Chapter 2, Problem 2.2.16P

A uniform bar AB of weight W = 25 N is supported by two springs, as shown in the figure. The spring on the left has a stiffness k_[= 300 N/m and natural length L_t=250 mm. The corresponding quantities for the spring on the right are k₂= 400 N/m and L^ = 200 mm. The distance between the springs is L = 350 mm, and the spring on the right is suspended from a support that is a distance it = SO mm below the point of support for the spring on the left. Neglect the weight of the springs.

(a) At what distance x from the left-hand spring (figure part a) should a load P = 18 N be placed in order to bring the bar to a horizontal position? (b) If P is now removed, what new value of k_{is required so that the bar (figure part a) will hang in a horizontal position underweight If?

(c) If P is removed and k_t= 300 N/m. what distance b should spring k_tbe moved to the right so that the bar (figure part a) will hang in a horizontal position under weight II"?

(d) If the spring on the left is now replaced by two springs in series (k_t= 300 N/m, k_t) with overall natural length L_t= 250 mm (see figure part b). what value of k_; is required so that the bar will hang in a horizontal position under weight IF?

  

To determine

Location of load $P$, to bring bar to horizontal position.

Answer to Problem 2.2.16P

Location of load $P$is, $x=134.7mm$

Explanation of Solution

Given:

Weight, $W=25\text{N}$

Spring stiffness on left and right,

  $k_1=0.300\frac{\text{N}}{mm},k_2=0.400\frac{\text{N}}{mm}$

Natural lengths of both springs,

  $L_1=250mm{L}_2=200mm$

Distance between the springs,

  $L=350mm$

Load, $P=18\text{N}$

Distance from support, $h=80mm$

We have to use statics to get forces in both springs.

  ${\displaystyle \sum M_A}=0$

  $F_2=\frac{1}{L}\left(W\frac{L}{2}+Px\right)$

  $F_2=\frac{W}{2}+P\frac{x}{L}$

  ${\displaystyle \sum F_V}=0$

  $F_1=W+P-F_2$

  $F_1=\frac{W}{2}+P\left(1-\frac{x}{L}\right)$

Now, we use constraint equation to define horizontal position, then solve for location $x$.

  $L_1+\frac{F_1}{K_1}=L_2+h+\frac{F_2}{K_2}$

We are required to substitute expressions for $F_1\&F_2$above into constraint equation & solve for $x$

  $x=\frac{-2L_{1}L{K}_1{K}_2-K_{2}WL-2K_{2}PL+2L_{2}L{K}_1{K}_2+2hL{K}_1{K}_2+K_{1}WL}{-2P(K_1+K_2)}$

  $x=\frac{\begin{array}{l}-2\left(250mm\right)\left(350mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)-\left(0.400\frac{\text{N}}{mm}\right)\left(\text{25}\text{N}\right)\left(350mm\right)\\ -2\left(0.400\frac{\text{N}}{mm}\right)\left(18\text{N}\right)\left(350mm\right)+\\ 2\left(200mm\right)\left(350mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)\\ +2\left(80mm\right)\left(350mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)+\left(0.300\frac{\text{N}}{mm}\right)\left(\text{25}\text{N}\right)\left(350mm\right)\end{array}}{-2\left(18\text{N}\right)(\left(0.300\frac{\text{N}}{mm}\right)+\left(0.400\frac{\text{N}}{mm}\right))}$

  $x=134.7mm$

To determine

New value of spring constant $K_1$, so that bar is horizontal underweight $W$,if load $P$is removed.

Answer to Problem 2.2.16P

New value of spring constant $K_1$is, $0.204\frac{\text{N}}{mm}$

Explanation of Solution

Given:

Weight, $W=25\text{N}$

Spring stiffness on left and right,

  $k_1=0.300\frac{\text{N}}{mm},k_2=0.400\frac{\text{N}}{mm}$

Natural lengths of both springs,

  $L_1=250mm{L}_2=200mm$

Distance between the springs,

  $L=350mm$

Load, $P=18\text{N}$

Distance from support, $h=80mm$

New value of spring constant $K_1$, so that bar is horizontal underweight $W$, if load P is removed.

