STRUCTURAL STABILITY - Half Course

STRUCTURAL STABILITY - Half Course

CE 579: STRUCTRAL STABILITY AND DESIGN Amit H. Varma Assistant Professor School of Civil Engineering Purdue University Ph. No. (765) 496 3419 Email: [email protected] Office hours: M-W-F 9:00-11:30 a.m. Chapter 1. Introduction to Structural Stability OUTLINE Definition of stability Types of instability Methods of stability analyses Examples small deflection analyses Examples large deflection analyses Examples imperfect systems Design of steel structures STABILITY DEFINITION

Change in geometry of a structure or structural component under compression resulting in loss of ability to resist loading is defined as instability in the book. Instability can lead to catastrophic failure must be accounted in design. Instability is a strength-related limit state. Why did we define instability instead of stability? Seem strange! Stability is not easy to define. Every structure is in equilibrium static or dynamic. If it is not in equilibrium, the body will be in motion or a mechanism. A mechanism cannot resist loads and is of no use to the civil engineer. Stability qualifies the state of equilibrium of a structure. Whether it is in stable or unstable equilibrium. STABILITY DEFINITION Structure is in stable equilibrium when small perturbations do not cause large movements like a mechanism. Structure vibrates about it equilibrium position. Structure is in unstable equilibrium when small perturbations produce large movements and the structure never returns to its original equilibrium position.

Structure is in neutral equilibrium when we cant decide whether it is in stable or unstable equilibrium. Small perturbation cause large movements but the structure can be brought back to its original equilibrium position with no work. Thus, stability talks about the equilibrium state of the structure. The definition of stability had nothing to do with a change in the geometry of the structure under compression seems strange! STABILITY DEFINITION BUCKLING Vs. STABILITY Change in geometry of structure under compression that results in its ability to resist loads called instability. Not true this is called buckling. Buckling is a phenomenon that can occur for structures under compressive loads. The structure deforms and is in stable equilibrium in state-1. As the load increases, the structure suddenly changes to deformation state-2 at some critical load Pcr.

The structure buckles from state-1 to state-2, where state-2 is orthogonal (has nothing to do, or independent) with state-1. What has buckling to do with stability? The question is - Is the equilibrium in state-2 stable or unstable? Usually, state-2 after buckling is either neutral or unstable equilibrium BUCKLING PPcr P P P P BUCKLING Vs. STABILITY Thus, there are two topics we will be interested in this course Buckling Sudden change in deformation from state-1 to state-2 Stability of equilibrium As the loads acting on the structure are

increased, when does the equilibrium state become unstable? The equilibrium state becomes unstable due to: Large deformations of the structure Inelasticity of the structural materials We will look at both of these topics for Columns Beams Beam-Columns Structural Frames TYPES OF INSTABILITY Structure subjected to compressive forces can undergo: 1. Buckling bifurcation of equilibrium from deformation state-1 to state-2. 2. Failure due to instability of equilibrium state-1 due to large deformations or material inelasticity Bifurcation buckling occurs for columns, beams, and symmetric frames under gravity loads only

Elastic instability occurs for beam-columns, and frames subjected to gravity and lateral loads. Inelastic instability can occur for all members and the frame. We will study all of this in this course because we dont want our designed structure to buckle or fail by instability both of which are strength limit states. TYPES OF INSTABILITY BIFURCATION BUCKLING Member or structure subjected to loads. As the load is increased, it reaches a critical value where: The deformation changes suddenly from state-1 to state-2. And, the equilibrium load-deformation path bifurcates. Critical buckling load when the load-deformation path bifurcates Primary load-deformation path before buckling Secondary load-deformation path post buckling Is the post-buckling path stable or unstable? SYMMETRIC BIFURCATION Post-buckling load-deform. paths are symmetric about load axis. If the load capacity increases after buckling then stable symmetric bifurcation. If the load capacity decreases after buckling then unstable

symmetric bifurcation. ASYMMETRIC BIFURCATION Post-buckling behavior that is asymmetric about load axis. INSTABILITY FAILURE There is no bifurcation of the load-deformation path. The deformation stays in state-1 throughout The structure stiffness decreases as the loads are increased. The change is stiffness is due to large deformations and / or material inelasticity. The structure stiffness decreases to zero and becomes negative. The load capacity is reached when the stiffness becomes zero. Neutral equilibrium when stiffness becomes zero and unstable equilibrium when stiffness is negative. Structural stability failure when stiffness becomes negative. INSTABILITY FAILURE FAILURE OF BEAM-COLUMNS P M K=0 M

K<0 K M P No bifurcation. Instability due to material and geometric nonlinearity INSTABILITY FAILURE Snap-through buckling P Snap-through INSTABILITY FAILURE Shell Buckling failure very sensitive to imperfections Chapter 1. Introduction to Structural Stability OUTLINE Definition of stability Types of instability Methods of stability analyses

Examples small deflection analyses Examples large deflection analyses Examples imperfect systems Design of steel structures METHODS OF STABILITY ANALYSES Bifurcation approach consists of writing the equation of equilibrium and solving it to determine the onset of buckling. Energy approach consists of writing the equation expressing the complete potential energy of the system. Analyzing this total potential energy to establish equilibrium and examine stability of the equilibrium state. Dynamic approach consists of writing the equation of dynamic equilibrium of the system. Solving the equation to determine the natural frequency () of the system. Instability corresponds to the reduction of to zero. STABILITY ANALYSES Each method has its advantages and disadvantages. In fact, you can use different methods to answer different questions

The bifurcation approach is appropriate for determining the critical buckling load for a (perfect) system subjected to loads. The deformations are usually assumed to be small. The system must not have any imperfections. It cannot provide any information regarding the post-buckling loaddeformation path. The energy approach is the best when establishing the equilibrium equation and examining its stability The deformations can be small or large. The system can have imperfections. It provides information regarding the post-buckling path if large deformations are assumed The major limitation is that it requires the assumption of the deformation state, and it should include all possible degrees of freedom. STABILITY ANALYSIS The dynamic method is very powerful, but we will not use it in this class at all. Remember, it though when you take the course in dynamics or earthquake engineering In this class, you will learn that the loads acting on a structure change its

stiffness. This is significant you have not seen it before. Ma Ma a P Mb 4E I a L Mb 2E I b L What happens when an axial load is acting on the beam. The stiffness will no longer remain 4EI/L and 2EI/L. Instead, it will decrease. The reduced stiffness will reduce the natural frequency and period elongation. You will see these in your dynamics and earthquake engineering class. STABILITY ANALYSIS FOR ANY KIND OF BUCKLING OR STABILITY ANALYSIS NEED TO DRAW THE FREE BODY DIAGRAM OF THE DEFORMED STRUCTURE.

WRITE THE EQUATION OF STATIC EQUILIBRIUM IN THE DEFORMED STATE WRITE THE ENERGY EQUATION IN THE DEFORMED STATE TOO. THIS IS CENTRAL TO THE TOPIC OF STABILITY ANALYSIS NO STABILITY ANALYSIS CAN BE PERFORMED IF THE FREE BODY DIAGRAM IS IN THE UNDEFORMED STATE BIFURCATION ANALYSIS Always a small deflection analysis To determine Pcr buckling load Need to assume buckled shape (state 2) to calculate Example 1 Rigid bar supported by rotational spring k P Rigid bar subjected to axial force P Rotationally restrained at end L

Step 1 - Assume a deformed shape that activates all possible d.o.f. L k P L cos L (1-cos) BIFURCATION ANALYSIS L k L sin L cos P L (1-cos) Write the equation of static equilibrium in the deformed state M o 0 k P L sin 0 k L sin For small deformations sin

k k Pcr L L P Thus, the structure will be in static equilibrium in the deformed state when P = Pcr = k/L When P

L L sin O k L sin L cos L (1-cos) M o 0 (k L sin ) L P L sin 0 k L2 sin P L sin For small deformations sin k L2 Pcr k L L Thus, the structure will be in static equilibrium in the deformed state when P = Pcr = k L. When P

k D C B L L L Assume deformed state that activates all possible d.o.f. Draw FBD in the deformed state P A 1 1 2) L sin 1 L B P k k 1 2) L sin 2 2 L C

Assume small deformations. Therefore, sin= D BIFURCATION ANALYSIS Write equations of static equilibrium in deformed state P 1 A 1 2 ) L sin 1 L 1 2 ) B P k k L sin 2 2 D L C P k

P 1 2) 1 A 2 L D C k(22-1) L sin 1 L L sin 2 1+(1-2) B k(21-2) M B 0 k ( 21 2 ) P L sin 1 0 M C 0 k (2 2 1 ) P L sin 2 0

k (21 2 ) P L 1 0 k (2 2 1 ) P L 2 0 BIFURCATION ANALYSIS Equations of Static Equilibrium k (21 2 ) P L 1 0 k ( 2 2 1 ) P L 2 0 k 1 0 2k PL 2k PL 2 0 k Therefore either 1 and 2 are equal to zero or the determinant of the coefficient matrix is equal to zero. When 1 and 2 are not equal to zero that is when buckling occurs the coefficient matrix determinant has to be equal to zero for equil. Take a look at the matrix equation. It is of the form [A] {x}={0}. It can also be rewritten as [K]-K]-]-[K]-I]){x}={0}]){x}={0}x}={0}}={x}={0}0} 2k L k L

k 0 1 0 1 L P 0 1 0 2k 2 L BIFURCATION ANALYSIS This is the classical eigenvalue problem. [K]-K]-]-[K]-I]){x}={0}]){x}={0}x}={0}}={x}={0}0}. We are searching for the eigenvalues () of the stiffness matrix [K]. These eigenvalues cause the stiffness matrix to become singular Singular stiffness matrix means that it has a zero value, which means that the determinant of the matrix is equal to zero. 2k PL k 0 k 2k PL (2k PL) 2 k 2 0 (2k PL k ) (2k PL k ) 0 (3k PL) (k PL) 0 3k k

