Structural Analysis II Course Code CIVL 322 Dr

Structural Analysis II Course Code: CIVL 322 Dr. Aeid A. Abdulrazeg 2005 Pearson Education South Asia Pte Ltd 1

Outline n n n Displacement method of analysis: general procedures Slope-deflection equations Analysis of beams Analysis of frames: No sidesway Analysis of frames: Sidesway 2005 Pearson Education South Asia Pte Ltd 2

Displacement Method of Analysis: General Procedures n n Displacement method requires satisfying equilibrium equations for the structures The unknowns displacement are written in terms of the loads by using the load- displacement relations These equations are solved for the displacement. Once the displacement are obtained, the unknown loads are determined from the compatibility equation using the load displacement relations. 2005 Pearson Education South Asia Pte Ltd 3

Displacement Method of Analysis: General Procedures n n When a structure is loaded, specified points on it called nodes, will undergo unknown displacement. These displacement are referred to as the degree of freedom The number of these unknowns is referred to as the degree in which the structure is kinematically indeterminate We will consider some e. g. s 2005 Pearson Education South Asia Pte Ltd 4

Displacement Method of Analysis: General Procedures n n n Any load applied to the beam in Fig 1(a) will cause node A to rotate Node B is completely restricted from moving Hence, the beam has only one unknown degree of freedom The beam in Fig 1(b) has nodes at A, B & C There are 4 degrees of freedom A, B, C and C 2005 Pearson Education South Asia Pte Ltd 5

Displacement Method of Analysis: General Procedures n Fig 1 2005 Pearson Education South Asia Pte Ltd 6

Slope-deflection Equations n n Slope deflection method requires less work both to write the necessary equation for the solution of a problem& to solve these equation for the unknown displacement & associated internal loads General Case ¡ To develop the general form of the slopedeflection equation, we will consider the typical span AB of the continuous beam as shown in Fig 2 when subjected to arbitrary loading 2005 Pearson Education South Asia Pte Ltd 7

Slope-deflection Equations n General Case (cont’d) ¡ The slope-deflection equation can be obtained using the principle of superposition by considering separately the moments developed at each support due to each of the displacement & then the loads ¡ Fig. 2 2005 Pearson Education South Asia Pte Ltd 8

Slope-deflection Equations n Angular Displacement ¡ Consider node A of the member shown in Fig 3(a) to rotate A while its while end node B is held fixed ¡ To determine the moment MAB needed to cause this displacement, we will use the conjugate beam method ¡ The conjugate beam is shown in Fig 3(b) 2005 Pearson Education South Asia Pte Ltd 9

Slope-deflection Equations ¡ Fig. 3 2005 Pearson Education South Asia Pte Ltd 10

Slope-deflection Equations n Angular Displacement (cont’d) 2005 Pearson Education South Asia Pte Ltd 11

Slope-deflection Equations n Angular Displacement (cont’d) ¡ From which we obtain the following: 2005 Pearson Education South Asia Pte Ltd 12

Slope-deflection Equations n Angular Displacement (cont’d) ¡ Similarly, end B of the beam rotates to its final position while end A is held fixed, Fig. 4 ¡ We can relate the applied moment MBA to the angular displacement B & the reaction moment MAB at the wall ¡ Fig 4 2005 Pearson Education South Asia Pte Ltd 13

Slope-deflection Equations n Angular Displacement (cont’d) ¡ The results are: 2005 Pearson Education South Asia Pte Ltd 14

Slope-deflection Equations n Relative linear displacement ¡ If the far node B if the member is displaced relative to A, so that the cord of the member rotates clockwise & yet both ends do not rotate then equal but opposite moment and shear reactions are developed in the member, Fig. 5(a) ¡ Moment M can be related to the displacement using conjugate beam method 2005 Pearson Education South Asia Pte Ltd 15

Slope-deflection Equations n Relative linear displacement (cont’d) ¡ The conjugate beam is free at both ends since the real member is fixed support, Fig. 5(b) ¡ The displacement of the real beam at B, the moment at end B’ of the conjugate beam must have a magnitude of as indicated 2005 Pearson Education South Asia Pte Ltd 16

Slope-deflection Equations n Fig. 5 2005 Pearson Education South Asia Pte Ltd 17

Slope-deflection Equations n Relative linear displacement (cont’d) 2005 Pearson Education South Asia Pte Ltd 18

Slope-deflection Equations n Fixed end moment ¡ In general, linear & angular displacement of the nodes are caused by loadings acting on the span of the member ¡ To develop the slope-deflection equation, we must transform these span loadings into equivalent moment acting at the nodes & then use the load-displacement relationships just derived 2005 Pearson Education South Asia Pte Ltd 19

Slope-deflection Equations n Slope-deflection equations ¡ If the end moments due to each displacement & loadings are added together, the resultant moments at the ends can be written as: 2005 Pearson Education South Asia Pte Ltd 20

Example # 1 n Determine the support moments for the continuous beam shown in figure. EI is constant. 20 k A B C 12. 5 2005 Pearson Education South Asia Pte Ltd 12. 5 15 10 21

Slope-deflection Equations Slope-deflection equations for simply support beam n A n B C Solve the two equations simultaneously 2005 Pearson Education South Asia Pte Ltd D

Example # 2 n Determine the support moments for the continuous beam shown in figure. EI is constant. 20 k 2 k/ft A B 25 ft 2005 Pearson Education South Asia Pte Ltd C 25 ft D 15 ft 10 ft 23

Example # 3 n Determine the support moments for the continuous beam shown in figure. EI is constant. 2005 Pearson Education South Asia Pte Ltd 24

Example # 4 n n n Determine the moment at Support B, assuming B settles 0. 25 n. E = 29 (106) psi I = 500 in 4 40 k A B 5 2005 Pearson Education South Asia Pte Ltd 5 10 C 25