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Principle of Superposition Often used to determine the stress or displacement at a point in a member when the member is subjected to a complicated loading. Resultant stress or displacement can be determined by algebraically summing the stress or displacement caused by each load component applied separately to the member.

Transcript of Principle of Superposition - WordPress.com · 05.07.2017 · Principle of Superposition ......

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Principle of Superposition

• Often used to determine the stress or displacement at a point in a member when the member is subjected to a complicated loading.

• Resultant stress or displacement can be determined by algebraically summing the stress or displacement caused by each load component applied separately to the member.

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Conditions for Superposition

1. The loading N must be linearly related to the stress S or displacement D that is to be determined.

2. The loading must not significantly change the original geometry or configuration of the member.

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Statically Indeterminate Axially Loaded Members• Consider the bar which is fixed supported

at both of its ends.

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Statically Indeterminate Axially Loaded Members• From its free-body diagram, there are two

unknown support reactions.

• Equilibrium requires

+ΣF = 0; FB + FA - 500 N = 0

• This type of problem is called statically indeterminate, since the equilibrium equation is not sufficient to determine both reactions on the bar.

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• In order to establish an additional equation needed for solution, it is necessary to consider how points on the bar are displaced.

• Specifically, an equation that specifies the conditions for displacement is referred to as a compatibility or kinematiccondition.

• This equation can be expressed in terms of the internal loads by using a load–displacement relationship.

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• 𝐹𝐴 + 𝐹𝐵 − 500 = 0

𝛿 =𝑁𝐿

𝐴𝐸

• 𝛿 =𝐹𝐴(2𝑚)

𝐴𝐸−

𝐹𝐵(3𝑚)

𝐴𝐸

• Two equations established

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Procedure for Analysis

• Equilibrium• Draw a free-body diagram • Write the equations of equilibrium for the member.

• Compatibility• Consider drawing a displacement diagram • Express the compatibility conditions in terms of the displacements

• Load–Displacement• Use a load–displacement relation, such as δ = NL/AE, to relate

the unknown displacements in the compatibility equation to the reactions.

• Solve all the equations for the reactions.

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Example 1

• The steel rod has a diameter of 10 mm. It is fixed to the wall at A, and before it is loaded, there is a gap of 0.2 mm between the wall at B’ and the rod. Determine the reactions on the rod if it is subjected to an axial force of P = 20 kN. Neglect the size of the collar at C. Take Est = 200 GPa.

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Example 2

• The aluminum post is reinforced with a brass core. If this assembly supports an axial compressive load of P = 9 kip, applied to the rigid cap, determine the average normal stress in the aluminum and the brass. Take Eal = 10(103) ksi and Ebr = 15(103) ksi.

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Example 3

• The three steel bars(E=29x103) are pin connected to a rigid member. If the applied load on the member is 15 kN, determine the force developed in each bar. Bars AB and EF each have a cross-sectional area of 50 mm2, and bar CD has a cross-sectional area of 30 mm2.

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Example 4

• The bolt is made of aluminum alloy, and it passes through the cylindrical tube made of magnesium alloy. The tube has an outer radius of 1/2 in., and it is assumed that both the inner radius of the tube and the radius of the bolt are 1/4 in. When the bolt is snug against the tube it produces negligible force in the tube. Using a wrench, the nut is then further tightened one-half turn. If the bolt has 20 threads per inch, determine the stress in the bolt.

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Thermal Stress

• A change in temperature can cause a body to change its dimensions.

• Generally, if the temperature increases, the body will expand, whereas if the temperature decreases, it will contract.

• Ordinarily this expansion or contraction is linearly related to the temperature increase or decrease that occurs.

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Thermal Stress

If the material is homogeneous and isotropic, it has been found from experiment that the displacement of the end of a member having a length L can be calculated using the formula

𝛿𝑇 = 𝛼∆𝑇𝐿

𝛼 = a property of the material, referred to as the linear coefficient of thermal expansion.

∆𝑇 = the algebraic change in temperature of the member

L = the original length of the member

𝛿𝑇 = the algebraic change in the length of the member

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Example 1

• The steel bar is constrained to just fit between two fixed supports when T1 = 60˚F. If the temperature is raised to T2 = 120˚F, determine the average normal thermal stress developed in the bar. 𝛼 =[6.60(10-6)/˚F], E= 29(103) kip/in2]

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Example 2

• The rigid beam shown is fixed to the top of the three posts made of steel and aluminum. The posts each have a length of 250 mm when no load is applied to the beam, and the temperature is T1= 20˚C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 80˚C.