Question

Problem 3 Analyze the frame structure subjected to the following support movements: a beam-end settlement at Node #4; . a column base rotation at (Node #1). All members have a constant bending stiffness El and are considered as axially rigid. a) Determine the degree of kinematic indeterminacy (DKI) and show the independent DOFs. b) Assemble the structure stiffness matrix Kg. c) Assemble the structure fixed-end force vector Po. Then solve for nodal displacement vector Us based on the equations of equilibrium Ky Ur=P;-Po. d) Draw the shear force and moment diagrams of the structure. e) Draw the deflected shape of the structure and indicate the inflection points. 2 L + + + (a) 3 (b) L (c) D. Ay = 1 EA00 El = Const. for all members 0 = 1/L

Answer 1

2 Bet 2 EL C3 04 L 48 = 1 I А Os I L Sol"Total mumber of Kinematic indeterminaciy = 2 (Oo, Oc) as shown in figure below. Pi D2 B Dy 02 TI Now comparibility equations. AOL = Pfo = fixed and force vector 'SD = AD-AD2 o Kft. Uf = Pf - Pfo ļ as shown in question} Kiffa Sa stiffness matrix Da Uf. unknown displacements Pf Ap force nector at displacement

Now calculate stiffness matrix (8]= [kat 2 to get first column of stiffness matrix, krouide unit dis flaument at D, and restrain all the other displacements ܠܠܠܠ 4EI l Sus 4ET e Sala LEI l to get second column of stiffness matrix prouide unit displacement in direction of Dy and reshain D, S22 = 1 512 S12= 2ET l S22 = #££ + the + tcf = 12 12 EL g l

Now 3 stiffness matrix Kft as= EI l 4 2 2 12. © Now force meetor AD = PFF :) = no forces given in don of o, fD2 Fixed end force vector gures is dish of P, fD2 Aa= Pfo = AME D c B g & Amen AMLA 07 AMLI А ADLI = Pfo I =0 = no effect at B ADLO =QEL Q EL 3665/62 e 27687 12 Alfatih 8 ADZO 8 EI ez 6 El A el 4EL XL = l l 4EI e2

4 Now (Ao - Aor] = [ff P to 0 -8 EI 72 SD Now from equilibriem equations CAO - ADI] = St [ao-Aor D C 12 -2 sol= 44 EI -2 4 12 -2 Da l 44EI -2 4 -8EF/0² Di . ] = l 4461 12x0–2X(-361/2) +*(-88472) D2 - 2x0 + 4 X B, Os = [ht] - A [-3 tys. The Dz = Oc= l 44E1

5 Now we have to calculate I draw shear force and bending moment diagrams. For that first we need to calculate member end action. AM3 Ama C D to Ama togen Ama In the about fegiene, Ama, Ama, Ama, Ama are the unknown member and actions. Ama moment at A Ama moment is member AC at e Ама moment in Member CO at C Ama moment at D from equations of equilibrium Ame Am = Amet Amo .D member end actions in restrained stuchere member and action is dish of du due to unit des placements, AMD

0 Now Ame mahix AMLI 46Ixt = 4E1 l l e2 Amira । 12 Amla = 6E1/12 Amez = 661/42 I (Am] matrix Ձ ԲY 1 l2 با ما an Now Amo matrix prouide unit displacement at Oy at AMD3I Amdal Amore "ANDRI Amosi AMDII AMD 21 AMD 31 Amoeiro Now frouide unit dis placement at Da -

Amo32 the AMOAL AMD22 AMDI2 Amo 12 QET AMD22 4617 l l AMD 32 = 4E1 e Amo 42 L BL l Now Ama Amet AmD.D. Ami 1 Ami QE1 4 I LET = + 2 l Tt AM3 e2 im 2 3 tama Ami 0 24 Ama 2 EI + O 2 V l 2 11 l Ama 2 3 1 Ama O 3

Ami 8 = 2 EI 84 + l2 [-2] - SEI 12 e 2 = EԸ 6 "I 12 0.54 EL Ama LEI 유 - + 8ET ez 11 12 -0.909 EL lL Ama 11 6 El + 8 EL [-4 3.09 BI Il 12 1112 ez Ama 6 EI + Petit C--) = 4.54 EI 12 e2 Now, equilibrium of diffrent members = m = for easy calculation put ET 3.09 -0.gog = 2,181m 3.09 m B с D 0.gog M 4.54m 4.5+ 0.54m A

9) Now individual members. B ç 2,18m 2.187 92.18 772 = 9.18 EL l3 с. Shear force diagram B 2.180 BMD 0.909m 0 369 m/ 0909 0.369 EL 12 B * с 01369 60 70.54m 0.54 SED BMO

3.09 4.54 0 27,63 Ef ze e2 7,63 RR e3 7,63 2016 SFD 4.54 BMD 3.09 r 7.63 Rf/13 218 Elle3 В D e & Stand hen मै Stand here Shear force deegra А 0.369 08/13 2.18 -3.49 C B D + 7.63 449ата 3м) A hoggen B.M (C7 0.54 A

0 # Deflected shape น of he inflection point ye op