Pcr or L L Smallest value of Pcr will govern. Therefore, Pcr=k/L BIFURCATION ANALYSIS Each eigenvalue or critical buckling load (Pcr) corresponds to a buckling shape that can be determined as follows Pcr=k/L. Therefore substitute in the equations to determine 1 and 2 k (21 2 ) P L 1 0 k (2 2 1 ) P L 2 0 Let P Pcr k Let P Pcr k L k (21 2 ) k1 0 L k (2 2 1 ) k 2 0 k1 k 2 0 k1 k 2 0 1 2 1 2

All we could find is the relationship between 1 and 2. Not their specific values. Remember that this is a small deflection analysis. So, the values are negligible. What we have found is the buckling shape not its magnitude. P k P k The buckling mode is such that 1=2 Symmetric =buckling modeD A 1 2 L B C 1 L BIFURCATION ANALYSIS Second eigenvalue was Pcr=3k/L. Therefore substitute in the equations to determine 1 and 2 k (21 2 ) P L 1 0 k (2 2 1 ) P L 2 0 Let P Pcr 3k

Let P Pcr 3k L k (21 2 ) 3k1 0 L k (2 2 1 ) 3k 2 0 k1 k 2 0 k1 k 2 0 1 2 1 2 All we could find is the relationship between 1 and 2. Not their specific values. Remember that this is a small deflection analysis. So, the values are negligible. What we have found is the bucklingCshape not its magnitude. The buckling mode is such that 1=-2 AntisymmetricL buckling mode P A k k 2=-1 P D 1 L B

BIFURCATION ANALYSIS Homework No. 1 Problem 1.1 Problem 1.3 Problem 1.4 All problems from the textbook on Stability by W.F. Chen Chapter 1. Introduction to Structural Stability OUTLINE Definition of stability Types of instability Methods of stability analyses Bifurcation analysis examples small deflection analyses Energy method

Examples small deflection analyses Examples large deflection analyses Examples imperfect systems Design of steel structures ENERGY METHOD We will currently look at the use of the energy method for an elastic system subjected to conservative forces. Total potential energy of the system depends on the work done by the external forces (We) and the strain energy stored in the system (U). =U - We. For the system to be in equilibrium, its total potential energy must be stationary. That is, the first derivative of must be equal to zero. Investigate higher order derivatives of the total potential energy to examine the stability of the equilibrium state, i.e., whether the equilibrium is stable or unstable ENERGY METHD The energy method is the best for establishing the equilibrium equation and examining its stability

The deformations can be small or large. The system can have imperfections. It provides information regarding the post-buckling path if large deformations are assumed The major limitation is that it requires the assumption of the deformation state, and it should include all possible degrees of freedom. ENERGY METHOD Example 1 Rigid bar supported by rotational spring Assume small deflection theory k P Rigid bar subjected to axial force P Rotationally restrained at end L Step 1 - Assume a deformed shape that activates all possible d.o.f. L k P L cos L (1-cos) ENERGY METHOD SMALL DEFLECTIONS L

k L sin L cos P L (1-cos) Write the equation representing the total potential energy of system U We 1 U k2 2 We P L (1 cos ) 1 k 2 P L (1 cos ) 2 d k P L sin d d For equilibrium; 0 d Therefore , k P L sin 0 For small deflections; k P L 0 k Therefore , Pcr L ENERGY METHOD SMALL DEFLECTIONS The energy method predicts that buckling will occur at the same load Pcr as the bifurcation analysis method.

At Pcr, the system will be in equilibrium in the deformed. Examine the stability by considering further derivatives of the total potential energy This is a small deflection analysis. Hence will be zero. In this type of analysis, the further derivatives of examine the stability of the initial state-1 (when =0) 1 k 2 P L (1 cos ) 2 d k P L sin k P L d d2 k PL 2 d When P Pcr When P Pcr When P Pcr d2 0 Stable equilibrium 2 d d2 0 Unstable equilibrium 2 d d2 0 Not sure d 2

ENERGY METHOD SMALL DEFLECTIONS In state-1, stable when PPcr No idea about state during buckling. No idea about post-buckling equilibrium path or its stability. P Unstable Indeterminate Pcr Stable ENERGY METHOD LARGE DEFLECTIONS Example 1 Large deflection analysis (rigid bar with rotational spring) U We 1 U k2 L 2 We P L (1 cos ) k 1 k 2 P L (1 cos ) 2 L cos d k P L sin d

d For equilibrium; 0 d Therefore , k P L sin 0 k Therefore , P for equilibrium L sin The post buckling P relationsh ip is given above P L sin L (1-cos) ENERGY METHOD LARGE DEFLECTIONS Large deflection analysis See the post-buckling load-displacement path shown below The load carrying capacity increases after buckling at Pcr Pcr is where 0 Rigid bar with rotational spring 1.2 k for equilibrium L sin P Pcr sin 1

P 0.8 Load P/Pcr 0.6 0.4 0.2 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 End rotation 0=00=0 0.4 0.6 0.8 1 ENERGY METHOD LARGE DEFLECTIONS

Large deflection analysis Examine the stability of equilibrium using higher order derivatives of 1 k 2 P L (1 cos ) 2 d k P L sin d d2 k P L cos d 2 k But , P L sin d2 k k L cos L sin d 2 d2 k ( 1 ) tan d 2 d2 0 Always (i.e., all values of ) d 2 Always STABLE d2

But , 0 for 0 2 d ENERGY METHOD LARGE DEFLECTIONS At =0, the second derivative of =0. Therefore, inconclusive. Consider the Taylor series expansion of at =0 0 d 1 d2 1 d3 1 d4 1 dn 2 3 4 ..... n 2 3 4 n d 0 2! d 0 3! d 0 4! d 0 n! d 0 Determine the first non-zero term of ,

1 k 2 P L (1 cos ) 2 d k P L sin d d2 k P L cos d 2 d3 P L sin d 3 d4 P L cos d 4 0 0 d 0 d 0 d2 0 d 2 0 1 d4 1 4 k4 0 4 4! d 0 24 d3 P L sin 0 d 3 0 d4 P L cos PL k

d 4 0 Since the first non-zero term is > 0, the state is stable at P=Pcr and =0 ENERGY METHOD LARGE DEFLECTIONS Rigid bar with rotational spring 1.2 1 0.8 Load P/Pcr STABL E STABL E 0.6 STABL E 0.4 0.2 0 -1 -0.8 -0.6 -0.4 -0.2 0

0.2 End rotation 0=00=0 0.4 0.6 0.8 1 ENERGY METHOD IMPERFECT SYSTEMS Consider example 1 but as a system with imperfections The initial imperfection given by the angle 0 as shown below k P L 0 L cos(0) The free body diagram of the deformed system is shown below L k(0=0 P L sin 0 L cos

L (cos0-cos) ENERGY METHOD IMPERFECT SYSTEMS L k(0=0 L sin 0 U We 1 U k ( 0 ) 2 2 We P L (cos 0 cos ) L cos 1 k ( 0 ) 2 P L (cos 0 cos ) 2 d k ( 0 ) P L sin d d For equilibrium; 0 d Therefore, k ( 0 ) P L sin 0 k ( 0 ) for equilibrium L sin The equilibrium P relationship is given above Therefore, P P

L (cos0-cos) ENERGY METHOD IMPERFECT SYSTEMS P k ( 0 ) L sin P 0 Pcr sin P relationsh ipswith for rotational different values Rigid bar springof 0 shown below : 1.2 1 Load P/Pcr 0.8 0.6 0.4 0.2 0 -1 -0.8 -0.6

-0.4 -0.2 0 0.2 0.4 0.6 End rotation 0=00=0 0=00=0.0=05 0=00=0.1 0=00=0. 0=00=0.3 0.8 1 ENERGY METHODS IMPERFECT SYSTEMS As shown in the figure, deflection starts as soon as loads are applied. There is no bifurcation of load-deformation path for imperfect systems. The load-deformation path remains in the same state through-out. The smaller the imperfection magnitude, the close the loaddeformation paths to the perfect system load deformation path

The magnitude of load, is influenced significantly by the imperfection magnitude. All real systems have imperfections. They may be very small but will be there The magnitude of imperfection is not easy to know or guess. Hence if a perfect system analysis is done, the results will be close for an imperfect system with small imperfections ENERGY METHODS IMPERFECT SYSTEMS Examine the stability of the imperfect system using higher order derivatives of 1 2 k ( 0 ) P L (cos 0 cos ) 2 d k ( 0 ) P L sin d d2 k P L cos d 2 Equilibrium path will be stable d2 if 0 2 d i.e., if k P L cos 0 k i.e., if P L cos k ( 0 ) k i.e., if

L sin L cos i.e., 0 tan Which is always true, hence always in STABLE EQUILIBRIUM ENERGY METHOD SMALL DEFLECTIONS Example 2 - Rigid bar supported by translational spring at end P k L Assume deformed state that activates all possible d.o.f. Draw FBD in the deformed state L P L sin O k L sin L cos L (1-cos) ENERGY METHOD SMALL DEFLECTIONS Write the equation representing the total potential energy of system U We 1 1 U k ( L sin ) 2 k L2 2 2 2 We P L (1 cos )

P L L sin O 1 k L2 2 P L (1 cos ) 2 d k L2 P L sin d d For equilibriu m; 0 d Therefore , k L2 P L sin 0 For small deflections; k L2 P L 0 Therefore , Pcr k L k L sin L cos L (1-cos) ENERGY METHOD SMALL DEFLECTIONS The energy method predicts that buckling will occur at the same load Pcr as the bifurcation analysis method. At Pcr, the system will be in equilibrium in the deformed. Examine the stability by considering further derivatives of the total potential energy

This is a small deflection analysis. Hence will be zero. In this type of analysis, the further derivatives of examine the stability of the initial state-1 (when =0) 1 k L2 2 P L (1 cos ) 2 d k L2 P L sin d d2 k L2 P L cos 2 d For small deflections and 0 2 d 2 k L PL 2 d When, P k L When, P k L When P kL d2 0 STABLE 2 d d2 0 UNSTABLE d 2 d2 0 INDETERMINATE d 2

