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HW#9 Problems 6.5 and 6.6, Chajes Using the energy method, determine the critical load for the one-degree-of- freedom model of a flat plate shown in Fig. 6-26. The model consists of four rigid bars pin connected to each other and to the supports. At the center of the model two linear rotational springs of stiffness C M / connect opposite bars to each other. Also, each of the two transverse bars contains a linear extensional springs of stiffness K. For small lateral deflections the energy in the extensional springs can be neglected. ( Pcr 4C / a ) Fig. 6-26 Using the same model, obtain and plot relationships for the load P versus the lateral deflection d when (a) the lateral deflection is large, (b) the lateral deflection is large and the loads are applied eccentrically to the plane of the undeformed model. Which fundamental buckling characteristics of an actual plate are demonstrated by these models? (Note: for large deflections the energy in the extensional springs must be considered.) 1 Model Analysis I Using the energy method, determine the critical load for the one-degree-of- freedom model of a flat plate shown in Fig. 6-26. The model consists of four rigid bars pin connected to each other and to the supports. At the center of the model two linear rotational springs of stiffness C M / connect opposite bars to each other ( is the total rotation of the rotational spring to cause the moment M.). Also, each of the two transverse bars contains a linear extensional spring of stiffness K. For small lateral deflections the energy in the extensional springs can be neglected. Solution 2 P d a 2a a d a Fig. 6-27 Strain energy of the deformed body is that associated with the rotationa l spring only as the strain energy stored in the extensional spring is neglected. 1 U C 2 2 4C 2 2 2 where is the deformation angle of the rigid bar at the support. The loss of potential energy of the applied load P is V P 2 where is the horizontal displacement of the load point given by 2a 2a cos 2a 1 cos V 2 Pa 1 cos U V 4C 2 2 Pa 1 cos 8C 2 Pa sin 8C 2Pa 0 for a small 4C Pcr a Model Analysis II Using the model, obtain and plot relationships for the lateral load P versus the lateral deflection d when (a) the lateral deflection is large, (b) the lateral deflection is large and the loads are applied eccentrically to the plane of the undeformed model. Which fundamental buckling characteristics of an actual plate are demonstrated by these models? (Note: For large deflections the energy in the extensional springs must be considered.) Solution (a) As the lateral deflection becomes large, the strain energy stored in the extensional spring must be considered as well as that of the rotational spring. It is noted that the horizontal projection of the rigid rods perpendicular to the direction of the applied load remains unchanged due to the boundary condition given in the model (pin-pin). Therefore, the angle of rotation of these bars with respect to the undeformed horizontal plane is given by the relationship tan d / a . 1 The total strain energy stored in the deformed body is U Ur Ue 3 The strain energy due to rotational springs is U r C 2 C 2 2C 2 2C 2 2C 2 2 1 2 1 2 (a) 2 2 where sin 1 d / a (b) From the Pythagoras theorem applied to the deformed model, the length of the extensional bar is given by d 2 2 2 a d 2 a 1 2 (c) 2 2a a where is given by a2 d 2 d 2 2a 1 cos 2a 2 a d 2a 1 2 2 2a 1 1 (d) a a 2 d d 2 2 a 1 1 1 (e) a a Hence, the extension of each spring is 2 2 2 d d a a 1+ 1 1 1 (f) a a 2 2 d If is neglected since it is small compared to , then 2a a 2 1 1 d a (g) a 1 Ue K 2 2 (h) 2 4 Using equation (f) for , 2 2 2 2 2 2 d d d d U e Ka 2 2 1 1 2 1 1 1 (i) a a a a Using equation (g) for , d 2 d 2 U e Ka 2 2 1 2 (j) a a The total strain energy is then U Ur Ue U =2C 2 2C 2 2 Ka 2 2 2 d d d d 2 2 2 2 (k) Ka 2 1 1 2 1 1 1 a a a a Or approximate total strain energy is then d 2 d 2 U 2C 2 2C 2 2 Ka 2 Ka 2 2 1 (l) a a Substituting the expressions for the deformed angle into equation (l), yields 1 d 2 1 d 2 d 2 d 2 U 2C tan 2C sin 2 Ka 2 Ka 2 2 1 (m) a a a a d 2 The loss of potential energy of the applied load is given by (since 2a 1 1 ) a d 2 V P 2 Pa 1 1 (n) a The total potential energy is the sum of equation (m) and equation. (n). 