CH 4: Deflection and Stiffness
CH 4: Deflection and Stiffness
Stress analyses are done to ensure that machine elements will not fail due to stress levels exceeding the allowable values. However, since we are dealing with deformable bodies (not rigid), deflections should be considered also where they are in many cases more limiting than stresses. Take for example shafts where excessive deflection will interfere with the function of the elements mounted on the shaft and might cause failure of the system, thus usually shafts are designed based on deflections rather than stresses.
Spring Rates
In most types of loading situations, the stress developed in the element (bar, shaft, beam, etc.) is linearly related to the loading. As long as the stress in the material remains within the linear elastic region, the stress is also linearly related to the deflection. Therefore, there is a linear relation between load and deflection and elements under loading behave similar to linear springs, and thus we can define the spring rate or spring constant for the element as:
Axial, lateral, moment, torque, etc. 퐿표푎푑𝑖푛푔 푘 = 퐷푒푓푙푒푐푡𝑖표푛 Axial, lateral, bending, twisting, etc.
Tension, Compression and Torsion For a bar with constant cross-section the deformation is found as: 퐹 퐿 훿 = 퐴 퐸 Thus, the spring constant for an axially loaded bar is: 퐴 퐸 푘 = 퐿
For a round shaft subjected to torque, the angular deflection is found as: 푇퐿 휃 = 휃 in radians 퐺퐽
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 1 of 23 Thus, the spring constant is:
퐺 퐽 푘 = 퐿
Deflection Due to Bending
The deflection of beams is much larger than that of axially loaded elements, and thus the problem of bending is more critical in design than other types of deformation.
- Shafts are treated as beams when analyzed for lateral deflection.
The beam governing equations are:
푞 푑4푦 Load intensity = 퐸 퐼 푑푥4
푉 푑3푦 Shear force = 퐸퐼 푑푥3
푀 푑2푦 Moment = 퐸 퐼 푑푥2
푑푦 Slope 휃 = 푑푥
Deflection 푦 = 푓(푥)
Knowing the load intensity function, we can integrate four times (using the known boundary conditions to evaluate the integration constants) to get the deflection. However, in most cases, the expression of bending moment is easy to find (using sections) and thus we start from the moment governing equation and integrate to get the slope and deflection.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 2 of 23 Example: The beam shown has constant cross-section and it is made from homogeneous isotropic material. Find the deflection, slope and the location and value of maximum slope.
Solution: 푑2푦 푀 푃 푥 = = − 푑푥2 퐸 퐼 퐸 퐼 푑푦 푃푥2 Integrating = − + 퐶 (1) 푑푥 2 퐸 퐼 1 푃푥3 Integrating 푦 = − + 퐶 푥 + 퐶 (2) 6 퐸 퐼 1 2
푑푦 B.C.’s: 1 푦 = 0 @ 푥 = 퐿 2 = 0 @ 푥 = 퐿 푑푥
Substituting B.C. 2 in equation (1) we get: 푃퐿2 퐶 = 1 2 퐸 퐼 푃푥3 푃퐿2 푦 = − + 푥 + 퐶 (3) 6퐸퐼 2퐸퐼 2 Substituting B.C. 1 in equation (3) we get: 푃퐿3 푃퐿3 0 = − + + 퐶 6퐸퐼 2퐸퐼 2 푃퐿3 퐶 = − 2 3퐸퐼 푃 Thus, 푦 = (−푥3 + 3퐿2푥 − 2퐿3) 6퐸퐼
푑푦 푃 and the slope = (퐿2 − 푥2) 푑푥 2퐸퐼
푑 푑푦 Maximum slope occurs when: ( ) = 0 푥 = 0 푑푥 푑푥
푑푦 푃퐿2 ( ) = 푑푥 푥=0 2퐸퐼
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 3 of 23 Beam Deflections by Superposition
Superposition resolves the effect of combined loading on a structure by determining the effects of each load separately and adding the results algebraically.
Superposition: Resolve, Find, Add.
The resulting reactions, deflections and slopes of all common types of loading and boundary conditions are available and can be found in specialized books such as “Roak’s Formulas for Stress and Strain”.
Table A-9 gives the results for some common cases. Note: superposition is valid as long as the deflections are small such that the deformation resulting from one type of loading will not alter the geometry. Also, each effect should be linearly related to the load that produces it.