ENERGY METHOD LARGE DEFLECTIONS Write the equation representing the total potential energy of system U We 1 U k ( L sin ) 2 2 We P L (1 cos ) P L L sin O 1 k L2 sin 2 P L (1 cos ) 2 d k L2 sin cos P L sin d d For equilibrium; 0 d Therefore, k L2 sin cos P L sin 0 Therefore, P k L cos for equilibrium The post buckling P relationship is given above L cos L (1-cos) ENERGY METHOD LARGE DEFLECTIONS Large deflection analysis

See the post-buckling load-displacement path shown below The load carrying capacity decreases after buckling at Pcr Pcr is where 0 Rigid bar with translational spring 1.2 P k L cos P cos Pcr 1 for equilibrium 0.8 Load P/Pcr 0.6 0.4 0.2 0 -1 -0.8 -0.6 -0.4 -0.2 0

0.2 End rotation 0.4 0.6 0.8 1 ENERGY METHOD LARGE DEFLECTIONS Large deflection analysis Examine the stability of equilibrium using higher order derivatives of 1 k L2 sin 2 P L (1 cos ) 2 d k L2 sin cos P L sin d d2 k L2 cos 2 P L cos 2 d For equilibrium P k L cos d2 2 2 2 k L cos 2

k L cos 2 d d2 2 2 2 2 2 k L (cos sin ) k L cos 2 d d2 k L2 sin 2 2 d d2 0 ALWAYS . HENCE UNSTABLE d 2 ENERGY METHOD LARGE DEFLECTIONS

At =0, the second derivative of =0. Therefore, inconclusive. Consider the Taylor series expansion of at =0 0 d 1 d2 1 d3 1 d4 1 dn 2 3 4 ..... n 2 3 4 n d 0 2! d 0 3! d 0 4! d 0 n! d 0 Determine the first non-zero term of , 1 k L2 sin 2 P L (1 cos ) 0 2 d 1 k L2 sin 2 P L sin 0 d 2 d2

2 k L cos 2 P L cos 0 2 d d3 2k L2 sin 2 P L sin 0 3 d d4 2 4 k L cos 2 P L cos 4 d d4 2 2 2 4 k L k L 3 k L

d 4 d4 0 4 d UNSTABLE at 0 when buckling occurs Since the first non-zero term is < 0, the state is unstable at P=Pcr and = ENERGY METHOD LARGE DEFLECTIONS Rigid bar with translational spring 1.2 UNSTABL E 1 UNSTABL E Load P/Pcr 0.8 0.6 UNSTABLE 0.4 0.2 0 -1 -0.8 -0.6 -0.4

-0.2 0 0.2 End rotation 0.4 0.6 0.8 1 ENERGY METHOD - IMPERFECTIONS Consider example 2 but as a system with imperfections The initial imperfection given by the angle 0 as shown below P L 0 k L cos(0) The free body diagram of the deformed system is shown below L P L sin O

L sin0=0 0 L cos L (cos0-cos) ENERGY METHOD - IMPERFECTIONS L P L sin O U We 1 U k L2 (sin sin 0 ) 2 2 We P L (cos 0 cos ) L sin0=0 0 L cos 1 k L2 (sin sin 0 ) 2 P L (cos 0 cos ) 2 d k L2 (sin sin 0 ) cos P L sin d d For equilibrium; 0 d Therefore , k L2 (sin sin 0 ) cos P L sin 0 sin 0

) for equilibrium sin The equilibrium P relationsh ip is given above Therefore , P k L cos (1 L (cos0-cos) ENERGY METHOD - IMPERFECTIONS P k L cos (1 Pmax}={0} 1.2 Pmax}={0} sin 0 ) sin P sin 0 cos (1 ) Pcr sin sin dP 0 k L( sin 2 0 ) 0 sin 0 sin 3 d sin k L cos3 Envelope of peak loads Pmax 1 Load P/Pcr

0.8 0.6 0.4 0.2 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 End rotation 0=00=0 0=00=0.0=05 0=00=0.1 0=00=0. 0=00=0.3 0.8 1

ENERGY METHOD - IMPERFECTIONS As shown in the figure, deflection starts as soon as loads are applied. There is no bifurcation of load-deformation path for imperfect systems. The load-deformation path remains in the same state through-out. The smaller the imperfection magnitude, the close the loaddeformation paths to the perfect system load deformation path. The magnitude of load, is influenced significantly by the imperfection magnitude. All real systems have imperfections. They may be very small but will be there The magnitude of imperfection is not easy to know or guess. Hence if a perfect system analysis is done, the results will be close for an imperfect system with small imperfections. However, for an unstable system the effects of imperfections may be too large. ENERGY METHODS IMPERFECT SYSTEMS Examine the stability of the imperfect system using higher order derivatives of 1 k L (sin sin ) P L (cos cos ) 2 2

2 0 0 d k L2 (sin sin 0 ) cos P L sin d d2 k L2 (cos 2 sin 0 sin ) P L cos 2 d sin 0 For equilibrium P k L 1 sin sin 0 d2 2 2 k L (cos 2 sin sin ) k L 1 cos 2

0 2 sin d sin 0 cos 2 d2 2 2 2 2 k L cos sin sin 0 sin cos sin d 2 sin 0 cos 2 d2 2 2 k L sin sin 0 sin sin d 2 3 2 2 d2 2 sin sin 0 (sin cos ) k L sin d 2 3

d2 2 sin sin 0 k L sin d 2 ENERGY METHOD IMPERFECT SYSTEMS 3 d2 2 sin sin 0 k L sin d 2 d2 0 when P Pmax}={0} Stable d 2 d2 0 when P Pmax}={0} Unstable d 2 P k L cos (1 sin 0 ) sin Pmax}={0} k L cos 3 and When P Pmax}={0} k L cos (1 sin 0

) k L cos 3 sin sin 0 cos 2 sin sin 0 1 1 sin 2 sin 1 3 sin 0 sin and 3 d2 2 sin 0 sin k L 0 sin d 2 When P Pmax}={0} k L cos (1 sin 0 ) k L cos 3 sin sin 0 cos 2 sin sin 0 1 1 sin 2

sin 1 3 sin 0 sin and 3 d2 2 sin 0 sin k L 0 sin d 2 Chapter 2. Second-Order Differential Equations This chapter focuses on deriving second-order differential equations governing the behavior of elastic members 2.1 First order differential equations 2.2 Second-order differential equations 2.1 First-Order Differential Equations Governing the behavior of structural members

Elastic, Homogenous, and Isotropic Strains and deformations are really small small deflection theory Equations of equilibrium in undeformed state Consider the behavior of a beam subjected to bending and axial forces 2.1 First-Order Differential Equations Assume tensile forces are positive and moments are positive according to the right-hand rule Longitudinal stress due to bending My P Mx y x A Ix Iy This is true when the x-y axis system is a centroidal and principal axis system. y dA x dA x y dA 0 A A dA A; A

Centroidal axis A 2 x dA I y ; A 2 y dA I x A I x and I y are principal moment of inertia My P Mx y x A Ix Iy 2.1 First-Order Differential Equations My Mx P y x The corresponding strain is A E E Ix E Iy

If P=My=0, then Plane-sections remain plane and perpendicular to centroidal axis before and after bending The measure of bending is curvature which denotes the change in the slope of the tan centroidal y axis between two point dz apart Mx y E Ix y For small deformations tan y y y M y x E Ix M x E I x y y and similarly M y E I y x 2.1 First-Order Differential Equations Shear Stresses due to bending t

Vy Ix s y t ds O Vx s t x t ds Iy O 2.1 First-Order Differential Equations Differential equations of bending Assume principle of superposition Treat forces and deformations in y-z and x-z plane seperately Both the end shears and qy act in a plane parallel to the y-z plane through the shear center S dV y q y dz dM x V y dz d 2M x

q y 2 dz d 2 (E I x y ) q y dz 2 E I x y q y 2.1 First-Order Differential Equations Differential equations of bending E I x y q y y v 2 3/ 2 1 (v) For small deflection s y v E I x v iv q y Similarly E I y u iv q x u deflection in positive x direction v deflection in positive y direction Fourth-order differential equations using firstorder force-deformation theory Torsion behavior Pure and Warping Torsion Torsion behavior uncoupled from bending behavior Thin walled open cross-section subjected to torsional moment

This moment will cause twisting and warping of the cross-section. The cross-section will undergo pure and warping torsion behavior. Pure torsion will produce only shear stresses in the section Warping torsion will produce both longitudinal and shear stresses The internal moment produced by the pure torsion response will be equal to Msv and the internal moment produced by the warping torsion response will be equal to Mw. The external moment will be equilibriated by the produced internal moments MZ=MSV + MW Pure and Warping Torsion MZ=MSV + MW Where, MSV = G KT and MW = - E Iw " MSV = Pure or Saint Venants torsion moment KT = J = Torsional constant =

is the angle of twist of the cross-section. It is a function of z. IW is the warping moment of inertia of the cross-section. This is a new cross-sectional property you may not have seen before. MZ = G KT - E Iw " (3), differential equation of torsion Pure Torsion Differential Equation Lets look closely at pure or Saint Venants torsion. This occurs when the warping of the cross-section is unrestrained or absent dz r d d r r dz G r M SV r dA G r 2 dA A A M SV G K T where , K T J r 2 dA A For a circular cross-section warping is absent. For thin-walled open cross-sections, warping will occur. The out of plane warping deformation w can be calculated using an equation I will not show. Pure Torsion Stresses The torsional shear stresses vary linearly about the center of the thin plate

SV G r SV max}={0} G t sv Warping deformations The warping produced by pure torsion can be restrained by the: (a) end conditions, or (b) variation in the applied torsional moment (non-uniform moment) The restraint to out-of-plane warping deformations will produce longitudinal stresses (w) , and their variation along the length will produce warping shear stresses (w) . Warping Torsion Differential Equation Lets take a look at an approximate derivation of the warping torsion differential equation. This is valid only for I and C shaped sections. h 2 where u f flange lateral displacement u f M f moment in the flange V f Shear force in the flange E I f u f M f borrowing d .e. of bending E I f u f V f M W V f h M W E I f u f h MW

