Chapter 14, Problem 1P

(a)

To determine

Calculate the stiffness co-efficient K.

Answer to Problem 1P

The stiffness co-efficient K is $\underset{\_}{476.25\text{kips/in}\text{.}}$.

Explanation of Solution

Given information:

The modulus of elasticity (E) is 30,000 ksi.

The vertical displacement of joint A $\left(\Delta \right)$ is 1 inch.

The vertical load acting at joint A is 24 kips.

Calculation:

Show the free body diagram of the structure as in Figure (1).

Refer Figure (1),

Find the force in the member AC using the relation;

$\begin{array}{c}{F}_{AC}=\frac{AE\Delta }{L}\\ =\frac{1\times 30,000\times 1}{\left(16\text{ft}\times 12\right)\text{in}.}\\ =156.25\text{kips}\end{array}$

Find the angle in the member;

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{V}{H}\right)\\ ={\tan }^{-1}\left(\frac{16}{12}\right)\\ =53.13°\end{array}$

Find the inclined displacement of joint A;

$\begin{array}{c}{\Delta }_{AB}=\Delta \sin \theta \\ =1\times \sin 53.13°\\ =0.8\text{in}\text{.}\end{array}$

Find the force in the member AB using the relation;

$\begin{array}{c}{F}_{AB}=F_{AD}=\frac{AE{\Delta }_{AB}}{L}\\ =\frac{2\times 30,000\times 0.8}{\left(20\text{ft}\times 12\right)\text{in}.}\\ =200\text{kips}\end{array}$

Find vertical force in the member AB;

$\begin{array}{c}{F}_v=F\sin \theta \\ =200\sin 53.13°\\ =160\text{kips}\end{array}$

Find horizontal force in the member AB;

$\begin{array}{c}{F}_h=F\cos \theta \\ =200\cos 53.13°\\ =120\text{kips}\end{array}$

Show the free body diagram of the joint A as in Figure (2).

Refer Figure (2),

Consider the upward force is positive and downward force is negative.

Find the stiffness coefficient;

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}\sum F_y=0\\ 160+160+156.25-K=0\\ K=476.25\text{kips/in}\text{.}\end{array}$

Thus, the stiffness co-efficient K is $\underset{\_}{476.25\text{kips/in}\text{.}}$.

(b)

To determine

Calculate the vertical displacement at A.

Answer to Problem 1P

The vertical displacement at A is $\underset{\_}{0.050\text{in}\text{.}}$.

Explanation of Solution

Find the vertical displacement at A;

$\begin{array}{c}\Delta =\frac{F}{K}\\ =\frac{24}{476.25}\\ =0.050\text{in}\text{.}\end{array}$

Thus, the vertical displacement at A is $\underset{\_}{0.050\text{in}\text{.}}$.

(c)

To determine

Calculate the axial forces in all bars.

Answer to Problem 1P

The axial force in the member AC is $\underset{\_}{7.88\text{kips}}$.

The axial force in the member AB and AD is $\underset{\_}{10.08\text{kips}}$.

Explanation of Solution

Find the axial force in the member AC using the relation;

$\begin{array}{c}{F}_{AC}=\frac{AE\Delta }{L}\\ =\frac{1\times 30,000\times 0.0504}{\left(16\text{ft}\times 12\right)\text{in}.}\\ =7.88\text{kips}\end{array}$

Thus, the axial force in the member AC is $\underset{\_}{7.88\text{kips}}$.

Find the force in the member AB and AD using the relation;

$\begin{array}{c}{F}_{AB}=F_{AD}=\frac{AE\Delta }{L}\\ =\frac{2\times 30,000\times 0.0504}{\left(20\text{ft}\times 12\right)\text{in}.}\\ =10.08\text{kips}\end{array}$

Thus, the axial force in the member AB and AD is $\underset{\_}{10.08\text{kips}}$.