Initial Stress In A Composite Member Under Tensile Loading Given Allowable Stresses

Initial Stress In A Composite Member Under Tensile Loading Given Allowable Stresses
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A composite member consists of a steel rod shaft in an aluminum tube. The members are fastened together by adjustable nuts. Dimensions are:
Sectional area of steel rod = 1.5 in²; modulus of elasticity = 30,000,000; allowable stress = 15,000 lb/in².
Sectional area of aluminum tube = 2 in²; modulus of elasticity = 10,000,000; allowable stress = 10,000 lb/in²

Determine initial stresses (prestressed) in the members so that under tensile loading both members will attain their allowable stresses simultaneously.

The strings: S₇P₃A₃₁ (Force - Pull).

The math:
Pj Problem of Interest is of type force (pull).

Assumptions: consistent deformation is applicable, that is, load caried by each member of the composite member is proportional to its stiffness if members are uniform.
Where stiffness = AE/l (A is area, E is modulus of elasticity, l is length).
Allowable stress same as working stress usually less than damaging stress.
Steel rod and aluminum tube are of equal length.

Equations of Interest:
Given n members of a composite member with respective area, A₁, A₂...A_n; moduli of elasticity, E₁, E₂ ...E_n and lengths, l₁, l₂ ...l_n.
Load carried by member k is P_k = P[_total(A_kE_k)/l_k]/[ⁿΣ_n=1 (A_nE_n)/l_n]-------(1).
Where k = 1,2...n.
When composite member is prestressed (i.e.has initial stress), then P_k represents the increment of force in each member due to the applied load.

Force in memeber = (Allowable Stress)Area -----------(2)

So, force in steel rod, P₁ = 1.5(15000) = 22,500 lb.
So, force in aluminum = 2(10000) = 20,000 lb.
So, total load on composite member = 22,500 + 20,000 = 42,5000 lb.

In the prestress case P_k of equation (1) is an increment.
Let P_i denote initial tension or compression in the members. Then:
For Steel rod:
P_isteel + 42,500[(1.5)(30)/((2)(10) + (1.5)(30))] = 22,500
So, P_isteel = -6920 lb (compression if we consider compression to be -ve).

For aluminum tube:
P_ialuminum + 42,500[(2)(10)/((2)(10) + (1.5)(30))] = 20,000
So, P_ialuminum = +6920 lb (tension if we consider tension to be +ve).

Math

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