Now,

  $F_1=\frac{W}{2}$

  $F_2=\frac{W}{2}$

Since, $P=0$

Same constant equation as above but now:

  $P=0$

  $L_1+\frac{\left(\frac{W}{2}\right)}{K_1}-\left(L_2+h\right)-\frac{\left(\frac{W}{2}\right)}{K_2}=0$

Now, solve for $K_1$

  $K_1=\frac{-W{K}_2}{\left[2K_2\left[L_1-\left(L_2+h\right)\right]\right]-W}$

  $K_1=\frac{-25\text{N}\left(0.400\frac{\text{N}}{mm}\right)}{\left[2\left(0.400\frac{\text{N}}{mm}\right)\left[250mm-\left(200mm+80mm\right)\right]\right]-25\text{N}}$

  $K_1=0.204\frac{\text{N}}{mm}$

To determine

Distance moved by spring $K_1$to the right so that bar will hang in a horizontal position underweight $W$.

Answer to Problem 2.2.16P

Distance moved by spring $K_1$is, $b=74.1mm$

Explanation of Solution

Given:

New position for $k_1$is shown as:

Weight, $W=25\text{N}$

Spring stiffness on left and right,

  $k_1=0.300\frac{\text{N}}{mm},k_2=0.400\frac{\text{N}}{mm}$

Natural lengths of both springs.

  $L_1=250mm{L}_2=200mm$

Distance between the springs,

  $L=350mm$

Load, $P=18\text{N}$

Distance from support, $h=80mm$

Use $K_1=0.300\text{N}/mm$

But relocate spring,

  $K_1(x=b)$

So, that bar ends up in horizontal position underweight $W$.

  $b=\frac{2L_1{K}_1{K}_{2}L+WL{K}_2-2L_2{K}_1{K}_{2}L-2h{K}_1{K}_{2}L-W{K}_{1}L-W{K}_{1}L}{\left(2L_1{K}_1{K}_2\right)-2L_2{K}_1{K}_2-2h{K}_1{K}_2-2W{K}_1}$

Statics are as follows:

  ${\displaystyle \sum M_{K_1}}=0$

  $F_2=\frac{W\left(\frac{L}{2}-b\right)}{L-b}$

  ${\displaystyle \sum F_V}=0$

  $F_1=W-F_2$

  $F_1=W-\frac{W\left(\frac{L}{2}-b\right)}{L-b}$

  $F_1=\frac{WL}{2(L-b)}$

Now, we have the constraint equation − substitute above expression for $F_1$

  $\&$

$F_2$and solve for $b$:

  $L_1+\frac{F_1}{K_1}-\left(L_2+h\right)-\frac{F_2}{K_2}=0$

Use the following data:

Spring stiffness on left and right.

  $k_1=0.300\frac{\text{N}}{mm},k_2=0.400\frac{\text{N}}{mm}$

Natural lengths of both springs:

  $L_1=250mm{L}_2=200mm$

Distance between the springs.

  $L=350mm$

By substituting $F_1$

  $\&$

  $F_2$and above given data in constraint equation, we get:

  $b=74.1mm$

To determine

Value of $K_3$required so that the bar will hang in horizontal position.

Answer to Problem 2.2.16P

The required value is, $K_3=0.638\frac{\text{N}}{mm}$

Explanation of Solution

Given:

Weight, $W=25\text{N}$

Spring stiffness on left and right,

  $k_1=0.300\frac{\text{N}}{mm},k_2=0.400\frac{\text{N}}{mm}$

Natural lengths of both springs,

  $L_1=250mm{L}_2=200mm$

Distance between the springs:

  $L=350mm$

Load, $P=18\text{N}$

Distance from support, $h=80mm$

Value of $K_3$required, so that the bar will hang in horizontal position:

  $F_1=\frac{W}{2},F_2=\frac{W}{2}$

New constraint equation is as follows:

  $L_1+\frac{F_1}{K_1}+\frac{F_1}{K_3}-\left(L_2+h\right)-\frac{F_2}{K_2}=0$

  $L_1+\frac{\frac{W}{2}}{K_1}+\frac{\frac{W}{2}}{K_3}-\left(L_2+h\right)-\frac{\frac{W}{2}}{K_2}=0$

  $K_3=\frac{W{K}_1{K}_2}{-2L_1{K}_1{K}_2-W{K}_2+2L_2{K}_1{K}_2+2h{K}_1{K}_2+W{K}_1}$

  $K_3=\frac{\left(\text{25}\text{N}\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)}{\begin{array}{l}-2\left(250mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)-\left(25\text{N}\right)\left(0.400\frac{\text{N}}{mm}\right)\\ +2\left(200mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)+\\ 2\left(80mm\right)\left(0.300\frac{\text{N}}{mm}\right)\left(0.400\frac{\text{N}}{mm}\right)+\left(25\text{N}\right)\left(0.300\frac{\text{N}}{mm}\right)\end{array}}$

  $K_3=0.638\frac{\text{N}}{mm}$