Answer to Question #200954 in Mechanical Engineering for chandler

Question

The bar shown in the figure is fixed at A and an axial force of 150 kN is acting at C. The

cross-sectional areas of each segment are given in the figure. The design requires that there

should not be a support reaction at B after displacement along the z axis occurs. To provide this

a linearly distributed axial load q(z) is applied between C and D . If the value of this distribution

is supposed to be zero at D , then obtain the expression of the function of this distribution. The

Elasticity modulus of the material is 3500 MPa. (40 Points)

Answer 1

Given,

As we are moving from A to C, then the load of the rod gets increase.

Compression in the small part of the rod, $dx=d\delta_x$

$d\delta_x= \frac{kx}{2AE}dx$

Total compression in CD,

$=\int d\delta_x$

$=\int_0^3\frac{kx^2\times 10^4 dx}{2\times 20\times 3500\times 10^6}$

$=-\frac{k\times 10^{-3}}{14} [ \frac{x^3}{3}]_0^3$

$=\frac{-9k}{14}$

Let the reaction force at the hinge A

,

$R_A=150[-\frac{kx^2}{2}]_0^3$

$=150-4.5k$

Extension in the bar AC,

$\Delta AC=\frac{R_A \times 10^4 \times 2}{30 \times 3500 \times 10^6}$

$=\frac{4}{21}[150-4.5k]$

Total change in length $, \Delta AC + \Delta CD \leq 5$

$k=15.71 KN/m^2$

$q(x)= 15.71 x$

$q(z)=az+b$

$2a+b=\frac{330}{7}$

$5a+b=0$

Now, subtracting,

$a= \frac{-110}{7}$

$b= \frac{550}{7}$

Hence, $q(z)=\frac{-110}{7}(z)+\frac{550}{7}$