Chapter 1, Problem 1.3.30P

Space frame A BCD is clamped at A, except it is Free to translate in the .v direction. There is also a roller support at D, which is normal to line CDE. A triangularly distributed Force with peak intensity q₀ = 75 N/m acts along AB in the positive - direction. Forces P_x= 60 N and Pz = = 45 N are applied at joint C, and a concentrated moment My = 120 N . m acts at the mid-span of member BC.

(a) Find reactions at supports A and I).

(b) Find internal stress resultants N. E’I T, and .11 at the mid-height of segment AB.

  

To determine

Reactions at support A,D.

Answer to Problem 1.3.30P

The correct answers are:

$\begin{array}{l}{A}_y=-120\text{ N, }{A}_z=-60\text{ N, }\\ M_{Ax}=-7\text{0 N}\cdot \text{m, }{M}_{Ay}=-142.5\text{ N}\cdot \text{m, }{M}_{Az}=-18\text{0 N}\cdot \text{m,}\\ D_x=-60\text{ N, }{D}_y=120\text{ N, }{D}_z=30\text{ N}\text{.}\end{array}$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

and $q_0=75\text{ N/m, }{P}_x=60\text{ N, }{P}_z=-45\text{ N, and }{M}_y=120\text{ N}\cdot \text{m}\text{.}$

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

The forces and corresponding position vectors are,

Force	Position vector
$\overrightarrow{A}=(0,A_y,A_z)$	$\overrightarrow{r_A}=(0,0,0)$
$\overrightarrow{Q}=(0,0,q_0)=(0,0,75)$	$\overrightarrow{r_Q}=\left(0,\frac{4}{3},0\right)$
$\overrightarrow{P}=(P_x,0,P_z)=(60,0,-45)$	$\overrightarrow{r_C}=(1.5,2,0)$
$\begin{array}{c}\overrightarrow{D}=(D_x,D_y,D_z)\\ =\frac{D}{\sqrt{{(-0.5)}^2+2^2+{(0.5)}^2}}(-1,2,0.5)\\ =D(-0.5,1,0.25)\end{array}$	$\begin{array}{c}\overrightarrow{r_D}=(1.5,2,0)-\frac{0.75}{\sqrt{{(-0.5)}^2+2^2+{(0.5)}^2}}(-1,2,0.5)\\ =(1.5,2,0)-0.75(-0.5,1,0.25)\\ =(1.87,1.25,-0.18)\end{array}$

Take equilibrium of forces vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{A}+\overrightarrow{Q}+\overrightarrow{P}+\overrightarrow{D}=0\\ \Rightarrow (0,A_y,A_z)+(0,0,75)+(60,0,-45)+D(-0.5,1,0.25)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}60-0.5D=0\mathrm{....}\text{(1)}\\ A_y+D=0\mathrm{....}\text{(2)}\\ A_z+30+0.25D=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $D=120\text{ N, }{A}_y=-120\text{ N, }{A}_z=-60\text{ N}\text{.}$

Now, calculate components of D as,

  $\begin{array}{c}\overrightarrow{D}=(D_x,D_y,D_z)\\ ==D(-0.5,1,0.25)\\ =120(-0.5,1,0.25)\\ =(-60\text{ N},120\text{ N},30\text{ N})\end{array}$

Now take equilibrium of moments about A in vector form as,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{M_A}}=0\\ \Rightarrow \overrightarrow{M}+\overrightarrow{M_y}+\overrightarrow{r_Q}\times \overrightarrow{Q}+\overrightarrow{r_C}\times \overrightarrow{P}+\overrightarrow{r_D}\times \overrightarrow{D}=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(0,M_y,0)\\ +\left(0,\frac{4}{3},0\right)\times (0,0,75)+(1.5,2,0)\times (60,0,-45)\\ +(1.87,1.25,-0.18)\times (-60,120,30)=0\end{array}$

Evaluate the cross products to get,

  $\begin{array}{l}\Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(0,M_y,0)+(100,0,0)+(-90,67.5,-120)+(60,-45,300)=0\\ \Rightarrow \left[\left(M_o{}_x+70\right),\left(M_o{}_y+142.5\right),\left(M_o{}_z+180\right)\right]=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)=(-70\text{ N}\cdot \text{m}\text{.},-142.5\text{ N}\cdot \text{m},-180\text{ N}\cdot \text{m})\end{array}$

Conclusion:

Therefore the forces and moments are:

$\begin{array}{l}{A}_y=-120\text{ N, }{A}_z=-60\text{ N, }\\ M_{Ax}=-7\text{0 N}\cdot \text{m, }{M}_{Ay}=-142.5\text{ N}\cdot \text{m, }{M}_{Az}=-18\text{0 N}\cdot \text{m,}\\ D_x=-60\text{ N, }{D}_y=120\text{ N, }{D}_z=30\text{ N}\text{.}\end{array}$

To determine

Internal stress resultants N,V,T and M at mid height of AB.