5 1 d 2 1 d 2 2 d 2 2 d 2C tan 2C sin 2 Ka Ka 2 1 2 a a a a (o) d 2 2 Pa 1 1 a For the principle of minimum potential energy, the derivative of the total potential energy with respect to the displacement amplitude must be equal to zero. Hence, d 1 1 1 d 1 1 2C 2sin 1 a 2C 2 tan a 2 d a d 2 d a 1 1 a a d 1 2 2 2 d 1 d 1 a a 2 Pa a a 0 Ka 2 2 2 a a 2 d 2 1 2 1 d a a Solving the above equation for P, gives 2 sin 1 d tan 1 d 2C d a a Kda 1 1 P 1 2 (p) d a d 2 d 2C d 2 1 1 a a 1 a It should be noted here that the approximate expression for the extension of the spring results in a slightly smaller value of extension. However, neglecting the quantity / 2 of the spring extension results in a slightly greater rotation of the extension bars, thereby alleviating the effect of the approximation. All subsequent derivation will be based on this approximate expression. The error involves would be less than 0.5% for a value of d up to 50% of the plate dimension. 6 It should be of interest to note that the bifurcation buckling load, Pcr 4C / a , can be obtained by taking the limiting value of equation (p) (applying the L’Hpital rule) when d approaches to zero. 1 Kda 1 2 d 1 d tan 1 d 1 d sin a a a 2 2 d d 1 d 2C 2 2 a 1 1 a a a Pcr 2C lim d 0 1 d d sin 1 a 1 a 1 d 3/2 2 d 2 d 2 2 a a 1 a a 1 2 2 a d 1 a 2d tan 1 d a Ka 1 2 1 Kd 3/2 2 d 2 2 2C 2 d d 2 1 (q) a 1 2aC 1 a a a 1 1 4C Pcr 2C as expected a a a Solution (b) 2 1 P e 7 Fig. 6-28 From Fig. 6-28, it becomes evident that d 2 1 2 a a cos 2 a a d 2 2 2a 1 1 a 2ed 2 2e sin 2e (This is justified for a small e, i.e., if e 0.2a , error < 0.7%. It is a assumed here that the eccentricity occurs at both ends.) In the case when e is introduced to account for geometric imperfection of the structure, the eccentricity remains a very small quantity. 1 2 The loss of potential energy of applied load is d 2 d 2 d V P 2 Pa 1 1 2 Pe sin 2 Pa 1 1 2 Pe (r) a a a Equation (r) can be derived alternately as ed M end Pe and the loss of potential energy is VM end 2M end 2 Pe 2 P a The strain energy stored in the deformed structure is the same as in equation (m). Hence, the total potential energy is 1 d 2 1 d 2 2 d 2 2 d U V 2C tan 2C sin 2 Ka Ka 2 2 1 a a a a (s) d 2 ed 2 Pa 1 1 2 P a a 8 In order to minimize the total potential energy , set the derivative of with respect to the arbitrary displacement d, equal to zero and solve the resulting equation for P. d 2 1 4C 1 d 4C 1 d a Ka 2 2 Kad 0 sin tan d a 2 a d 2 a 2 1 d 1 d 1 a a a d 2 a P 2P e d 2 a 1 a sin 1 d tan 1 d adK a a 1 2 1 1 d 2 d 2C d 2 1 1 4C a a a Pcr (t) a 2d a 2e 2 a d 1 a Equations (p) and (t) are plotted comparatively in Fig. 6-29a It can be shown by applying the L’Hpital rule that the limit value of equation (t) yields 4C / a , the bifurcation buckling load, when d approaches to zero. Discussion Examination of the elastic responses of this model clearly illustrates the different load versus deflection characteristics between columns and plates. The elastic response of this model is analogous to that of a column when the spring constant K is small, and to that of a plate when the spring constant K is stiff enough. When K is small, the force developed in the extensional spring is not strong enough to restore the deformed model to the stable state (original state). Instability 9 of the model is further accelerated due to the finite lateral displacement of the model when K is small (illustrated in Fig. 6-30a for small K). As a result, the column buckles. The load carrying capacity of the column is considered exhausted. Additional load carrying capacity may exist in plate structures beyond the elastic buckling loads