MW h2 E I f 2 E I W where I W is warping moment of inertia new sec tion property Torsion Differential Equation Solution Torsion differential equation MZ=MSV+MW = G KT - E IW This differential equation is for the case of concentrated torque G K T E I w M Z G KT MZ E IW E IW MZ E IW 2 C1 C 2 cosh z C 3 sinh z Mz z 2 E I W Torsion differential equation for the case of distributed torque dM Z

dz G K T E I w iv m Z m Z G KT m iv Z E IW E IW iv 2 mZ E IW mz z 2 C 4 C 5 z C 6 cosh z C 7 sinh z 2 G KT The coefficients C1 .... C6 can be obtained using end conditions Torsion Differential Equation Solution 0 Torsionally fixed end conditions are given by These imply that twisting and warping at the fixed end are fully restrained. Therefore, equal to zero. Torsionally pinned or simply-supported end conditions given by: 0 These imply that at the pinned end twisting is fully restrained ( =0) and

warping is unrestrained or free. Therefore, W=0 =0 Torsionally free end conditions given by = = = 0 These imply that at the free end, the section is free to warp and there are no warping normal or shear stresses. Results for various torsional loading conditions given in the AISC Design Guide 9 can be obtained from my private site Warping Torsion Stresses Restraint to warping produces longitudinal and shear stresses W E Wn W t E SW where, Wn Normalized Unit Warping Section Pr operty SW Warping Statical Moment Section Pr operty The variation of these stresses over the section is defined by the section property Wn and Sw The variation of these stresses along the length of the beam is defined by the derivatives of . Note that a major difference between bending and torsional behavior is

The stress variation along length for torsion is defined by derivatives of , which cannot be obtained using force equilibrium. The stress variation along length for bending is defined by derivatives of v, which can be obtained using force equilibrium (M, V diagrams). Torsional Stresses Torsional Stresses Torsional Section Properties for I and C Shapes and derivatives for concentrated torque at midspan Summary of first order differential equations E I x v M x E I y u M y (1) G K T E I W M z (3) (2) NOTES: (1) Three uncoupled differential equations (2) Elastic material first order force-deformation theory (3) Small deflections only (4) Assumes no influence of one force on other deformations (5) Equations of equilibrium in the undeformed state. HOMEWORK # 3 Consider the 22 ft. long simply-supported W18x65 wide flange beam shown in Figure 1 below. It is subjected to a uniformly distributed load of 1k/ft that is placed with an eccentricity of 3 in. with respect to the centroid (and shear center).

At the mid-span and the end support cross-sections, calculate the magnitude and distribution of: Normal and shear stresses due to bending Shear stresses due to pure torsion Warping normal and shear stresses over the cross-section. Provide sketches and tables of the individual normal and shear stress distributions for each case. Superimpose the bending and torsional stress-states to determine the magnitude and location of maximum stresses. HOMEWORK # 2 22 ft. Span 3in. W18x65 Cross-section Chapter 2. Second-Order Differential Equations This chapter focuses on deriving second-order differential equations governing the behavior of elastic members 2.1 First order differential equations

2.2 Second-order differential equations 2.2 Second-Order Differential Equations Governing the behavior of structural members Elastic, Homogenous, and Isotropic Strains and deformations are really small small deflection theory Equations of equilibrium in deformed state The deformations and internal forces are no longer independent. They must be combined to consider effects. Consider the behavior of a member subjected to combined axial forces and bending moments at the ends. No torsional forces are applied explicitly because that is very rare for CE structures. Member model and loading conditions Member is initially straight and prismatic. It has a thin-walled open cross-section Member ends are pinned and prevented from translation. The forces are applied only at the member ends

These consist only of axial and bending moment forces P, MTX, MTY, MBX, MBY Assume elastic behavior with small deflections Right-hand rule for positive moments and reactions and P assumed positive. Member displacements (cross-sectional) Consider the middle line of thinwalled cross-section x and y are principal coordinates through centroid C Q is any point on the middle line. It has coordinates (x, y). Shear center S coordinates are (xo, y0) Shear center S displacements are u, v, and Member displacements (cross-sectional)

Displacements of Q are: uQ = u + a sin vQ = v a cos where a is the distance from Q to S But, sin = (y0-y) / a cos = (x0-x) / a Therefore, displacements of Q are: uQ = u + (y0-y) vQ = v (x0 x) Displacements of centroid C are: uc = u + (y0) vc = v - (x0) Internal forces second-order effects Consider the free body diagrams of the member in the deformed state. Look at the deformed state in the x-z and y-z planes in this Figure. The internal resisting moment at a distance z from the lower end are: Mx = - MBX + Ry z + P vc My = - MBY + Rx z - P uc

The end reactions Rx and Ry are: Rx = (MTY + MBY) / L Ry = (MTX + MBX) / L Internal forces second-order effects Therefore, z M TX M BX P v x0 L z M TY M BY P u y0 L M x M BX M y M BY Internal forces in the deformed state In the deformed state, the cross-section is such that the principal coordinate systems are changed from x-y-z to the system uc vc x z y P MBx

Ry MBY x uc vc z y Rx M +dd a M M z Mx Rx My Ry P

Internal forces in the deformed state The internal forces Mx and My must be transformed to these new axes Since the angle is small MMx + My M = M y Mx z M TX M BX P v x0 L z M TY M BY P u y0 L M x M BX M y M BY z M TX M BX P v P x0 M BY z M TY M BY L L z z M TY M BY P u P y0 M BX M TX M BX L L

M M BX M M BY Twisting component of internal forces Twisting moments M are produced by the internal and external forces There are four components contributing to the total M (1) Contribution from Mx and My M1 (2) Contribution from axial force P M (3) Contribution from normal stress M3 (4) Contribution from end reactions Rx and Ry M The total twisting moment M = M1 + M + M3 + M Twisting component 1 of 4 u v Twisting moment due to Mx & My M1 = Mx sin (du/dz) + Mysin (dv/dz) Therefore, due to small angles, M1 = Mx du/dz + My dv/dz

M1 = Mx u + My v Twisting component 2 of 4 u v The axial load P acts along the original vertical direction In the deformed state of the member, the longitudinal axis is not vertical. Hence P will have components producing shears. These components will act at the centroid where P acts and will have values as shown above assuming small angles Twisting component 2 of 4 These shears will act at the centroid C, which is eccentric with respect to the shear center S. Therefore, they will produce secondary twisting. M = P (y0 du/dz x0 dv/dz) Therefore, M = P (y0 u x0 v) Twisting component 3 of 4 The end reactions (shears) Rx and Ry act at the shear center S

at the ends. But, along the member ends, the shear center will move by u, v, and . Hence, these reactions will also have a twisting effect produced by their eccentricity with respect to the shear center S. M + Ry u + Rx v = 0 Therefore, M = (MTY + MBY) v/L (MTX + MBX) u/L Twisting component 4 of 4 Wagners effect or contribution complicated. Two cross-sections that are d apart will warp with respect to each other. The stress element dA will become inclined by angle (a d/ d with respect to d axis. Twist produced by each stress element about S is equal to

d dM 3 a dA a d d M 3 a 2 dA d A Twisting component 4 of 4 Let , a 2 dA K A d d d K for small angles dz M 3 K M 3 Twisting component 4 of 4 Let , a 2 dA K A d d d K for small angles dz M 3 K M 3 x}={0} x}={0}

y y Total Twisting Component M = M1 + M + M3 + M M1 = Mx u + My v M = P (y0 u x0 v) M = (MTY + MBY) v/L (MTX + MBX) u/L M3= -K Therefore, MMx u + My v+ P (y0 u x0 v) (MTY + MBY) v/L (MTX + MBX) u/L-K z M TX M BX P v P x0 M BY z M TY M BY L L z z M TY M BY P u P y0 M BX M TX M BX L L M While, M BX

M M BY Total Twisting Component M = M1 + M + M3 + M M1 = Mx u + My v M = P (y0 u x0 v) M3= -K M = (MTY + MBY) v/L (MTX + MBX) u/L Therefore, v u M TX M BX K L L v u M BY ) ( M TX M BX ) K L L M M x u M y v P y0 u x0 v M TY M BY M ( M x P y0 ) u ( M y P x0 ) v ( M TY z ( M BX M TX ) P (v x0 ) L z ( M BY M TY ) P (u y0 ) L But , M x M BX and , M y M BY z

z ( M BX M TX ) P y0 ) u ( M BY ( M BY M TY ) P x0 ) v L L v u M BY ) ( M TX M BX ) K L L M ( M BX ( M TY Internal moments about the axes Thus, now we have the internal moments about the axes for the deformed member cross-section. z z M TX M BX P v P x0 M BY M TY M BY L L z z M M BY M MTY M P u P y

M MTX M BX M TX+MBY BX 0 BX TY +M BY L L z z M ( M BX ( M BX M TX ) P y0 ) u ( M BY ( M BY M TY ) P x0 ) v L L v u ( M TY M BY ) ( M TX M BX ) K L L M M BX x z y Internal Moment Deformation Relations

The internal moments M, M, and M will still produce flexural bending about the centroidal principal axis and twisting about the shear center. The flexural bending about the principal axes will produce linearly varying longitudinal stresses. The torsional moment will produce longitudinal and shear stresses due to warping and pure torsion. The differential equations relating moments to deformations are still valid. Therefore, M = - E I v ..(I = Ix) M = E I u ..(I= Iy) M = G KT E Iw Internal Moment Deformation Relations Therefore, z z M TX M BX P v P x0 M BY M TY M BY L L z z M E I y u M BY M M TX M P u

P y M M M MTX BX 0 BX TY +MBY TY+MBY BX L L z M G KT E I w ( M BX ( M BX M TX ) P y0 ) u L z v u ( M BY ( M BY M TY ) P x0 ) v ( M TY M BY ) ( M TX M BX ) K L L L M E I x v M BX Second-Order Differential Equations You end up with three coupled differential equations that relate the applied forces and moments to the deformations u, v, and . Therefore,

