Chapter 1, Problem 1.3.26P

A plane frame with a pin support at A and roller supports at C and £ has a cable attached at E. which runs over Frictionless pulleys al D and B (see figure). The cable force is known to be 400 N. There is a pin connection just Lo the left of joint C.

(a) Find reactions at supports^, C, and E.

(b) Find internal stress, resultants N, V, and M just to the right of joint C.

(c) Find resultant force in the pin near C.

  

To determine

Reactions at supports A, C and E.

Answer to Problem 1.3.26P

The correct answers are:

$A_x=320\text{ N, }{A}_y=-240\text{ N, }{C}_y=192\text{ N, }{E}_y=-192\text{ N}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information:

  

Calculation:

Consider the following free body diagram:

  

Member AB :

Take equilibrium of moments about point B:

  $\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -4A_y-3F_{CY}=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x-400\times \frac{4}{5}+400\times \frac{3}{5}+F_{CX}=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of forces in $y-$direction in:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+400\times \frac{3}{5}+400\times \frac{4}{5}-F_{CY}=0\mathrm{....}\text{(3)}\end{array}$

Member DCE :

Take equilibrium of moments about point C:

  $\begin{array}{l}{\displaystyle \sum M_C=0}\\ \Rightarrow 5E_y+4\times 400\times \frac{3}{5}=0\\ \Rightarrow E_y=-\frac{1}{5}\times 4\times 400\times \frac{3}{5}=-192\text{ N }\mathrm{....}\text{(4)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow 400\times \frac{3}{5}+F_{CX}=0\\ \Rightarrow F_{CX}=-400\times \frac{3}{5}=-240\text{ N}\text{. }\mathrm{....}\text{(5) }\end{array}$

Take equilibrium of forces in $y-$direction in,

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow C_y-400\times \frac{4}{5}+F_{CY}+E_y=0\mathrm{....}\text{(6)}\end{array}$

Now, solve equations (1-6) to get

  $A_x=320\text{ N, }{A}_y=-240\text{ N, }{C}_y=192\text{ N, }{E}_y=-192\text{ N, }{F}_{CX}=-240\text{ N, }{F}_{CY}=320\text{ N}\text{.}$

Conclusion:

Thus, the reaction forces are: $A_x=320\text{ N, }{A}_y=-240\text{ N, }{C}_y=192\text{ N, }{E}_y=-192\text{ N}\text{.}$

To determine

Internal stress resultants N, V and M just to the right of C.

Answer to Problem 1.3.26P

The correct answers are:

$N=-312\text{ N, }V=-57.9\text{ N, }M=289\text{ N}\cdot \text{m}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information:

  

Calculation of external forces:

Consider the following free body diagram:

  

Member AB :

Take equilibrium of moments about point B:

  $\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -4A_y-3F_{CY}=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x-400\times \frac{4}{5}+400\times \frac{3}{5}+F_{CX}=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of forces in $y-$direction in,

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+400\times \frac{3}{5}+400\times \frac{4}{5}-F_{CY}=0\mathrm{....}\text{(3)}\end{array}$

Member DCE :

Take equilibrium of moments about point C:

  $\begin{array}{l}{\displaystyle \sum M_C=0}\\ \Rightarrow 5E_y+4\times 400\times \frac{3}{5}=0\\ \Rightarrow E_y=-\frac{1}{5}\times 4\times 400\times \frac{3}{5}=-192\text{ N }\mathrm{....}\text{(4)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow 400\times \frac{3}{5}+F_{CX}=0\\ \Rightarrow F_{CX}=-400\times \frac{3}{5}=-240\text{ N}\text{. }\mathrm{....}\text{(5) }\end{array}$

Take equilibrium of forces in $y-$direction in:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow C_y-400\times \frac{4}{5}+F_{CY}+E_y=0\mathrm{....}\text{(6)}\end{array}$

Now, solve equations (1-6) to get

  $A_x=320\text{ N, }{A}_y=-240\text{ N, }{C}_y=192\text{ N, }{E}_y=-192\text{ N, }{F}_{CX}=-240\text{ N, }{F}_{CY}=320\text{ N}\text{.}$

Calculation of internal forces:

Consider the following free body diagram:

  

Consider the right hand side of the above free body diagram.

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow -N-400\times \frac{5}{\sqrt{4^2+5^2}}=0\\ \Rightarrow N=-312.3\text{ N}\text{.}\end{array}$

Take equilibrium of forces in $y-$direction in:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow V+E_y+400\times \frac{4}{\sqrt{4^2+5^2}}=0\\ \Rightarrow V+(-192)+400\times \frac{4}{\sqrt{4^2+5^2}}=0\\ \Rightarrow V=-57.9\text{ N}\text{.}\end{array}$

Take equilibrium of moments about point E:

  $\begin{array}{l}{\displaystyle \sum M_E=0}\\ \Rightarrow -M-V\times 5=0\\ \Rightarrow M=-5V=-5\times (-57.9)=289.5\text{ N}\cdot \text{m}\text{.}\end{array}$

Conclusion:

Thus, the internal forces are: $N=-312\text{ N, }V=-57.9\text{ N, }M=289\text{ N}\cdot \text{m}\text{.}$

To determine

Resultant force in the pin near C.

Answer to Problem 1.3.26P

The correct answers are:

${\text{Resultant}}_C=400\text{ N}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information:

  

Calculation:

Consider the following free body diagram;

  

Member AB :

Take equilibrium of moments about point B:

  $\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -4A_y-3F_{CY}=0\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x-400\times \frac{4}{5}+400\times \frac{3}{5}+F_{CX}=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of forces in $y-$direction in:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+400\times \frac{3}{5}+400\times \frac{4}{5}-F_{CY}=0\mathrm{....}\text{(3)}\end{array}$

Member DCE :

Take equilibrium of moments about point C,

  $\begin{array}{l}{\displaystyle \sum M_C=0}\\ \Rightarrow 5E_y+4\times 400\times \frac{3}{5}=0\\ \Rightarrow E_y=-\frac{1}{5}\times 4\times 400\times \frac{3}{5}=-192\text{ N }\mathrm{....}\text{(4)}\end{array}$

Take equilibrium of forces in $x-$direction:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow 400\times \frac{3}{5}+F_{CX}=0\\ \Rightarrow F_{CX}=-400\times \frac{3}{5}=-240\text{ N}\text{. }\mathrm{....}\text{(5) }\end{array}$

Take equilibrium of forces in $y-$direction in:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow C_y-400\times \frac{4}{5}+F_{CY}+E_y=0\mathrm{....}\text{(6)}\end{array}$

Now, solve equations (1-6) to get:

  $A_x=320\text{ N, }{A}_y=-240\text{ N, }{C}_y=192\text{ N, }{E}_y=-192\text{ N, }{F}_{CX}=-240\text{ N, }{F}_{CY}=320\text{ N}\text{.}$

Resultant force at pin near C is:

  $\begin{array}{c}{\text{Resultant}}_C=\sqrt{F_{CX}{}^2+F_{CY}{}^2}\\ =\sqrt{{(-240)}^2+{320}^2}\\ =400\text{ N}\text{.}\end{array}$

Conclusion:

Thus, the resultant force at pin near C is: ${\text{Resultant}}_C=400\text{ N}\text{.}$