(0.866\right)\\ \approx 6.15\text{ft}\end{array}$

Substitute $7.1\text{ft}$for $b$, and $60°$for ${\theta }_A$in Equation (III).

  $\begin{array}{c}AD=\left(7.1\text{ft}\right)\cos \left(60°\right)\\ =\left(7.1\text{ft}\right)\left(0.5\right)\\ =3.55\text{ft}\end{array}$

Substitute $10\text{ft}$for $L$, $6.15\text{ft}$for $CD$in Equation (IV).

  $\begin{array}{l}\sin {\theta }_B=\frac{6.15\text{ft}}{10\text{ft}}\\ \sin {\theta }_B=0.615\\ {\theta }_B=37.95°\end{array}$

Substitute $10\text{ft}$for $L$, $6.15\text{ft}$for $CD$in Equation (V).

  $\begin{array}{c}DB=\sqrt{{\left(10\text{ft}\right)}^2-{\left(6.15\text{ft}\right)}^2}\\ =\sqrt{\left(100{\text{ft}}^2\right)-\left(37.8225{\text{ft}}^2\right)}\\ \approx 7.88\text{ft}\end{array}$

Substitute $3.55\text{ft}$for $AD$, and $7.88\text{ft}$for $DB$in Equation (VI).

  $\begin{array}{c}AB=3.55\text{ft}+7.88\text{ft}\\ =11.43\text{ft}\end{array}$

Substitute $3.55\text{ft}$for $AD$, $49\text{kips}$for $P$, $6.15\text{ft}$for $CD$and $11.43\text{ft}$for $AB$in Equation (VII).

  $\begin{array}{l}\left(-49\text{kips}\right)\left(3.55\text{ft}\right)-2\left(49\text{kips}\right)\left(6.15\text{ft}\right)+\left(R_{By}\right)\left(11.43\text{ft}\right)=0\\ -173.95\text{kips}\cdot \text{ft}-602.7\text{kips}\cdot \text{ft}+\left(R_{By}\right)\left(11.43\text{ft}\right)=0\\ R_{By}\approx 67.95\text{kips}\end{array}$

Substitute $49\text{kips}$for $P$in Equation (VIII).

  $\begin{array}{l}{R}_{Ax}+2\left(49\text{kips}\right)=0\\ R_{Ax}=-98\text{kips}\end{array}$

Substitute $49\text{kips}$for $P$, $67.95\text{kips}$for $R_{By}$in Equation (IX).

  $\begin{array}{l}{R}_{Ay}+67.95\text{kips}-49\text{kips}=0\\ R_{Ay}=-18.95\text{kips}\end{array}$

Substitute $-18.95\text{kips}$for $R_{Ay}$, and $60°$for ${\theta }_A$in Equation (X).

  $\begin{array}{l}{F}_{AC}\sin \left(60°\right)=-\left(-18.95\text{kips}\right)\\ F_{AC}\sin \left(60°\right)=\left(18.95\text{kips}\right)\\ F_{AC}=21.88\text{kips}\end{array}$

Substitute $-98\text{kips}$for $R_{Ax}$, $21.88\text{kips}$for $F_{AC}$and $60°$for ${\theta }_A$in Equation (XI).

  $\begin{array}{l}-98\text{kips}+\left(21.88\text{kips}\right)\cos \left(60°\right)+F_{AB}=0\\ -98\text{kips}+\left(10.94\text{kips}\right)+F_{AB}=0\\ F_{AB}=87.06\text{kips}\end{array}$

Substitute $87.06\text{kips}$for $F_{AB}$and $37.95°$for ${\theta }_B$in Equation (XII).

  $\begin{array}{l}87.06\text{kips}-F_{BC}\cos \left(37.95°\right)=0\\ F_{BC}\cos \left(37.95°\right)=87.06\text{kips}\\ F_{BC}=110.40\text{kips}\end{array}$

Refer to table F-3 (a), “Properties of channel sections” to obtain the area of cross-section of $\text{C}10\times 20$as $5.87{\text{in}}^2$.

Substitute $21.88\text{kips}$for $F_{AC}$and $5.87{\text{in}}^2$for $A$in Equation (XIII).

  $\begin{array}{c}{\tau }_{AC}=\frac{21.88\text{kips}}{2\left(5.87{\text{in}}^2\right)}\\ =\frac{21.88\text{kips}}{\left(11.74{\text{in}}^2\right)}\\ =1.86\text{ksi}\end{array}$

Substitute $87.06\text{kips}$for $F_{AB}$and $5.87{\text{in}}^2$for $A$in Equation (XIV).

  $\begin{array}{c}{\tau }_{AB}=\frac{87.06\text{kips}}{2\left(5.87{\text{in}}^2\right)}\\ =\frac{87.06\text{kips}}{\left(11.74{\text{in}}^2\right)}\\ \approx 7.42\text{ksi}\end{array}$

Substitute $110.40\text{kips}$for $F_{BC}$and $5.87{\text{in}}^2$for $A$in Equation (XV).

  $\begin{array}{c}{\tau }_{BC}=\frac{110.40\text{kips}}{2\left(5.87{\text{in}}^2\right)}\\ =\frac{110.40\text{kips}}{\left(11.74{\text{in}}^2\right)}\\ \approx 9.40\text{ksi}\end{array}$

Conclusion:

The maximum shear stress in the member AC is = $1.86\text{ksi}$.

The maximum shear stress in the member AB is = $7.42\text{ksi}$.

The maximum shear stress in the member BC is = $9.40\text{ksi}$.

To determine

The maximum permissible load.

Answer to Problem 2.6.11P

The maximum permissible load is = $36.5\text{kip}$.

Explanation of Solution

Write the expression for the maximum permissible load on member $BC$.

  ${\left(P\right)}_{BC}=\frac{{\sigma }_{BC}A\sin {\theta }_B}{1.385}$....... (XVI)

Here, permissible normal stress is ${\sigma }_{BC}$.

Write the expression for the maximum permissible load on member $AC$.

  ${\left(P\right)}_{AC}=\frac{{\sigma }_{AC}A\sin {\theta }_A}{0.385}$....... (XVII)

Here, permissible normal stress is ${\sigma }_{AC}$.

Calculation:

Substitute $14\text{ksi}$, for ${\sigma }_{BC}$, $5.87{\text{in}}^{\text{2}}$for $A$, and $37.95°$for ${\theta }_B$in Equation (XVI).

  $\begin{array}{c}{\left(P\right)}_{BC}=\frac{\left(14\text{ksi}\right)\left(5.87{\text{in}}^{\text{2}}\right)\sin \left(37.95°\right)}{1.385}\\ =\frac{50.538\text{kip}}{1.385}\\ \approx 36.5\text{kip}\end{array}$

Substitute $14\text{ksi}$, for ${\sigma }_{AC}$, $5.87{\text{in}}^{\text{2}}$for $A$, and $60°$for ${\theta }_B$in Equation (XVII).

  $\begin{array}{c}{\left(P\right)}_{AC}=\frac{\left(14\text{ksi}\right)\left(5.87{\text{in}}^{\text{2}}\right)\sin \left(60°\right)}{0.385}\\ =\frac{71.169\text{kip}}{0.385}\\ \approx 184.85\text{kip}\end{array}$

The lowest value of the maximum permissible load is = $36.5\text{kip}$.

Conclusion:

The maximum permissible load is = $36.5\text{kip}$.