1 2 3 z z E I x v P v P x0 M BY M TY M BY M BX M TX M BX L L z z E I y u P u P y0 M BX M MM +M M BYB M BY MTY MBYBX +dM TYTX TX+M TX B L L X Xz E I w (G KT K ) u ( M BX ( M BX M TX ) P y0 ) L z v u

v ( M BY ( M BY M TY ) P x0 ) ( M TY M BY ) ( M TX M BX ) 0 L L L These differential equations can be used to investigate the elastic behavior and buckling of beams, columns, beam-columns and also complete frames that will form a major part of this course. Chapter 3. Structural Columns 3.1 Elastic Buckling of Columns 3.2 Elastic Buckling of Column Systems Frames 3.3 Inelastic Buckling of Columns 3.4 Column Design Provisions (U.S. and Abroad) 3.1 Elastic Buckling of Columns Start out with the second-order differential equations derived in Chapter 2. Substitute P=P and MTY = MBY = MTX = MBX = 0 Therefore, the second-order differential equations simplify to: 1 E I x v P v P x0 0 2 E I y u P u P y0 0

3 E I w (G KT K ) u ( P y0 ) v ( P x0 ) 0 This is all great, but before we proceed any further we need to deal with Wagners effect which is a little complicated. Wagners effect for columns K a 2 dA A where, P M y M x E Wn A Ix Iy M P (v x0 ) M P (u y0 ) P P (v x0 ) y P (u y0 ) x K E Wn a 2 dA A Ix Iy A P P (v x0 ) y P (u y0 ) x K

E Wn a 2 dA Ix Iy A A P Neglecting higher order terms; K a 2 dA A A Wagners effect for columns But , a 2 ( x0 x) 2 ( y0 y ) 2 a 2 dA ( x0 x) 2 ( y0 y ) 2 dA A A a 2 dA x02 y02 x 2 y 2 2 x0 x 2 y0 y dA A A a 2 dA x02 y02 dA x 2 dA y 2 dA 2 x0 x dA 2 y0 y dA A A A A a 2 dA ( x02 y02 ) A I x I y A Finally , P ( x02 y02 ) A I x I y A I I K P ( x02 y02 ) x y A

Ix Iy 2 2 2 Let r0 ( x0 y0 ) A K K P r02 A A Second-order differential equations for columns Simplify to: 1 E I x v P v P x0 0 2 E I y u P u P y0 0 3 E I w ( P r02 G KT ) u ( P y0 ) v ( P x0 ) 0 Where 2 2 0 2 0

r0 x y Ix I y A Column buckling doubly symmetric section For a doubly symmetric section, the shear center is located at the centroid xo= y0 = 0. Therefore, the three equations become uncoupled 1 2 3 E I x v P v 0 E I y u P u 0 E I w ( P r02 G KT ) 0 Take two derivatives of the first two equations and one more derivative of the third equation. 1 E I x viv P v 0 2 E I y u iv P u 0 3 P Let , Fv2 E Ix E I w iv ( P r02 G KT ) 0 P Fu2 E Iy

2 P r G KT F2 0 E Iw Column buckling doubly symmetric section 1 v iv Fv2 v 0 2 u iv Fu2 u 0 3 iv F2 0 All three equations are similar and of the fourth order. The solution will be of the form C1 sin z + C2 cos z + C3 z + C4 Need four boundary conditions to evaluate the constant C1..C4 For the simply supported case, the boundary conditions are: u= u=0; v= v=0; = =0 Lets solve one differential equation the solution will be valid for all three. Column buckling doubly symmetric section v iv Fv2 v 0 Solution is

v C1 sin Fv z C2 cos Fv z C3 z C4 v C1 Fv2 sin Fv z C2 Fv2 cos Fv z The coefficient matrix 0 Boundary conditions : v(0) v(0) v( L) v( L) 0 Fv2 sin Fv L 0 sin Fv L 0 C2 C4 0 v(0) 0 v(0) 0 C2 0 C1 sin Fv L C2 cos Fv L C3 L C4 C1 Fv2 sin Fv L C2 Fv2 cos Fv L 0 0 Fv L sin Fv2 sin Fv L 1 1 cos Fv L Fv2 cos Fv L v( L) 0 v( L) 0 0 0 L 0 1 C1 0

0 C2 0 1 C3 0 0 C4 0 Fv L n Fv P n E Ix L n2 2 Px 2 E I x L Smallest value of n 1: 2 E Ix Px L2 Column buckling doubly symmetric section sin Fu L 0 Similarly, sin F L 0 Fu L n F L n P n Fu E Iy L

P r02 G KT n F E Iw L n2 2 Py 2 E I y L n2 2 1 P 2 E I w G KT 2 L r0 Similarly , Smallest value of n 1: Summary 2 E Iy Py L2 2 E Ix Px L2 2 E Iy Py L2 2 E Iw 1 P G K T 2 2 L

r0 Smallest value of n 1: n2 2 1 P 2 E I w G KT 2 L r0 1 2 3 Column buckling doubly symmetric section Thus, for a doubly symmetric cross-section, there are three distinct buckling loads Px, Py, and Pz. The corresponding buckling modes are: v = C1 sin(z/L), u =C2 sin(z/L), and = C3 sin(z/L). These are, flexural buckling about the x and y axes and torsional buckling about the z axis. As you can see, the three buckling modes are uncoupled. You must compute all three buckling load values. The smallest of three buckling loads will govern the buckling of the column. Column buckling boundary conditions Consider the case of fix-fix boundary conditions:

viv Fv2 v 0 Solution is v C1 sin Fv z C2 cos Fv z C3 z C4 v C1 Fv cos Fv z C2 Fv sin Fv z C3 Boundary conditions : v(0) v(0) v( L) v( L) 0 C2 C4 0 The coefficient matrix 0 Fv L sin Fv L 2 cos Fv L 2 0 2 sin Fv L Fv L Fv L F L cos 2sin 0 v 2 2 2 Fv L n 2 C1 Fv C3 0 2n F C1 sin Fv L C2 cos Fv L C3 L C4 v( L ) 0 v L C1 Fv cos Fv L C2 Fv sin Fv L C3 v( L) 0 4 n2 2 Px E Ix L2

0 1 0 1 C1 0 Smallest value of n 1: Fv 0 1 0 C2 0 2 E Ix 2 E Ix Px cos Fv L L 1 C3 2 2 sin Fv L 0 0.5 L K L Fv cos Fv L Fv sin Fv L 1 0 C4 0 v(0) 0 v(0) 0 Column Boundary Conditions The critical buckling loads for columns with different boundary conditions can be expressed as: Px Py

2 E Ix Kx L 2 1 2 E Iy K L 2 2 y 2 E I 1 w P G KT 2 2 K z L r0 3 Where, Kx, Ky, and Kz are functions of the boundary conditions: K=1 for simply supported boundary conditions K=0.5 for fix-fix boundary conditions

K=0.7 for fix-simple boundary conditions Column buckling example. Consider a wide flange column W27 x 84. The boundary conditions are: v=v=u=u===0 at z=0, and v=v=u=u===0 at z=L For flexural buckling about the x-axis simply supported K x=1.0 For flexural buckling about the y-axis fixed at both ends K y = 0.5 For torsional buckling about the z-axis pin-fix at two ends - K z=0.7 Px 2 E Ix Kx L 2 2 E Iy 2 E A rx 2 Kx L 2 2 E A ry 2

2 E A L K x rx 2 2 2 E A ry Py 2 2 2 r K y L K y L K y L x rx 2 E I 1 2 E I A 2 w w P G K

G K r T T x 2 2 2 rx2 I x I y K z L r0 L K z rx Column buckling example. Px 2 E A 1 2 E 5823.066 2 2 2 PY A Y L

L L Y Kx Kx r r x x rx 2 2 2 Py 2 E A (ry / rx ) E (ry / rx ) 791.02 2 2 2 PY A Y L L L Y K y Ky r rx x rx

P 2 E I w A 1 2 G K r T x 2 r2 I I PY A Y L K x x y z rx P 2 E I w 1 2 G K r T x 2 PY rx2 I x I y Y

L K z rx P 578.26 0.2333 2 PY L rx Column buckling example. 2 Flexural buckling about y-axis Critical buckling load / yield load (Pcr/PY) 1.8 Flexural buckling about x-axis 1.6 1.4 1.2 Yield load PY Cannot be exceeded 1

0.8 0.6 Torsional buckling about z-axis governs Torsional buckling about z-axis 0.4 Flexural buckling about y-axis governs 0.2 0 0 10 20 30 40 50 60 70 80 L-rx (Slenderness Ratio) Px - flexural buckling Py - flexural buckling

Pz - torsional buckling 90 100 Column buckling example. When L is such that L/rx < 31; torsional buckling will govern rx = 10.69 in. Therefore, L/rx = 31 L=338 in.=28 ft. Typical column length =10 15 ft. Therefore, typical L/rx= 11.2 16.8 Therefore elastic torsional buckling will govern. But, the predicted load is much greater than PY. Therefore, inelastic buckling will govern. Summary Typically must calculate all three buckling load values to determine which one governs. However, for common steel buildings made using wide flange sections the minor (y-axis) flexural buckling usually governs. In this problem, the torsional buckling governed because the end conditions for minor axis flexural buckling were fixed. This is very rarely achieved in common building construction. Column Buckling Singly Symmetric Columns

x Well, what if the column has only one axis of symmetry. Like the xaxis or the y-axis or so. As shown in this figure, the y axis is the axis of symmetry. The shear center S will be located on this axis. C S y Therefore x0= 0. The differential equations will simplify to: 1 E I x v P v 0 2 E I y u P u P y0 0 3 E I w ( P r02 G KT ) u ( P y0 ) 0 Column Buckling Singly Symmetric Columns

The first equation for flexural buckling about the x-axis (axis of non-symmetry) becomes uncoupled. E I x v P v 0 (1) Equations (2) and (3) are still coupled in terms of u and . iv E I x v P v 0 v iv Fv 2 v 0 where, Fv 2 P E Ix v C1 sin Fv z C2 cos Fv z C3 z C4 Boundary conditions sin Fv L 0 2 E Ix Px ( K x Lx ) 2 Buckling mod v C1 sin Fv z 2 E I y u P u P y0 0 3 E I w ( P r02 G KT ) u ( P y0 ) 0