that are known as postbuckling strengths of plates if the boundary conditions of such structures are conducive to develop a considerable degree of tensile stresses within the structures. Consider the model with a large spring constant K. As the lateral displacement increases, a considerable amount of restoring force develops in the extensional spring. This large restoring force in the spring enables the model to resist the load beyond the elastic buckling load, P=4C/a. Note the boundary condition of the two rigid rods with extensional springs. They are supported by pins at their ends enabling the springs to stretch. If there is a roller, the spring will not be stretched and no postbuckling strength is realized. Also, note that the model is still within the elastic range, and no material nonlinearity is involved. Hence, this postbuckling strength is governed by geometric nonlinearity. In a real structure, however, the displacement simply cannot be infinitely large before encountering material nonlinearity. Hence, postbuckling behavior in a real structure is governed by both geometric and material nonlinearity, thereby rendering a formidable computational challenge. An incremental nonlinear analysis algorithm implemented in modern finite element codes, such as NASTRAN, ABAQUS, ADINA, etc., makes such an analysis possible. When the load is applied with an eccentricity “e” (frequently introduced to account for initial imperfection of structural elements), the model exhibits deflection amplification type behavior with no bifurcation type buckling behavior. For a very small value of initial imperfection, the ultimate load of the model asymptotically approaches to the elastic buckling load if very large 10 deflections are permitted. This illustrates the fact that the elastic bifurcation buckling load indeed establishes an upper-bound of the ultimate load of a real structure made of components with code permitted geometric imperfections. C PLATE MODEL ANALYSIS-CHAJES BOOK PROBLEM 6.5 AND 6.6 (A) AND (B) C ECCENTRICALLY APPLIED LOAD C SOLUTION BASED ON LARGE DISPLACEMENT THEORY IMPLICIT REAL*8 (A-H,O-Z) DIMENSION D(100),P(100) 99 READ(5,*,END=98) A,C,EK,E WRITE(6,20)A,C,EK,E 20 FORMAT(/' ','P-D RELATIONSHIP(',' A =',F5.2,' C =',F5.2,' EK =', 1 F5.2,' ECC =',F5.2,')') WRITE(6,21) 21 FORMAT(/' ',' DEFLECTION',' APPLIED LOAD') 10 FORMAT(3X,D15.5,3X,D15.5) D(1)=A/20000.0 DO 100 I=1,50 IF (I.EQ.1) GO TO 12 D(I)=A/100.0*(I-1) 12 DA=D(I)/A DA2=(DA)**2 OMDA2=1.-DA2 OPDA2=1.+DA2 DUM1=DASIN(DA)/DSQRT(OMDA2) DUM2=DATAN(DA)/OPDA2 DUM3=EK*D(I)*A/(2.*C)*(1.-1./DSQRT(OPDA2)) P(I)=4.*C/A*(DUM1+DUM2+DUM3)/(2.*D(I)/A/DSQRT(OMDA2)+2.*E/A) WRITE(6,10) D(I),P(I) 100 CONTINUE GO TO 99 98 STOP END The same problem may be solved based on the small displacement theory so that simplified relationships permitted in the microgeometry can be utilized. Model Analysis II Using the model, obtain and plot relationships for the lateral load P versus the lateral deflection d when 11 (a) the lateral deflection is large, (b) the lateral deflection is large and the loads are applied eccentrically to the plane of the undeformed model. Which fundamental buckling characteristics of an actual plate are demonstrated by these models? (Note: For large deflections the energy in the extensional springs must be considered.) Solution (a) As the lateral deflection becomes large, the strain energy stored in the extensional spring must be considered as well as that of the rotational spring. It is noted that the horizontal projection of the rigid rods perpendicular to the direction of the applied load remains unchanged due to the boundary condition given in the model (pin-pin). Therefore, the angle of rotation of these bars with respect to the undeformed horizontal plane is given by the relationship tan 1 d / a d / a (assuming small displacement theory). The total strain energy stored in the deformed body is U Ur Ue The strain energy due to rotational springs is U r C 2 C 2 2C 2 