Example:
Solved by integration
Table A-9, Case 6
Table A-9, Case 8
Note that: Table A-9, Case 10 RA=R1+R3+R5+R7 RB=R2+R4+R6+R8
See Examples 4-3 & 4-4 from text
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 4 of 23 Beam Deflections by Singularity Functions
As shown in Ch. 3, singularity functions can be used to write an expression for loading over a range of discontinuities.
The loading intensity function 푞(푥) can be integrated four times to obtain the deflection equation푦(푥).
(Recall the singularity functions integration and evaluation rules which has been introduced in Ch.3)
See Examples 4-5 & 4-6 from text
Strain Energy
When loads are applied to an elastic member, the member will elastically deform, thus transferring the work done by the load into potential energy called strain energy.
If a load “퐹” is applied to a member and as a result the member deformed a distance “푦”, then the work done by the load is (if the relation between the load and deformation is linear):
1 퐹 푈 = 퐹푦 but, 푦 = 2 푘 spring rate 퐹2 푈 = 2푘
This equation defines the strain energy in general where the load can also mean a torque or moment provided that consistent units are used for “푘”.
Therefore, the strain energy for different types of loading can be defined as follows:
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 5 of 23 Factors Strain Energy Strain Energy Loading Type (All factors are Involved (General expression) constant with 풙)
퐿 Axial Force 퐹2퐿 퐹2 F, E, A 푈 = 푈 = ∫ 푑푥 2퐸퐴 2퐸퐴 0
퐿 Bending Moment 푀2퐿 푀2 M, E, I 푈 = 푈 = ∫ 푑푥 2퐸퐼 2퐸퐼 0
퐿 Torque 푇2퐿 푇2 T, G, J 푈 = 푈 = ∫ 푑푥 2퐺퐽 2퐺퐽 0
퐿 Transverse Shear 푉2퐿 푉2 V, G, A 푈 = 퐶 푈 = ∫ 퐶 푑푥 2퐺퐴 2퐺퐴 0
- Where 퐶 is the cross-section correction factor (Table 4-1 in text) : 퐶 = 1.2 for rectangular sections 퐶 = 1.11 for circular sections 퐶 = 2 for thin-walled circular sections 퐶 = 1 for box and structural sections (where only the area of the web is used)
Example: The cantilever beam shown has a stepped square cross-section and it is made of steel (퐸 = 210 퐺푃푎, 퐺 = 75 퐺푃푎). Find the stain energy in the beam.
Solution: 푀 = 500 푥 푁. 푚푚 & 푉 = 500 푁
250 푀2 500 푀2 Due to moment 푈푀 = 푈푀1 + 푈푀2 = ∫ 푑푥 + ∫ 푑푥 = 0 2퐸퐼1 250 2퐸퐼2 250 500 1 250 25×104 푥2 500 25×104 푥2 1 25 푥3 25 푥3 [∫ 5 푑푥 + ∫ 5 푑푥] = 3 [| | + | | ] 2퐸 0 4×10 250 8×10 2(210×10 ) 120 0 240 250
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 6 of 23 푈푀 = 34.877 푚퐽
2 2 푉 퐿1 푉 퐿2 Due to shear 푈푉 = 푈푉1 + 푈푉2 = 퐶 + 퐶 2퐺퐴1 2퐺퐴2 Very small compared to the 5002 250 250 strain energy due to moment = 1.2 ( + ) = 0.389 푚퐽 2(75×103) 2.2×103 3.1×103
푈푇표푡푎푙 = 푈푀 + 푈푉 = 35.266 푚퐽 Usually the strain energy due to shear is neglected for long beams
Castigliano’s Theorem
Castigliano’s theorem is one of the energy methods (based on strain energy) and it can be used for solving a wide range of deflection problems.
Castigliano’s theorem states that when a body is elastically deformed by a system of loads, the deflection at any point “푃” in any direction “푎” is equal to the partial derivative of the strain energy with respect to a load at “푃” in the direction “푎”.
The theory applies to both linear and rotational deflections 휕푈 훿𝑖 = 휕퐹𝑖
Where 훿𝑖 is the displacement of the point of application of the force 퐹𝑖 in the direction of 퐹𝑖 휕푈 표푟, 휃𝑖 = 휕푀𝑖
Where 휃𝑖 is the rotation (in radians) at the point of application of the moment 푀𝑖 in the direction of 푀𝑖 It should be clear that Castiglione’s theorem finds the deflection at the point of application of the load in the direction of the load. Q: What about if there is no load at the desired point in the desired direction?