These equations will be satisfied by the solutions of the form u=C2 sin (z/L) and =C3 sin (z/L) Column Buckling Singly Symmetric Columns E I y u P u P y0 0 (2) E I w ( P r02 G KT ) u ( P y0 ) 0 (3) E I y u iv P u P y0 0 E I w iv ( P r02 G KT ) u ( P y0 ) 0 z z ; C3 sin L L Therefore, substituting these in equations 2 and 3 Let , u C2 sin 4 2 2 z z z E I y C2 sin P C2 sin P y0 C3 sin 0 L L L L

L L 4 2 2 z z z E I w C3 sin ( P r02 G KT ) C3 sin P y0 C2 sin 0 L L L L L L Column Buckling Singly Symmetric Columns 2 E I y P C2 P y0 C3 0 L 2 and E I w ( P r02 G KT ) C3 P y0 C2 0

L 2 E Iy Let , Py L2 and 2 E Iw 1 P G K T 2 2 L r0 Py P C2 P y0 C3 0 P P r02C3 P y0 C2 0 P y0 C2 Py P 0 2 C P y ( P P ) r 0

0 3 Py P P y0 0 P y0 ( P P ) r02 Column Buckling Singly Symmetric Columns ( Py P)( P P) r02 P 2 y02 0 Py P P ( Py P ) P 2 r02 P 2 y02 0 P 2 (1 2 0 2 0 y ) P( Py P ) Py P 0 r y02 ( Py P ) ( Py P ) 4 Py P (1 2 ) r0 P y02 2 (1 2 ) r0 2 y02 4 Py P (1 2 ) ( Py P ) r0 P 1 1

y02 ( Py P ) 2 2 (1 2 ) r0 Thus, there are two roots for P Smaller value will govern y02 2 4 Py P (1 r 2 ) y 0 0 4 Py P (1 2 ) ( Py P ) ( Py P ) 2 1 2 ( P P ) r0 ( Py P ) P P y 1 1 y02 ( Py P ) 2 2 (1 2 ) P 2 r 0

y 2 (1 02 ) r0 Column Buckling Singly Symmetric Columns The critical buckling load will the lowest of Px and the two roots shown on the previous slide. If the flexural torsional buckling load govern, then the buckling mode will be C2 sin (z/L) x C3 sin (z/L) This buckling mode will include both flexural and torsional deformations hence flexural-torsional buckling mode. Column Buckling Asymmetric Section No axes of symmetry: Therefore, shear center S (xo, yo) is such that neither xo not yo are zero. E I x v P v P x0 0 (1) E I y u P u P y0 0 (2) E I w ( P r02 G KT ) u ( P y0 ) v ( P x0 ) 0 (3)

For simply supported boundary conditions: (u, u, v, v, , =0), the solutions to the differential equations can be assumed to be: u = C1sin (z/L) v = C2 sin (z/L) = C3 sin (z/L) These solutions will satisfy the boundary conditions noted above Column Buckling Asymmetric Section Substitute the solutions into the d.e. and assume that it satisfied too: 2 z z z E I x C1 sin P C1 sin P x0 C3 sin 0 L L L L

2 z z z E I y C2 sin P C2 sin P y0 C3 sin 0 L L L L 3 z

z z z 2 E I w C3 cos ( P r0 G KT ) C3 cos P y0 C1 cos P x0 C2 cos 0 L L L L L L L L 2 E Ix P L 0

P x0 0 2 E Iy P L P y0 z C1 sin L 0 z P y0 0 C2 sin L 0 2 z 2 E I w ( P r0 G K T ) C3 cos L L L

P x0 Column Buckling Asymmetric Section P P x 0 Px 0 0 Py P P y0 z C sin 1 L P x0 0 z P y0 C2 sin L 0 0 P P r02 C cos z

L 3 L where, 2 2 Px EI x L Py EI y L 2 E Iw 1 P G K T 2 L2 r0 Either C1, C2, C3 = 0 (no buckling), or the determinant of the coefficient matrix =0 at buckling. Therefore, determinant of the coefficient matrix is: y2 x2 2 2 o P Px P Py P P P P Px 2 P P Py o2

ro ro 0 Column Buckling Asymmetric Section P Px P Py y2 x2 2 o P P P P Px 2 P P Py o2 r r o o 2

0 This is the equation for predicting buckling of a column with an asymmetric section. The equation is cubic in P. Hence, it can be solved to obtain three roots Pcr1, Pcr2, Pcr3. The smallest of the three roots will govern the buckling of the column. The critical buckling load will always be smaller than Px, Py, and P The buckling mode will always include all three deformations u, v, and . Hence, it will be a flexural-torsional buckling mode. For boundary conditions other than simply-supported, the corresponding Px, Py, and P can be modified to include end condition effects Kx, Ky, and K Homework No. 4 See word file

Problem No. 1 Problem No. 2 Consider a column with doubly symmetric cross-section. The boundary conditions for flexural buckling are simply supported at one end and fixed at the other end. Solve the differential equation for flexural buckling for these boundary conditions and determine the eigenvalue (buckling load) and the eigenmode (buckling shape). Plot the eigenmode. How the eigenvalue compare with the effective length approach for predicting buckling? What is the relationship between the eigenmode and the effective length of the column (Refer textbook). Consider an A992 steel W14 x 68 column cross-section. Develop the normalized buckling load (Pcr/PY) vs. slenderness ratio (L/rx) curves for the column crosssection. Assume that the boundary conditions are simply supported for buckling about the x, y, and z axes. Which buckling mode dominates for different column lengths? Is torsional buckling a possibility for practical columns of this length? Will elastic buckling occur for most practical lengths of this column? Problem No. 3 Consider a C10 x 30 column section. The length of the column is 15 ft. What is the buckling capacity of the column if it is simply supported for buckling about the yaxis (of non-symmetry), pin-fix for flexure about the x-axis (of symmetry) and simply supported in torsion about the z-axis. Which buckling mode dominates?

Column Buckling - Inelastic A long topic Effects of geometric imperfection Pv 0 EIx v Pu 0 EIy u Leads to bifurcation buckling of perfect doubly-symmetric columns P M x P(v v o ) 0 P(v v o ) 0 EIxv Fv2 (v v o ) 0 v v o o sin z L vo v v Mx Fv2v Fv2v o v Fv2v Fv2 (o sin v

z ) L Solution v c v p P v c A sin(Fv z) Bcos(Fv z) z z v p C sin Dcos L L Effects of Geometric Imperfection Solve for C and D first z L 2 z z z z z C sin Dcos Fv2 C sin Dcos Fv2o sin 0 L L L L L L 2 2 z z 2

2 2 sin C Fv C Fv o cos D Fv D0 L L L L 2 2 2 2 2 C Fv C Fv o 0 and D Fv D0 L L v Fv2v p Fv2o sin p Fv2o C 2 2 Fv L Solution becomes and D 0 Fv2o z v A sin(Fv z) Bcos(Fv z) sin L 2 2 Fv L

Geometric Imperfection Solve for A and B Boundary conditions v(0) v(L) 0 v(0) B 0 v(L) A sin Fv L 0 A 0 Solution becomes Fv2 o z v sin 2 L 2 Fv L Fv2 o 2 P z PE o z L v sin sin Fv2 L 1 P L 1 2 PE L P

z P v E o sin P L 1 PE Total Deflection P z z P v v o E o sin o sin P L L 1 PE P P z 1 z E 1o sin o sin L 1 P L 1 P PE PE z AFo sin L AF = amplification factor

Geometric Imperfection 1 AF amplification factor P 1 PE M x P(v v o ) z M x AF (Po sin ) L i.e., M x AF (moment due to initial crooked ) 12 10 Amplification Factor AF Increases exponentially Limit AF for design Limit P/PE for design 8 6 Value used in the code is 0.877 This will give AF = 8.13 Have to live with it. 4 2 0 0 0.2 0.4

0.6 P/P E 0.8 1 Residual Stress Effects Residual Stress Effects History of column inelastic buckling Euler developed column elastic buckling equations (buried in the million other things he did). Take a look at: http://en.wikipedia.org/wiki/EuleR An amazing mathematician In the 1750s, I could not find the exact year. The elastica problem of column buckling indicates elastic buckling occurs with no increase in load. dP/dv=0 History of Column Inelastic Buckling Engesser extended the elastic column buckling theory in 1889.

He assumed that inelastic buckling occurs with no increase in load, and the relation between stress and strain is defined by tangent modulus Et Engessers tangent modulus theory is easy to apply. It compares reasonably with experimental results. PT=ETI / (KL)2 History of Column Inelastic Buckling In 1895, Jasinsky pointed out the problem with Engessers theory. If dP/dv=0, then the 2nd order moment (Pv) will produce incremental strains that will vary linearly and have a zero value at the centroid (neutral axis). The linear strain variation will have compressive and tensile values. The tangent modulus for the incremental compressive strain is equal to Et and that for the tensile strain is E. History of Column Inelastic Buckling In 1898, Engesser corrected his original theory by accounting for the different tangent modulus of the tensile increment.

