Initial Stress In A Composite Member Under Tensile Loading Given Allowable Stresses
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Initial Stress In A Composite Member Under Tensile Loading Given Allowable Stresses 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...An; moduli of elasticity, E1, E2 ...En and lengths, l1, l2 ...ln.
Load carried by member k is Pk = P[total(AkEk)/lk]/[nΣn=1 (AnEn)/ln]-------(1).
Where k = 1,2...n.
When composite member is prestressed (i.e.has initial stress), then Pk 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 Pk of equation (1) is an increment.
Let Pi denote initial tension or compression in the members. Then:
For Steel rod:
Pisteel + 42,500[(1.5)(30)/((2)(10) + (1.5)(30))] = 22,500
So, Pisteel = -6920 lb (compression if we consider compression to be -ve).

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

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