A: We apply a fictitious “dummy” load “푄” at that point in the desired direction.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 7 of 23 Procedure for using Castigliano’s theorem:
- Obtain an expression for the total strain due to all loads acting on the member (including the dummy load if one is used). - Find the deflection of the desired point (in the desired direction) as:
휕푈 If the deflection turns out to be 훿 = negative, it means that it is in the 𝑖 휕퐹 𝑖 opposite direction of the load
or if a dummy load 푄𝑖 is used:
휕푈 After taking the derivative, set 훿𝑖 = [ ] 휕푄 푄𝑖 = 0 to find the deflection 𝑖 푄푖=0 Note: If the strain energy is in the form of integral, it is better to take the partial
derivative with respect to the load “퐹𝑖” before integrating.
See Examples 4-8 & 4-9 from text
Example: For the cantilevered bar subjected to the horizontal load “P” as shown. Find the vertical deflection at the free end “Point A”. (Neglect the transverse shear). Solution: The components that define the total strain energy are:
- Bending in AB, where MAB = P y - Bending in BC, where MBC = P h+ Q x - Axial load in AB = Q Since there is no vertical - Axial load in BC = P load acting at point ”A”, we add a dummy load 푄 Thus, the total strain energy is: Since the axial loads are of constant magnitude ℎ 퐿 푃2푦2 (푃ℎ + 푄푥)2 푄2ℎ 푃2퐿 푈 = ∫ 푑푦 + ∫ 푑푥 + + 2퐸퐼 2퐸퐼 2퐸퐴 2퐸퐴 0 0
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 8 of 23 The vertical deflection at “A” is:
퐿 휕푈 푥(푃ℎ + 푄푥) 푄ℎ 푃ℎ푙2 푄푙3 푄ℎ 훿퐴 = [ ] = ∫ 푑푥 + = + + 휕푄 푄=0 퐸퐼 퐸퐴 2퐸퐼 3퐸퐼 퐸퐴 0 Set 푄 = 0
푃ℎ푙2 훿 = 퐴 2퐸퐼
The expressions for finding deflections using Castigliano’s theorem can be simplified and written as: 휕푈 1 휕퐹 훿𝑖 = = ∫ (퐹 )푑푥 푓표푟 푇푒푛푠𝑖표푛 표푟 퐶표푚푝푟푒푠푠𝑖표푛 휕퐹𝑖 퐴퐸 휕퐹𝑖
휕푈 1 휕푀 훿𝑖 = = ∫ (푀 ) 푑푥 푓표푟 퐵푒푛푑𝑖푛푔 휕퐹𝑖 퐸퐼 휕퐹𝑖
휕푈 1 휕푇 휃𝑖 = = ∫ (푇 ) 푑푥 푓표푟 푇표푟푠𝑖표푛 휕푇𝑖 퐺퐽 휕푇𝑖
See Example 4-11 from text
Deflection of Curved Beams
Examples of curved beams include machine frames, springs, clips, etc. The deflection of curved beams can be found using Castigliano’s theorem.
