WORCESTER POLYTECHNIC INSTITUTE

WORCESTER POLYTECHNIC INSTITUTE

MECHANICAL ENGINEERING DEPARTMENT STRESS ANALYSIS ES-2502, C’2012 Lecture 12: 02 February 2012

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General information Instructor: Cosme Furlong HL-151 (508) 831-5126 [email protected] http://www.wpi.edu/~cfurlong/es2502.html Teaching Assistants: Morteza Khaleghi HL-150 (508) 831-5125 [email protected]

Tatiana Popova [email protected] Mechanical Engineering Department

Statically indeterminate axially loaded member Axially loaded member

In this case, only one equilibrium equation:

FA



F

y

 0;

FB  FA  P  0 y FB

(1)

 Statically indeterminate problem

Need additional equations!!

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Statically indeterminate axially loaded member Additional equations are obtained by applying:

Axially loaded member

Compatibility or kinematic equations 

Load-displacement equations

y

 A/ B  0

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Statically indeterminate axially loaded member

Compatibility or kinematic equations:

A

 AC FA LAC FB LCB  0 AE AE

C

 CB B

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( 2)

Statically indeterminate axially loaded member Axially loaded member

Forces are obtained by solving system of equations:

FA

Equilibrium

FB  FA  P  0 y

FA LAC FB LCB  0 AE AE

FB Compatibility

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(1)

( 2)

Axial load: example F The 304 stainless steel post A has a diameter of d = 2 in and is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the average normal stresses developed in the post and the tube. Approach: 1) Apply equilibrium equations 2) Apply compatibility equations 3) Solve for stresses

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Axial load: example G The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 5 kip is applied to the rigid cap, determine the required diameter d of the steel post so that the load is shared equally between the post and tube. Approach: 1) Apply equilibrium equations 2) Apply compatibility equations 3) Solve for diameter

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Axial load: example H Two A-36 steel wires are used to support the 650-lbf engine. Originally, AB is 32 in long and A’B’ is 32.008 in long. Determine the forces supported by each wire when the engine is suspended from them. Each wire has a cross sectional area of 0.01 in2. Approach: 1) Apply equilibrium equations 2) Apply compatibility equations 3) Solve for forces

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Axial load: example I The concrete post is reinforced using six steel reinforcing rods, each having a diameter of 20 mm. Determine the stress in the concrete and the steel if the post is subjected to an axial load of 900 kN. Est = 200 GPa, Ec = 25 GPa.

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Principle of superposition Applied when a component is subjected to complicated loading conditions  break a complex problem into series of simple problems

P  P1  P2

Can only be applied for: (a) small deformations; (b) deformations in the elastic (linear) range of the  diagram

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Axial load: example J If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force P = 5 kN is applied. The assembly is made of A36 steel.

Approach: 1) Apply equilibrium equations 2) Apply compatibility equations 3) Solve for reactions

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Axial load: example K The two pipes are made of the same material and are connected as shown. If the cross-sectional area of BC is A and that of CD is 2A, determine the reactions at B and D when a force P is applied at the junction C.

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Axial load: example L The assembly consists of two red brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid cap G. If the supports at A, C and F are rigid, determine the average normal stress developed in rods AB, CD and EF.

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Thermal stresses: example Components are design to account for thermal expansions

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Thermal stresses: example In electronic components

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Thermal stresses: example In electronic components: J-lead attachment

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Thermal stresses: example In electronic components: J-lead attachment

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Thermal stresses: example In electronic components: J-lead attachment

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Thermal stresses: uniaxial effects

 T   T

 T   T L   T  L

(Thermal strains)

(Thermal deformations)



= linear coefficient of thermal expansion, 1/oC, 1/oF T = temperature differential L = original length of component

T , T

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Reading assignment

• Chapters 3 and 4 of textbook • Review notes and text: ES2001, ES2501

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Homework assignment

• As indicated on webpage of our course

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