This is known as the reduced modulus or double modulus The assumptions are the same as before. That is, there is no increase in load as buckling occurs. The corrected theory is shown in the following slide History of Column Inelastic Buckling The buckling load PR produces critical stress R=Pr/A During buckling, a small curvature d is introduced The strain distribution is shown. The loaded side has dL and dL The unloaded d ( y side y has y) dd U and dU L 1 dU ( y y y1 ) d d L E t ( y y1 y) d dU E( y y y1 ) d History of Column Inelastic Buckling

d v d L E t ( y y1 y) v dU E( y y y1 ) v But, the assumption is dP 0 y y y1 y y1 (d y ) dU dA d L dA 0 y y y1 y y1 (d y ) E( y y y1 ) dA E t ( y y1 y) dA 0 ES1 E t S2 0 y where, S1 ( y y y1 ) dA y y1 and S 2 y y1 ( y y1 y) dA

(d y ) History of Column Inelastic Buckling S1 and S2 are the statical moments of the areas to the left and right of the neutral axis. Note that the neutral axis does not coincide with the centroid any more. The location of the neutral axis is calculated using the equation derived ES1 - EtS2 = 0 M Pv y y y1 y y1 (d y ) M dU ( y y y1) dA d L ( y y1 y) dA M Pv v ( EI1 E t I2 ) y where, I1 ( y y y1 ) 2 dA y y1 and I 2 y y1

( y y1 y) 2 dA (d y ) History of Column Inelastic Buckling M Pv v ( EI1 E t I2 ) 0 Pv ( EI1 E t I2 )v P v v 0 EI1 E t I2 Fv2v 0 v where, Fv2 and E E 2 EI x PR (KL) 2 P P EI1 E t I2 EIx I1 I Et 2 Ix Ix E is the reduced or double modulus PR is the reduced modulus buckling load History of Column Inelastic Buckling For 50 years, engineers were faced with the dilemma that the

reduced modulus theory is correct, but the experimental data was closer to the tangent modulus theory. How to resolve? Shanley eventually resolved this dilemma in 1947. He conducted very careful experiments on small aluminum columns. He found that lateral deflection started very near the theoretical tangent modulus load and the load capacity increased with increasing lateral deflections. The column axial load capacity never reached the calculated reduced or double modulus load. Shanley developed a column model to explain the observed phenomenon History of Column Inelastic Buckling History of Column Inelastic Buckling History of Column Inelastic Buckling History of Column Inelastic Buckling Column Inelastic Buckling Three different theories P Tangent modulus

Reduced modulus Shanley model dP/dv=0 Tangent modulus theory assumes Perfectly straight column Ends are pinned Small deformations No strain reversal during buckling Slope is zero at buckling P=0 with increasing v v Elastic buckling analysis PT Tangent modulus theory Assumes that the column buckles at the tangent modulus load such that there is an increase in P (axial force) and M (moment). The axial strain increases everywhere and there is no strain reversal. Strain and stress state just before buckling PT T Mx - Pv = 0 Strain and stress state just after buckling

v v T=PT/A Mx T T T T=ETT Curvature = = slope of strain diagram T h h T y where y dis tan ce from centroid 2 h T y E T 2 PT Tangent modulus theory Deriving the equation of equilibrium M x ydA A T T

T ( y h / 2) E T M x T ( y h / 2)E T ydA A M x T y dA E T y 2 dA h / 2)E T y dA A A A M x 0 E T Ix 0 M x E T Ix v The equation Mx- PTv=0 becomes -ETIxv - PTv=0 Solution is PT= 2ETIx/L2 Example - Aluminum columns Consider an aluminum column with Ramberg-Osgood stressE 10100 ksi strain curve 40.15 ksi 0.2 n 0.002 E 0.2 1 0.002 n 1 n n

E 0.2 0.002 1 n nE n 1 0.2 E n 1 0.002 1 nE 0.2 0.2 E E n 1 E T 0.002 1 nE 0.2 0.2 n 0.000E+d00 1.980E-04 3.960E-04 5.941E-04

7.921E-04 9.901E-04 1.188E-03 1.386E-03 1.584E-03 1.782E-03 1.980E-03 2.178E-03 2.376E-03 2.575E-03 2.775E-03 2.979E-03 3.198E-03 3.458E-03 3.829E-03 4.483E-03 5.826E-03 8.771E-03 1.529E-02 2.949E-02 5.967E-02 1.221E-01 18.55 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

34 36 38 40 42 44 46 48 50 ET ET differences equation 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10100.0 10099.9 10099.8 10099.5 10098.8 10097.6 10094.2 10088.7 10075.1 10054.2 10005.7 9934.0

9779.8 9563.7 9142.0 8602.6 7697.4 6713.6 5394.2 4251.9 3056.9 2218.6 1488.8 1037.0 679.2 468.1 306.9 212.4 140.8 98.5 66.3 46.9 32.1 23.0 Tangent Modulus Buckling Ramberg-Osgood Stress-Strain Stress-tangent modulus relationship 60 12000 Tangent Modulus (ksi) Stress (ksi) 50 40 30 20 10 0 0.000

10000 8000 6000 4000 2000 0 0 0.010 0.020 0.030 Strain (in./in.) 0.040 0.050 10 20 30 40 Stress (ksi) ET differences ET equation 50 Tangent Modulus Buckling (KL/r) cr 223.2521046 157.8630771

128.8946627 111.6260523 99.84137641 91.1422898 84.3813604 78.93150275 74.41710153 70.59690679 67.3048795 64.4113691 61.77857434 59.17430952 56.09208286 51.5097656 44.14566415 34.1419685 24.00464013 15.9961201 10.48827475 6.902516144 4.596633406 3.105440361 2.129145204 Column Inelastic Buckling Curve 60 Tangent Modulus Buckling Stress 0 2 4 6 8 10 12 14 16 18 20 22 24

26 28 30 32 34 36 38 40 42 44 46 48 50 2 E T Ix PT L2 PT 2 E T Ix 2 ET T 2 A AL2 KL / r 50 40 30 KL / rcr 20 2 ET T 10

0 0 30 60 90 KL/r 120 150 Residual Stress Effects Consider a rectangular section with a simple residual stress distribution Assume that the steel material has elastic-plastic stress-strain curve. Assume simply supported end conditions Assume triangular distribution for residual stresses b x d y rc

rc rt 0=0.5y 0=0.5y y/b 0=0.5y y E Residual Stress Effects One major constrain on residual stresses is that they must be such that dA 0 r 2 0.5 y y b b / 2 b / 2 2 y x d dx 0.5 y

x d dx b 0 2d y b 2 2d y b 2 0.5 y d b 2 0.5 y d b 2 b 8 b 8 0 0 Residual stresses are produced by uneven cooling but no load is present Residual Stress Effects b Response will be such that elastic behavior when x d 0.5 y 2 EIy 2 EIx Px 2 and Py 2 L L Yielding occurs when 0.5 y i.e., P 0.5PY y x

Inelastic buckling will occur after 0.5 y y Y Y Y/b 2 Y b Y (1 2 ) Y b b b Residual Stress Effects Total axial force corresponding to the yielded sec tion Y (1 2 ) Y b 2bd Y bd 2 2 Y 1 2 bd Y (2 2 )bd Y bd 2bd Y 2 Y bd 2 2bd Y Y bd(1 2 2 ) PY (1 2 2 ) If inelastic buckling were to occur at this load Pcr PY (1 2 2 ) 1 Pcr

1 2 PY If inelastic buckling occurs about x axis 2E d3 Pcr PTx 2 (2b) L 12 2 EI x PTx 2 2 L 1 Pcr PTx Px 2 1 2 PY PTx Px 2 1 PTx 1 2 PY PTx Px 1 PTx 2 1 PY PY 2 PY PTx 1 1 PTx 2 1

PY 2x 2 PY PTx 2 1 PY 2 x PTx PY b b x y Pcr PTx 2 P 1 E r Let, x 2 2 x PY x Y K x Lx If inelastic buckling occurs about y axis 2E d Pcr PTy 2 (2b) 3 L 12 2 EIy 3 PTy 2 2

L 3 1 P PTy Py 2 1 cr 2 PY P 3 PTy Py 2 1 Ty PY 3 PTy Py PTy 2 1 PY PY PY 3 P P 1 Ty 2 2 1 Ty PY y PY 3

P 21 Ty PY 2 y PTy PY b b x y Pcr PTy P 1 E Let, y 2 2 PY y Y r 2 y K L y y Residual Stress Effects x 2.236 2.000 1.826 1.690 1.581

1.491 1.414 1.313 1.221 1.135 1.052 0.971 0.889 0.803 0.705 0.577 0.317 y 2.236 2.000 1.826 1.690 1.581 1.491 1.414 1.246 1.092 0.949 0.815 0.687 0.562 0.440 0.315 0.182 0.032 Column I nelastic Buckling Normalized column capacity P/P Y 0.200 0.250 0.300 0.350 0.400

0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950 0.995 1.200 1.200 1.000 1.000 0.800 0.800 0.600 0.600 0.400 0.400 0.200 0.200 0.000 0.000 0.0

0.5 1.0 Lambda 1.5 2.0 Tangent modulus buckling - Numerical 1 Discretize the cross-section into fibers Think about the discretization. Do you need the flange To be discretized along the length and width? 2 For each fiber, save the area of fiber (A fib), the distances from the centroid y fib and xfib, Ix-fib and Iy-fib the fiber number in the matrix. Afib yfib Centroidal axis 3 Discretize residual stress distribution 4 Calculate residual stress (r-fib) each fiber 5 Check that sum(r-fib Afib)for Section = zero Tangent Modulus Buckling - Numerical 6

7 Calculate effective residual strain (r) for each fiber r=r/E Assume centroidal strain 14 13 Calculate the critical (KL)X and (KL)Y for the (KL)X-cr = sqrt [(EI)Tx/P] (KL)y-cr = sqrt [(EI)Ty/P] Calculate the tangent (EI)TX and (EI)TY for the (EI)TX = sum(ET-fib{yfib2 Afib+Ix-fib}) (EI)Ty = sum(ET-fib{xfib2 Afib+ Iy-fib}) Calculate average stress = = P/A 8 9 Calculate total strain for each fiber tot=+r Assume a material stress-strain curve for each fiber Calculate Axial Force = P Sum (fibAfib) Calculate stress in each fiber fib 12 11

10 Tangent modulus buckling - numerical Section Dimension b d y 12 4 50 No. of fibers 20 A Ix Iy 48 64 576.00 fiber no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

18 19 20 Afib 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 2.4 xfib -5.7 -5.1 -4.5 -3.9 -3.3 -2.7 -2.1 -1.5 -0.9 -0.3 0.3 0.9 1.5 2.1 2.7

3.3 3.9 4.5 5.1 5.7 yfib 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 r-fib -22.5 -17.5 -12.5 -7.5 -2.5 2.5 7.5 12.5 17.5 22.5 22.5 17.5 12.5