Consider a curved beam subjected to “in-Plane” loading as shown: - Note that the load “퐹” is purely shear at θ = 0, and purely normal at θ = 90°. Thus, polar coordinates are used to represent the internal loading.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 9 of 23 The internal loading acting at any angle θ is:
Shear: Fr = F cos θ
Axial: Fθ = F sin θ
Bending Moment: M = F R sin θ
The total strain energy in the beam consists of four components: Strain energy due to bending moment 푀:
푀2 푈 = ∫ 푑휃 1 2퐴푒퐸
Where “푒” is the eccentricity (푒 = 푅 − 푟푛) However, when the radius “푅” is large compared to the height “ℎ” the strain energy due to moment can be approximated as:
푀2푅 푅 푈 = ∫ 푑휃 푓표푟 > 10 1 2퐸퐼 ℎ
The strain energy due to normal “axial” force 퐹휃 has two components: - Due to the axial force 퐹휃: 퐹2푅 푈 = ∫ 휃 푑휃 2 2퐴퐸
- Due to the moment produced by the axial force 퐹휃 (which has an opposite direction to the moment 푀):
푀퐹 푈 = − ∫ 휃 푑휃 3 퐴퐸
Because it causes deflection in the direction opposite to the force” 퐹”
The strain energy due to shear force 퐹푟
2 퐹푟 푅 푈 = ∫ 퐶 푑휃 4 2퐴퐺
Where “퐶” is the cross-section correction factor
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 10 of 23 According to Castigliano’s theorem, the deflection in the direction of the force “퐹” is found as:
휕푈 휕푈1 휕푈2 휕푈3 휕푈4 훿 = = + + + 휕퐹 휕퐹 휕퐹 휕퐹 휕퐹 Taking the partial derivatives and integrating between 0 to π, then simplifying we get:
휋퐹푅2 휋퐹푅 휋퐶퐹푅 훿 = − + 2퐴푒퐸 2퐴퐸 2퐴퐺 Note that the first term (which is due to moment) will be much larger than the other two terms when 푅 is large, since it has 푅2, and thus the other terms can be neglected and we can use the approximate expression of strain energy for 푅/ℎ > 10 which gives an approximate result:
휋퐹푅2 훿 ≅ 2퐸퐼
Example: For the beam shown, find the vertical deflection at point “A” considering the moment only and knowing that R/h>10 for the curved portion. Solution: Since the moment only is considered, the total strain energy has two components:
Due to moment in AB :
MAB = Fx
Due to moment in BC :
MBC = FL + FR ( 1 - cos θ )
퐿 휋⁄2 푀2 푀2 푈 = 푈 + 푈 = ∫ 퐴퐵 푑푥 + ∫ 퐵퐶 푅 푑휃 퐴퐵 퐵퐶 2퐸퐼 2퐸퐼 0 0
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 11 of 23 퐿 휋/2 휕푈 푥 (퐿 + 푅(1 − 푐표푠휃)) 훿 = = ∫(퐹푥) 푑푥 + ∫ (퐹퐿 + 퐹푅(1 − cos θ)) 푅푑휃 휕퐹 퐸퐼 퐸퐼 0 0 퐹 훿 = {4퐿3 + 3푅[2휋퐿2 + 4(휋 − 2)퐿푅 + (3휋 − 8)푅2]} 12퐸퐼
Statically Indeterminate Problems
A statically indeterminate problem has more unknown reactions than what can be found using static equilibrium equations (i.e., number of unknown is more than the number of available equations).
The additional supports that are not necessary to keep the member in equilibrium are called redundant supports and their reactions are called “redundant reactions”. The number of redundant reactions is defined as the degree of indeterminacy, and the same number of additional equations is needed to solve the problem.
How to solve statically indeterminate problems?
Using integration or singularity functions: 1. Write the equations of static equilibrium in terms of applied loads and the unknown reactions. 2. Obtain the expressions for slope and deflection of the beam. 3. Apply the boundary conditions consistent with the restraints “reactions” 4. Solve the equations obtained from steps 1 and 3.
Using Superposition: - Choose the redundant reaction(s). - Write the equations of static equilibrium for the remaining reactions in terms of applied loads and the redundant reaction.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 12 of 23 - Using superposition divide the loading where once the applied loading is present and the redundant reaction is removed, and the other time the applied load is removed and the redundant reaction is present. - Find the deflection (can be found from tables or using Castigliano’s theorem) at the location of the redundant reaction in both cases and use that as an additional compatibility equation where the sum of the deflections is zero. (note that if the redundant reaction is a moment, the corresponding deflection will be slope) - Solve the compatibility equation(s) for the redundant reaction, and then find the remaining reactions from the equilibrium equations.
