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Strength of Materials I EGCE201 กำลังวัสดุ 1

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1 Strength of Materials I EGCE201 กำลังวัสดุ 1
Instructor: ดร.วรรณสิริ พันธ์อุไร (อ.ปู) ห้องทำงาน: 6391 ภาควิชาวิศวกรรมโยธา โทรศัพท์: 66(0) ต่อ 6391 Copyright © 2006 by Wonsiri Punurai

2 Deformations Under Uniform Axial Loading
From Hooke’s Law From the definition of strain Equating and solving for the deformation, ; ; ;

3 Example Steel bar (E=??) Cross-sectional area A=256 mm2
Find the elongation of the bar. Method: Identify segments Use equilibrium for internal axial Forces in each segment Use constitutive relation to relate Forces to displacements Use compatibility to get net displacement

4 Continuously varying bars (load, area, modulus could be varying with position)

5 Composite bars (Compound loading or discretely varying load/area/modulus )

6 Statically Indeterminate Problems
“Statically Indeterminate” means the # of unknowns exceeds the number of available equations of equilibrium. Statics (equilibrium analysis) alone cannot solve the problem nR = # of reactions (or unknowns) nE = # of equilibrium equations If nR = nE: statically determinate - forces in each member only depend on equilibrium. If nR > nE: statically indeterminate - too many unknowns, must invoke a constraint such as a deformation relation.

7 Statically Indeterminate Examples
Free body diagram Free body diagram

8 Statically Indeterminate: Load sharing Problem
A free body diagram Not sufficient to determine P1 and P2!!! If we consider the displacement of each member, additional information relating P1 and P2 becomes available In order for the rigid end plate to remain vertical, d1 =d2

9 Statically Indeterminate: Load sharing Problem

10 Statically Indeterminate: Reaction
Rod AB fixed between 2 rigid walls (-) compression (+) tension

11 Statically Indeterminate: Reaction

12 Statically Indeterminate: Superposition
The method of superposition is an effective way to determine reactions when a structure is statically indeterminate (held by more supports than required to maintain equilibrium causing “redundant reaction” ) The solution to the statically indeterminate using this method is (1) Considering the deformations caused by the given load and redundant reaction separately. (2) Superimposing the solutions for each upon one another to obtain the solution

13 Statically Indeterminate: Superposition
The reaction at C is redundant

14 Statically Indeterminate: Superposition

15 Temperature effect

16 Initial Stress/Strain and Temperature effect

17 Deformations under Thermal Effect
Coefficient of thermal expansion (CTE) - m/m/oF m/m/oC

18 Deformations under Thermal Effect

19

20 Thin Walled Pressure Vessels

21 Spherical Vessels

22 Torsion of shafts (แรงบิดภายในที่เกิดขึ้นด้ามเพลา)
Shafts are members with length greater than the largest cross sectional dimension used in transmitting torque from one plane to another Learning objectives Understand the theory and its limitation and its applications for design and analysis of Torsion of circular shafts.

23 Internal Torque (แรงบิดภายใน)

24 dF is related to the shearing stress: dF = tdA
Consider circular shaft AC subjected to equal and opposite torques T and T’. A cutting plane is passed through the shaft at B. The FBD for section BC must include the applied torque and elementary shearing forces dF. These forces are perpendicular To radius of the shaft and must balance to maintain equilibrium. The axis of the shaft is denoted as r. Taking moments about the Axis of the shaft results in dF is related to the shearing stress: dF = tdA So the applied torque can be related to the shearing stress as

25 Shear stress can’t exist on one plane only.
The applied torque produces a shear stress to the axis of the shaft. The equilibrium require equal stresses on the faces This results in the shear stress distribution shown.

26 Theory of Circular Shafts
Theory Objectives To obtain a formula for the relative rotation f2-f1 in terms of the internal torque T. To obtain a formula for the shear stress txq in terms of the internal torque T. f - angle of twist

27 Shearing Strain

28 Assume: Material is linearly elastic and isotropic
Shearing Stress Assume: Material is linearly elastic and isotropic Recall Define Polar moment of inertia for the cross section

29 Torsion formula Sign Convention Circular solid shaft
Circular hollow shaft with outer radius R, inner radius r Sign Convention

30 Relative rotation f2-f1 in terms of the internal torque T.
Shear stress txq in terms of the internal torque T. Maximum occurs at shaft’s outer radius

31 Direction of Shearing

32


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