Answer to Question #200954 in Mechanical Engineering for chandler

Question #200954

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)


1
Expert's answer
2021-06-01T03:47:51-0400

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}"


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