Example: Choosing R2 & M2 as the two redundant reactions
Compatibility equations: Solved for the 훿1 + 훿2 + 훿3 = 0 redundant reactions R2 & M2 휃1 + 휃2 + 휃3 = 0
where:
Can be used to M1 = M1’+M1’’+M1’’’
find R1 & M1 R1 = R1’+R1’’+R1’’’
Example: Determine the reactions for the beam shown using superposition.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 13 of 23 Choosing the reaction 푀1to Solution: be the redundant reaction From equilibrium we get:
퐹 푀1 푅 = + (1) 1 2 퐿 퐹 푀1 푅 = − (2) 2 2 퐿 From Table A-9 (case 5):
2 푑푦 퐹 2 2 퐹퐿 휃1 = [ ] = [ (12푥 − 3퐿 )] = − 푑푥 푥=0 48퐸퐼 푥=0 16퐸퐼 From Table A-9 (case 8) and setting a = 0 we get:
푑푦 푀1 2 2 푀1퐿 휃2 = [ ] = [ (3푥 + 2퐿 )] = 푑푥 푥=0 6퐸퐼퐿 푥=0 3퐸퐼 Compatibility:
2 푀1퐿 퐹퐿 3퐹퐿 휃 + 휃 = 0 − = 0 푀 = 1 2 3퐸퐼 16퐸퐼 1 16 Substituting in equations 1 & 2 we get:
11퐹 5퐹 푅 = & 푅 = 1 16 2 16 See the solution in Example 4-14 in the text book
Compression Members
The analysis and design of compression members “Columns” is different from members loaded in tension. A column is a straight and long (relative to its cross-section) bar that is subjected to a compressive axial load. A column can fail due to buckling (a sudden, large lateral deflection) before the compressive stress in the column reach its allowable (yield) value.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 14 of 23 Buckling or “elastic instability” can cause catastrophic failure of structures. To understand why buckling happens, it is important to understand equilibrium regimes. There are three states of equilibrium: - Stable Equilibrium: when the member is moved it tends to return to its original equilibrium position. - Neutral Equilibrium: when the member is moved from its equilibrium position, it is still in equilibrium at the displaced position. - Unstable Equilibrium: when the member is moved from its equilibrium position it becomes imbalanced and accelerates away from its equilibrium position.
With increasing compressive loading, the state of equilibrium of a column changes from stable to neutral to unstable.
The load that causes a column to become unstable is called the “critical buckling load” where under unstable equilibrium condition; any small lateral movement will cause buckling (catastrophic failure).
According to geometry and loading, columns can be categorized as: - Long columns with central loading. - Intermediate length columns with central loading. - Columns with eccentric loading. - Stratus or short columns with eccentric loading.
Long Columns with Central Loading
The critical “buckling” load for long columns with central loading is predicted by Euler formula and it depends on: - End conditions: four types of end conditions; pinned-pinned, fixed-fixed, fixed- pinned and fixed-free. - The column material “퐸”. - Geometry: length and cross-section.
Consider a pinned-pinned column of length “퐿” and subjected to axial load “푃”. - Assume that the column became slightly bent.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 15 of 23 The bending moment developed in the column is: 푀 = −푃푦
Using the beam governing equation:
푑2푦 푀 = 푑푥2 퐸퐼 푑2푦 푃 = − 푦 푑푥2 퐸퐼 or 푑2푦 푃 + 푦 = 0 푑푥2 퐸퐼
Which is a homogeneous, second order differential equation. The general solution for the differential equation is: 푃 푃 푦 = 퐴 푠𝑖푛√ 푥 + 퐵 푐표푠√ 푥 (1) 퐸퐼 퐸퐼 The equation can be evaluated using the boundary conditions: 1. 푦 = 0 @ 푥 = 0 2. 푦 = 0 @ 푥 = 퐿
From B.C.1: 퐵 = 0
푃 From B.C.2: 0 = 퐴 푠𝑖푛√ 퐿 퐸퐼
The constant “퐴” cannot be zero because that is a trivia solution which means that the column will not be buckle. Thus, 푃 푠𝑖푛√ 퐿 = 0 퐸퐼
This is satisfied when:
푃 √ 퐿 = 푛휋 where 푛 = 1,2,3 … 퐸퐼
Solving for “푃” we get:
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 16 of 23 푛2휋2퐸퐼 푃 = 퐿2 where the smallest “푃” is obtained for 푛 = 1 (The first mode of buckling)
Thus, the critical load “푃푐푟” for a column with pinned ends is:
휋2퐸퐼 푃 = 퐸푢푙푒푟 푓표푟푚푢푙푎 푐푟 퐿2 Substituting back in equation (1) we get the deflection curve: 휋푥 푦 = 퐴 푠𝑖푛 퐿 which is a half-wave
It should be realized that for columns with non-circular cross- section, the column will buckle about the axis of the cross-section having the smallest moment of inertia (the weakest axis).