7.5 2.5 -2.5 -7.5 -12.5 -17.5 -22.5 r-fib -7.759E-04 -6.034E-04 -4.310E-04 -2.586E-04 -8.621E-05 8.621E-05 2.586E-04 4.310E-04 6.034E-04 7.759E-04 7.759E-04 6.034E-04 4.310E-04 2.586E-04 8.621E-05 -8.621E-05 -2.586E-04 -4.310E-04 -6.034E-04 -7.759E-04 Ix fib 3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2

3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2 3.2 Iy fib 78.05 62.50 48.67 36.58 26.21 17.57 10.66 5.47 2.02 0.29 0.29 2.02 5.47 10.66 17.57 26.21 36.58 48.67 62.50 78.05 Tangent Modulus Buckling - numerical Strain Increment Fiber no. -0.0003 1 2 3 4 5

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 tot fib -1.076E-03 -9.034E-04 -7.310E-04 -5.586E-04 -3.862E-04 -2.138E-04 -4.138E-05 1.310E-04 3.034E-04 4.759E-04 4.759E-04 3.034E-04 1.310E-04 -4.138E-05 -2.138E-04 -3.862E-04 -5.586E-04 -7.310E-04 -9.034E-04 -1.076E-03 -31.2 -26.2

-21.2 -16.2 -11.2 -6.2 -1.2 3.8 8.8 13.8 13.8 8.8 3.8 -1.2 -6.2 -11.2 -16.2 -21.2 -26.2 -31.2 Efib Tx-fib Ty-fib Pfib 29000 92800 2.26E+d06 -74.88 29000 92800 1.81E+d06 -62.88 29000 92800 1.41E+d06 -50.88 29000 92800 1.06E+d06 -38.88 29000 92800 7.60E+d05 -26.88 29000 92800 5.09E+d05 -14.88 29000

92800 3.09E+d05 -2.88 29000 92800 1.59E+d05 9.12 29000 92800 5.85E+d04 21.12 29000 92800 8.35E+d03 33.12 29000 92800 8.35E+d03 33.12 29000 92800 5.85E+d04 21.12 29000 92800 1.59E+d05 9.12 29000 92800 3.09E+d05 -2.88 29000 92800 5.09E+d05 -14.88 29000 92800 7.60E+d05 -26.88 29000 92800 1.06E+d06 -38.88 29000 92800 1.41E+d06 -50.88 29000 92800 1.81E+d06 -62.88 29000 92800 2.26E+d06 -74.88

Tangent Modulus Buckling - Numerical P 0=0.0=00=00=03 0=0.0=00=00=0 -0.0005 -0.0006 -0.0007 -0.0008 -0.0009 -0.001 -0.0011 -0.0012 -0.0013 -0.0014 -0.0015 -0.0016 -0.0017 -0.0018 -0.0019 -0.002 -0.0021 -0.0022 -0.0023 -0.0024 -0.00249 Tx -417.6 -556.8 -696 -835.2 -974.4 -1113.6 -1252.8 -1384.8 -1510.08 -1624.32 -1734.72 -1832.16 -1924.8

-2008.32 -2083.2 -2152.8 -2209.92 -2263.2 -2304.96 -2340.48 -2368.32 -2386.08 -2398.608 Ty 150=00=00=0 150=00=00=0 1856000 1856000 1856000 1856000 1856000 1670400 1670400 1484800 1299200 1299200 1113600 1113600 928000 928000 742400 556800 556800 371200 371200 185600 185600 10=00=00=00=0 10=00=00=00=0 16704000 16704000 16704000 16704000

16704000 12177216 12177216 8552448 5729472 5729472 3608064 3608064 2088000 2088000 1069056 451008 451008 133632 133632 16704 16704 KLx-cr KLy-cr T/ Y (KL/r) x (KL/r)y 209.4395102 628.3185307 0.174 181.3799364 181.3799364 181.3799364 544.1398093 0.232 157.0796327 157.0796327 162.231147 486.6934411 0.29 140.4962946 140.4962946 148.0960979 444.2882938 0.348 128.254983 128.254983 137.1103442 411.3310325 0.406 118.7410412 118.7410412 128.254983 384.764949 0.464 111.0720735 111.0720735 120.9199576 362.7598728 0.522 104.7197551 104.7197551 109.11051 294.5983771 0.577 94.49247352 85.04322617 104.4864889 282.1135199

0.6292 90.48795371 81.43915834 94.98347542 227.960341 0.6768 82.25810265 65.80648212 85.97519823 180.5479163 0.7228 74.45670576 52.11969403 83.65775001 175.681275 0.7634 72.44973673 50.71481571 75.56517263 136.0173107 0.802 65.44135914 39.26481548 73.97722346 133.1590022 0.8368 64.06615482 38.43969289 66.30684706 99.46027059 0.868 57.423414 28.711707 65.22619108 97.83928663 0.897 56.48753847 28.24376924 57.58118233 69.0974188 0.9208 49.86676668 19.94670667 49.27629185 44.34866267 0.943 42.67452055 12.80235616 48.8278711 43.94508399 0.9604 42.28617679 12.68585304 39.56410897 23.73846538 0.9752 34.26352344 6.852704688 39.33088015 23.59852809 0.9868 34.06154136 6.812308273 27.70743725 8.312231176 0.9942 23.99534453 2.399534453 27.63498414 8.290495243 0.99942 23.9325983 2.39325983 Tangent Modulus Buckling - Numerical Inelastic Column Buckling 1 ( T/ Y) Normalized critical stress 1.2

0.8 0.6 0.4 0.2 0 0 20 40 60 80 100 120 KL/r ratio (KL/r)x (KL/r)y 140 160 180 200 Normalized column capacity Column Inelastic Buckling 1.2

1.2 1 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0.0 0.0 0.5 Num-x Elastic 1.0 Num-y Lambda AISC-Design 1.5

Analytical-x Analytical-y 2.0 ELASTIC BUCKLING OF BEAMS Going back to the original three second-order differential equations: Therefore, 1 2 3 z z E I x v P v P x0 M BY M TY M BY M BX M TX M BX L L z z E I y u P u P y0 M BX (M M BY M M TX M BX M TYTX+M (M

) BY TY+M BY) BX L L z E I w (G KT K ) u ( M BX ( M BX M TX ) P y0 ) L z v u v ( M BY ( M BY M TY ) P x0 ) ( M TY M BY ) ( M TX M BX ) 0 L L L ELASTIC BUCKLING OF BEAMS Consider the case of a beam subjected to uniaxial bending only: 1 2 3 because most steel structures have beams in uniaxial bending Beams under biaxial bending do not undergo elastic buckling P=0; MTY=MBY=0

The three equations simplify to: z M TX M BX L z E I y u M BX M TX M BX L z u E I w (G KT K ) u M BX ( M BX M TX ) ( M TX M BX ) 0 L L E I x v M BX Equation (1) is an uncoupled differential equation describing inplane bending behavior caused by MTX and MBX ELASTIC BUCKLING OF BEAMS Equations (2) and (3) are coupled equations in u and that describe the lateral bending and torsional behavior of the beam. In fact they define the lateral torsional buckling of the beam. The beam must satisfy all three equations (1, 2, and 3). Hence, beam in-plane bending will occur UNTIL the lateral torsional buckling moment is reached, when it will take over. Consider the case of uniform moment (Mo) causing compression

in the top flange. This will mean that -MBX = MTX = Mo ELASTIC BUCKLING OF BEAMS For this case, the differential equations (2 and 3) will become: E I y u M o 0 E I w (G KT K ) u M o 0 where : K Wagner ' s effect due to warping caused by torsion K a 2 dA A Mo But , y neglecting higher order terms Ix M K o y ( xo x) 2 ( yo y ) 2 dA Ix A K K Mo Ix 2 2 2 2 y x x 2

xx y y 2 yy0 dA o 0 o Mo Ix 2 2 2 2 2 x y dA y x y dA x 2 xy dA y y dA 2

y y dA o 0 o o A A A A A A ELASTIC BUCKLING OF BEAMS Mo 2 2 K y x y dA 2 yo I x Ix A y x 2 y 2 dA A K M o 2 yo Ix

K M o x 2 2 dA y x y where, x A Ix 2 yo x is a new sec tional property The beam buckling differential equations become : (2) E I y u M o 0 (3) E I w (G KT M o x ) u M o 0 ELASTIC BUCKLING OF BEAMS Equation (2) gives u Mo E Iy Substituting u from Equation (2) in (3) gives : 2 M E I w iv (G KT M o x ) o 0 E Iy For doubly symmetric sec tion : x 0 2

M G K T iv 2 o 0 E Iw E I y Iw G KT Let , 1 E Iw and M o2 2 2 E I y Iw iv 1 2 0 becomes the combined d .e. of LTB ELASTIC BUCKLING OF BEAMS Assume solution is of the form e z 4 1 2 2 e z 0 4 1 2 2 0 1 12 42 2 2 1 12 42 2 Let , 1 , and , 12 42 1 2 1 12 42

, i 2 i 2 Above are the four roots for C1e1z C2e 1 z C3ei 2 z C4e i 2 z collecting real and imaginary terms G1 cosh(1 z ) G2 sinh(1 z ) G3 sin( 2 z ) G4 cos( 2 z ) ELASTIC BUCKLING OF BEAMS Assume simply supported boundary conditions for the beam: (0) (0) ( L) ( L) 0 Solution for must satisfy all four b.c. 1 0 0 1 G1 G 12 0 0 22 2 0 sinh(1 L) sin( 2 L) cos( 2 L) G3 cosh(1 L) 12 cosh(1 L) 12 sinh(1 L) 22 sin( 2 L) 22 cos( 2 L) G4 For buckling coefficient matrix must be sin gular : det er min ant of matrix 0 12 22 sinh(1 L) sinh( 2 L) 0

Of these : only sinh( 2 L) 0 2 L n ELASTIC BUCKLING OF BEAMS 2 n L 12 42 1 2 L 2 2 2 1 42 1 2 L 2 2 2 2 2 2 2 2 2 1 1 2 21 2 L L L 2 4 4 2 2 2 2 1 2 L L 2 G KT 2 M o2

2 2 2 E I y Iw L E I w L2 Mo 2 2 G KT E2 I yIw 2 2 L E I w L 2E I y Mo L2 2E Iw G KT 2 L

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