A bar with variable cross-section is subjected to a uniform axial load of p =100

Question

A bar with variable cross-section is subjected to a uniform axial load of p =1000 kg/m. The cross-sections at the support, mid-length and free end are 60*60 cm2, 30*30 cm2 ´ and 20*20 cm2, respectively. Assume that the cross-section varies quadratically in between these points on all 4 sides (Hint: Obtain a quadratic formula that passes through the given areas). Assume E = 2 *10^6 kg/cm2 and L = 5 m.

Determine the displacement at the free end of the bar using truss element with 2 nodes (linear interpolation functions). What are the minimum number of elements for convergence? Show the variation of displacement and stress along the member for at least 5 cases.

Determine the displacement at the free end of the bar using truss element with 3 nodes (quadratic interpolation functions). What are the minimum number of elements for convergence? Show the variation of displacement and stress along the member for at least 5 cases. (Hint: For convergence, let the stress distribution be your acceptance criteria).

Explanation / Answer

It is clearly shown that we are having a bar of variable cross section,

Therefore,elongation can be find out by

Displacement = Pl/AE

By this,we can find displacement in all the three stated regions.

1) Displacement = 1×1000×10^3×5/.36×2×10^6

=Displacement = 6.944 @pt1 = l = 1.944m

2) Displacement =1×10^6×2.5/.09×2×10^6

Therefore, it is equal to =13.88m

3) Similarly we can find by using above formula.

However,in a two node Bar the only interelement continuity requirements is Linear.The equation which uses Interpolation formula is given as N^eu^e.

where,N1 and N2 are called as Shape functions.

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