Using the relation 퐼 = 퐴푘2 where “퐴” is the cross-sectional area and “푘” is the radius of gyration, we can write Euler formula in a more convenient form:
푃 휋2퐸 The stress value that will C ritical buckling stress 휎 = 푐푟 = 푐푟 퐴 (퐿/푘)2 cause a column to be in unstable equilibrium condition
where the ratio (퐿/푘) is called the “slenderness ratio”
This ratio (rather than the actual length) is used to classify columns in to length categories.
The critical loads for columns with different end conditions can be obtained by solving the differential equation and it can be seen by comparing the buckled shapes.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 17 of 23 Euler formula for different end conditions can be written in the form:
휋2퐸퐼 푃푐푟 = 2 퐿푒
where 퐿푒 is the “effective length” Or alternatively Euler formula can be written as:
2 2 퐶휋 퐸퐼 푃푐푟 퐶휋 퐸 푃 = 표푟 휎 = = 푐푟 퐿2 푐푟 퐴 (퐿/푘)2
where 퐶 is the “end-condition constant”
In reality it is not possible to fix an end (even if it is welded); thus, recommended “realistic” values of 퐶 are given in Table 4-2.
Plotting the Euler critical stress vs. the slenderness ratio we get:
It is noted that for slenderness ratio less than that of point “T”; buckling occurs at a stress value less than that predicted by Euler formula (it follows the dashed line).
Thus, the Euler formula is used for slenderness ratios larger than (L/k)1 which is given as:
Limiting 퐿 퐿 퐿 2휋2퐶퐸 Long columns: ( ) > ( ) slenderness ratio ( ) = √ 푘 푘 1 푘 1 푆푦
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 18 of 23 Intermediate-Length Columns with Central Loading
For columns with slenderness ratio smaller than (L/k)1, Euler formula cannot predict the critical buckling load. For this range the Johnson formula is used which is a parabolic fit between “푆푦” and point “T”.
The Johnson formula predicts the critical stress as:
2 2 푃푐푟 푆푦 퐿 퐿 퐿 휎푐푟 = = 푆푦 − 2 ( ) 푓표푟 < ( ) 퐴 4휋 퐶퐸 푘 푘 푘 1
Or alternatively using the effective length “퐿푒” it is written as:
2 2 푆푦 퐿푒 휎 = 푆 − ( ) 푐푟 푦 4휋2퐸 푘
Example: A column with one end fixed and the other free is to be made of Aluminum 2 alloy 2014 (E=72GPa, Sy=97MPa). The column cross-sectional area is to be 600 mm and its length is 2.5 m. Find the critical buckling load for the following cross-sections: a) A solid round bar b) A solid square bar Solution: Fixed-Free: 퐶 = 1/4 a) Round bar 휋 4퐴 4(600) 퐴 = 푑2 푑 = √ = √ = 27.64 푚푚 4 휋 휋
휋 휋 Moment of Inertia: 퐼 = 푑4 = (27.64)4 = 28650 푚푚4 64 64
퐼 28650 Radius of gyration: 푘 = √ = √ = 6.91 푚푚 퐴 600
퐿 2500 Slenderness ratio: ( ) = = 361.8 푘 6.91
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 19 of 23 Limiting slenderness ratio:
퐿 2휋2퐶퐸 2휋2(0.25)(72 × 109) √ ( ) = √ = 6 = 60.52 푘 1 푆푦 97 × 10
Comparing: 361.8 > 60.52 Long column (푢푠푒 퐸푢푙푒푟 푓표푟푚푢푙푎)
퐶휋2퐸퐼 (0.25)휋2(72 × 109)(28.65 × 10−9) 푃 = = = 814.4 푁 푐푟 퐿2 2.52 b) Square bar 퐴 = 푏2 푏 = √퐴 = √600 = 24.5 푚푚
푏4 24. 54 퐼 = = = 30000 푚푚4 12 12
퐼 30000 푘 = √ = √ = 7.07 푚푚 퐴 600
퐿 2500 ( ) = = 353.6 푘 7.07 353.6 > 60.52 Long column (푢푠푒 퐸푢푙푒푟 푓표푟푚푢푙푎)
(0.25)휋2(72 × 109)(30 × 10−9) 푃 = = 852.7 푁 푐푟 2. 52
The example shows that for the same cross-sectional area the square section has slightly higher critical buckling load than the circular one.