Answer to Problem 1.3.30P

The correct answers are:

$\begin{array}{l}N=120\text{ N, }V=41.3\text{ N, }\\ T=142.5\text{ N}\cdot \text{m, }M=180.7\text{ N}\cdot \text{m}\text{.}\end{array}$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

and $q_0=75\text{ N/m, }{P}_x=60\text{ N, }{P}_z=-45\text{ N, and }{M}_y=120\text{ N}\cdot \text{m}\text{.}$

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

The forces and corresponding position vectors are,

Force	Position vector
$\overrightarrow{A}=(0,A_y,A_z)$	$\overrightarrow{r_A}=(0,0,0)$
$\overrightarrow{Q}=(0,0,q_0)=(0,0,75)$	$\overrightarrow{r_Q}=\left(0,\frac{4}{3},0\right)$
$\overrightarrow{P}=(P_x,0,P_z)=(60,0,-45)$	$\overrightarrow{r_C}=(1.5,2,0)$
$\begin{array}{c}\overrightarrow{D}=(D_x,D_y,D_z)\\ =\frac{D}{\sqrt{{(-0.5)}^2+2^2+{(0.5)}^2}}(-1,2,0.5)\\ =D(-0.5,1,0.25)\end{array}$	$\begin{array}{c}\overrightarrow{r_D}=(1.5,2,0)-\frac{0.75}{\sqrt{{(-0.5)}^2+2^2+{(0.5)}^2}}(-1,2,0.5)\\ =(1.5,2,0)-0.75(-0.5,1,0.25)\\ =(1.87,1.25,-0.18)\end{array}$

Take equilibrium of forces vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{A}+\overrightarrow{Q}+\overrightarrow{P}+\overrightarrow{D}=0\\ \Rightarrow (0,A_y,A_z)+(0,0,75)+(60,0,-45)+D(-0.5,1,0.25)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}60-0.5D=0\mathrm{....}\text{(1)}\\ A_y+D=0\mathrm{....}\text{(2)}\\ A_z+30+0.25D=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $D=120\text{ N, }{A}_y=-120\text{ N, }{A}_z=-60\text{ N}\text{.}$

Now, calculate components of D as,

  $\begin{array}{c}\overrightarrow{D}=(D_x,D_y,D_z)\\ ==D(-0.5,1,0.25)\\ =120(-0.5,1,0.25)\\ =(-60\text{ N},120\text{ N},30\text{ N})\end{array}$

Now take equilibrium of moments about A in vector form as,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{M_A}}=0\\ \Rightarrow \overrightarrow{M}+\overrightarrow{M_y}+\overrightarrow{r_Q}\times \overrightarrow{Q}+\overrightarrow{r_C}\times \overrightarrow{P}+\overrightarrow{r_D}\times \overrightarrow{D}=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(0,M_y,0)\\ +\left(0,\frac{4}{3},0\right)\times (0,0,75)+(1.5,2,0)\times (60,0,-45)\\ +(1.87,1.25,-0.18)\times (-60,120,30)=0\end{array}$

Evaluate the cross products to get,

  $\begin{array}{l}\Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(0,M_y,0)+(100,0,0)+(-90,67.5,-120)+(60,-45,300)=0\\ \Rightarrow \left[\left(M_o{}_x+70\right),\left(M_o{}_y+142.5\right),\left(M_o{}_z+180\right)\right]=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)=(-70\text{ N}\cdot \text{m}\text{.},-142.5\text{ N}\cdot \text{m},-180\text{ N}\cdot \text{m})\end{array}$

Calculation internal stress resultants at mid point of AB :

Consider the following free body diagram,

Analyze the lower part of the above free body diagram,

Take equilibrium of forces in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{A}+\overrightarrow{N}+\overrightarrow{V}+(0,0,\frac{q_0}{4})=0\\ \Rightarrow (0,A_y,A_z)+(0,N,0)+(V_x,0,V_z)+(0,0,\frac{75}{4})=0\\ \Rightarrow (0,-120,-60)+(0,N,0)+(V_x,0,V_z)+(0,0,18.75)=0\end{array}$

Solve the above equation in component form to get,

$N=120\text{ N, }{V}_x=0,V_z=41.3\text{ N}\text{.}$

Take equilibrium of moments about A in vector form as,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{M_A}}=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(M_x,0,M_z)+(0,T,0)\\ +(0,\frac{2}{3},0)\times (0,0,18.75)+(0,1,0)\times (V_x,0,V_z)+(0,1,0)\times (0,N,0)=0\\ \Rightarrow (M_o{}_x,M_o{}_y,M_o{}_z)+(M_x,0,M_z)+(0,T,0)\\ +(0,\frac{2}{3},0)\times (0,0,18.75)+(0,1,0)\times (0,0,41.25)+(0,1,0)\times (0,120,0)=0\end{array}$

Evaluate the cross products and solve to get,

  $\Rightarrow M_x=16.25\text{ N}\cdot \text{m, }{M}_z=180\text{ N}\cdot \text{m, }T=142.5\text{ N}\cdot \text{m}\text{.}$

Equivalent moment,

$M=\sqrt{M_x{}^2+M_z{}^2}=\sqrt{{16.25}^2+{180}^2}=180.7\text{ N}\cdot \text{m}$

Conclusion:

Thus the internal stress resultants are:

$\begin{array}{l}N=120\text{ N, }V=41.3\text{ N, }\\ T=142.5\text{ N}\cdot \text{m, }M=180.7\text{ N}\cdot \text{m}\text{.}\end{array}$