2C 2 2C 2 2 1 2 1 2 (a) 2 2 where sin 1 d / a d / a (b) From the Pythagoras theorem applied to the deformed model, the length of the extensional bar is given by d 2 2 2 a d 2 a 1 2 (c) 2 2a a where is given by 12 2 d2 2a 1 cos 2a 1 1 a 2 (d) 2 a Since is so small, it is neglected in equation (c). Hence, d 2 1 d 2 a 1 a 1 (e) a 2 a Hence, the extension of each spring is 2 ad a (f) 2 a 1 Ue K 2 2 (g) 2 Using equation (g) for , 4 a2 d K d4 Ue K K 2 (h) 4 a 4 a2 The total strain energy is then U Ur Ue K d4 K d4 U =2C 2C 2 4C 2 2 (i) 4 a2 4 a2 Substituting the expressions for the deformed angle into equation (i), yields K d 4 4Cd 2 K d 4 U 4C 2 2 (j) 4 a2 a 4 a2 d 2 The loss of potential energy of the applied load is given by (since 2a 1 1 ) a 13 1d 2 d2 V P 2 Pa 1 1 P (k) 2 a a The total potential energy is the sum of equation (j) and equation (k). 4Cd 2 K d 4 d2 P (l) a2 4 a2 a For the principle of minimum potential energy, the derivative of the total potential energy with respect to the displacement amplitude must be equal to zero. Hence, 8Cd 4 Kd 3 2 Pd 2 0 d a 4a 2 a Solving the above equation for P, gives 4C Kd 2 P (m) a 2a It should be noted here that the approximate expression for the extension of the spring results in a slightly smaller value of extension. However, neglecting the quantity / 2 of the spring extension results in a slightly greater rotation of the extension bars, thereby alleviating the effect of the approximation. All subsequent derivation will be based on this approximate expression. The error involves would be less than 0.5% for a value of d up to 50% of the plate dimension. It should be of interest to note that the bifurcation buckling load, Pcr 4C / a , can be obtained by taking the limiting value of equation (m) when d approaches to zero. Solution (b) From the sketch, it becomes evident that 2 d2 1 2 a a cos 2a 1 1 2 a 14 2ed 2 2e sin 2e 2e (This is justified for a small e, i.e., if e 0.2a , error < 0.7%. It a is assumed here that the eccentricity occurs at both ends.) In the case when e is introduced to account for geometric imperfection of the structure, the eccentricity remains a very small quantity. 1 2 The loss of potential energy of applied load is d2 d V P P 2Pe (n) a a Equation (n) can be derived alternately as ed M end Pe and the loss of potential energy is VM end 2M end 2 Pe 2 P a The strain energy stored in the deformed structure is the same as in equation (j). Hence, the total potential energy is 4Cd 2 K d 4 d2 d U V 2 2 P 2Pe (o) a 4 a a a In order to minimize the total potential energy , set the derivative of with respect to the arbitrary displacement d, equal to zero and solve the resulting equation for P. 8Cd Kd 3 2 Pd 2 Pe 2 2 0 d a a a a e 4C Kd 2 4C Kd 2 1 P 1 P (p) d a 2a a 2a 1 e / d Equations (m) and (p) are plotted comparatively in Figs. 6-29b and Fig. 6-30b. It can be shown that the limit value of equation (t) yields 4C / a , the bifurcation buckling load, when d approaches to zero. 15 P-D RELATIONSHIP (LARGE DISPL) WHEN K IS LARGE 6 5 FORCE (*C/a) 4 a=1.0 in. K=30.0 kip/in. 3 C=1.0 k-in./in. 2 e=0.00*a, prob 6.6a e=0.01*a, prob 6.6b 1 e=0.02*a, prob 6.6b 0 0.0 0.1 0.2 0.3 0.4 0.5 DEFLECTION (*a) Fig. 6-29a Postbuckling of plate P-D RELATIONSHIP (SMALL DISPL) WHEN K IS LARGE 6 5 FORCE (*C/a) 4 a=1.0 in. K=30.0 kip/in. 3 C=1.0 k-in./in. e=0.00*a, prob 6.6a 2 e=0.01*a, prob 6.6b 1 e=0.02*a, prob 6.6b 0 0.0 0.1 0.2 0.3 0.4 0.5 DEFLECTION (*a) Fig. 6-29b Postbuckling of plate 16 P-D RELATIONSHIP (LARGE DISPL) WHEN K IS SMALL 6 a=1.0 in. 5 K=1.0 kip/in. FORCE (*C/a) 4 C=1.0 k-in./in. 3 e=0.00*a, prob 6.6a 2 e=0.01*a, prob 6.6b 1 e=0.02*a, prob 6.6b 0 0.00 0.10 0.20 0.30 0.40 0.50 DEFLECTION (*a) Fig. 6-30a Postbuckling of plate P-D RELATIONSHIP (SMALL DISPL) WHEN K IS SMALL 6 a=1.0 in. 5 K=1.0 kip/in. C=1.0 k-in./in. FORCE (*C/a) 4 3 2 e=0.00*a, prob 6.6a e=0.01*a, prob 6.6b 1 e=0.02*a, prob 6.6b 0 0.00 0.10 0.20 0.30 0.40 0.50 DEFLECTION (*a) Fig. 6-30b Postbuckling of plate 17