To increase the critical buckling load without increasing the cross-sectional area, hollow tube sections are used (since they have higher moment of inertia).
When thin-walled hollow tubes are used, wall buckling is considered too. For this reason circular tubes are better than square tubes because the buckling load of curved walls is higher than that of flat walls. The critical stress causing wall buckling of thin-wall circular tubes can be found as: 퐸푡 휎푐푟 = 푅√3(1 − 푣2) 퐸 & 푣 are young’s modulus and poisson’s ratio and 푅 & 푡 are the tube radius and wall thickness.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 20 of 23 Columns with Eccentric Loading
In most applications the load is not applied at the centroid of the cross-section, but rather it is eccentric.
The distance between the centroidal axis and the point of load application is called the eccentricity “푒”.
Considering a pinned-pinned column with eccentric load “푃”.
The bending moment developed in the column is:
푀 = −푃(푒 + 푦)
푑2푦 푀 Using the beam governing equation = we get: 푑푥2 퐸퐼 푑2푦 푃 푃푒 + 푦 = − 푑푥2 퐸퐼 퐸퐼 Using the same procedure used before to solve the differential equation, and the same B.C.s we can obtain the deflection equation. The maximum deflection occurs at mid-span (x=L/2) and it is found to be:
퐿 푃 훿 = 푦 = 푒 [푠푒푐 ( √ ) − 1] 푚푎푥 2 퐸퐼
and thus, maximum moment at mid-span is:
퐿 푃 푀 = 푃(푒 + 훿) = 푃푒 푠푒푐 ( √ ) 푚푎푥 2 퐸퐼
The maximum compressive stress at mid-span has two components; axial and bending:
푃 푀푐 푃 푀푐 where “c” is the distance from the 휎푐 = + = + 퐴 퐼 퐴 퐴푘2 neutral axis to the outer surface
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 21 of 23 Substituting the value of the maximum moment at mid-span we get:
Max compressive 푃 푒푐 퐿 푃 stress at mid span 휎 = [1 + 푠푒푐 ( √ )] 푐 퐴 푘2 2푘 퐸퐴
where the term (ec/k2) is called the eccentricity ratio
If we set the maximum value of compressive strength to be equal to the
compressive yield strength of the column material “푆푦푐” we can write:
퐴 푆푦푐 푷 = 푒푐 퐿 푷 The secant equation 1 + ( ) 푠푒푐 ( 푒 √ ) 푘2 2푘 퐸퐴
Where 퐿푒 is the effective length for any end condition
Note that “푷” appears twice in the equation, thus this equation needs an iterative process to find the critical buckling load. For the first iteration the critical buckling load obtained from Euler formula
“푃푐푟” is used, and the iterations continue until both sides converge.
Critical load decreases with increasing eccentricity ratio
Struts, or Short Compression Members
When a short bar is loaded in compression by a load “푃” acting along the centroidal axis, the stress increases uniformly as σ = P/A until it reaches the yield strength of the material.
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 22 of 23 However, if the load is eccentric an additional component of stress, due to the bending moment, is introduced:
푃 푀푦 푃 (푃푒)푦 푃 푒푦 휎 = + = + = (1 + ) 푐 퐴 퐼 퐴 퐼 퐴 푘2 The maximum compressive stress occurs at point “B” on the surface where 푦 = 푐
푃 푒푐 휎 = (1 + ) 푐 퐴 푘2
If the maximum stress value is 푆푦, then we can solve for the critical load 푃푐푟.
To distinguish short columns (or struts) from long columns with eccentric loading the slenderness ratio is used, where the limiting value is:
퐿 퐴퐸 Limiting value of slenderness ratio ( ) = 0.282√ for columns with eccentric loading 푘 2 푃
퐿 퐿 if > ( ) 퐿표푛푔 푐표푙푢푚푛 (푢푠푒 푠푒푐푎푛푡 푓표푟푚푢푙푎) 푘 푘 2 퐿 퐿 if < ( ) 푆푡푟푢푡 (푠ℎ표푟푡 푐표푙푢푚푛) 푘 푘 2
See Example 4-20 from text
Shigley’s Mechanical Engineering Design, 10th Ed. Class Notes by: Dr. Ala Hijazi
CH 4 (R1) Page 23 of 23