Structural Analysis II Course Code: CIVL322 Dr. Aeid A. Abdulrazeg

Outline Displacement method of analysis: general procedures Slope-deflection equations Analysis of beams Analysis of frames: No sidesway Analysis of frames: Sidesway

Displacement Method of Analysis: General Procedures 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.

Displacement Method of Analysis: General Procedures 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

Displacement Method of Analysis: General Procedures 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

Displacement Method of Analysis: General Procedures Fig 1

Slope-deflection Equations 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 slope-deflection equation, we will consider the typical span AB of the continuous beam as shown in Fig 2 when subjected to arbitrary loading

Slope-deflection Equations 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

Slope-deflection Equations 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)

Slope-deflection Equations Fig.3

Slope-deflection Equations Angular Displacement (cont’d)

Slope-deflection Equations Angular Displacement (cont’d) From which we obtain the following:

Slope-deflection Equations 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

Slope-deflection Equations Angular Displacement (cont’d) The results are:

Slope-deflection Equations 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

Slope-deflection Equations 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

Slope-deflection Equations Fig.5

Slope-deflection Equations Relative linear displacement (cont’d)

Slope-deflection Equations 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

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:

Example # 1 Determine the support moments for the continuous beam shown in figure. EI is constant. 15 20 k A 10 C 12.5 B

Slope-deflection Equations Slope-deflection equations for simply support beam A B C D Solve the two equations simultaneously

Example # 2 Determine the support moments for the continuous beam shown in figure. EI is constant. 20 k 2k/ft A B C D 25 ft 25 ft 15 ft 10 ft

Example # 3 Determine the support moments for the continuous beam shown in figure. EI is constant.

Example # 4 Determine the moment at Support B, assuming B settles 0.25 n. E = 29 (106) psi I = 500 in4 40 k A B C 5 5 10