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# ELEM STRUC I CEE 379

UW

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This 6 page Class Notes was uploaded by Mason Hackett on Wednesday September 9, 2015. The Class Notes belongs to CEE 379 at University of Washington taught by Marc Eberhard in Fall. Since its upload, it has received 32 views. For similar materials see /class/191978/cee-379-university-of-washington in Civil and Environmental Engineering at University of Washington.

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Date Created: 09/09/15

CEE 379 ELEMENT STIFFNESS MATRIX FOR PLANAR TRUSS Element Forces QNy X T 0059 xFDFx yFDFy I x 9L QNX p Yp DFy 9N Element Displacements xN DMyN DNy Jew1y DN Initial Length g xF xN2 yF WM 1 A COS 0x EXx1 xN 1 A cosgy 51110 yN Final Length z Ag cos 0x thSC DNX2 cos 0y DFy DNy 212 z M W cosz 0x 2g cosngNx 2g cos 0ka ZDFXDNX D DMZ 2 cos2 1 Moos 19wa Moos QyDFy ZDFyDNy DFy2 DNy212 z M w1 ZMXDNX 2ampka ZDFXDNX D DM2 2 2 2 2 12 y 2MyDNy 2MyDFy 2DFyDNy DFy DNy CEE 379 Divide by X and note that cos2 0x cos2 0y l 2 2 Ag l ZCOSQXDTNXZCOSQXD7 2DF2M le 2 le X 1 z D D D D D 2 D 2 200s0y7w200s0y7Fy 2 Z 2 2 12 1 2AXD M2AxD M 2DF le 1 X X X X z D D D D D 2 D 2 21 Ny2l 2 Rf y X y X X X X 2 2 FoerD ltlt then Dr amp and Dy ltlt y 2 ltlt g 2 g D D So l zl 200s0 D N 200s0 DJ Zcosg Ny200s0 if X I X I X y X y X D D z1 2AXD7M2AXD7M 2Ayfwlyfy m r 0 0 Us1ng Binomial Theorem 1r lECCf AK 1 Giving 7 Z cos t9xDNX cos QXDFX cos 9wa cos QyDFy EampDmampDk DMDW Axial Force Q AE cos QXDM cos QDEC cos 9wa cos QyDFy eampDMampD DMDW QM Q0089x Qk QM Qcos 0y Qly QR QCOSQ Qk Each component of axial force is QFyZ QCOSQyZ Qly ccos0x Zx Substituting in for Q and writing the equations in matrix form one nally obtains S cos 9 Sin 9 A y X y QM c2 cs c2 30 DM 1 xly A ylx DNX QM AE cs 32 cs sz DNy AE fix1y A xly 1 DM QtSC c2 cs c2 cs DEC A xly A xly DFX Q sc 32 cs 32 D5 flylx A xly A DFy Q k D CEE 379 ELEMENT STIFFNESS MATRIX FOR PLANAR TRUSS Element End Displacements Tquot I 9 xFDFxyFDFy AL xF yF DF Element End Forces Q g sm QNy 9 gt 39 cos gt T my BM 39 L QN 9N If you decide a priori that the element forces Q M Q Ft are linearly related to the displacements DNX DFSC then Q 211ij 171 and QM D M 0 0 0 Q 0 D o 0 M k k M k k Q o 0 DR 0 QW o o 0 DW ie if kl are constant then ku1 u2 ku1 ku2 Superposition only works because if we can neglect the higherorder terms of D M D k 19 Imtml if the strains are small ie 039 E 8 and if the initial and nal orientations are about the same gfmal CEE 379 Al7D QNx QNy QFX QFy M cos 9quot Casel gcos HXDNX cos 2 HXDNX cos 0x cos HyDNX cos 2 HXDNX cos 0x cos HyDNX ELEMENT STIFFNESS MATRIX FOR PLANAR TRUSS Case 2 AE Tcos HyDNy cos 0x cos HyDNy AE 2 Tcos HyDNy gcos 0x cos 2 HyDNy AE 2 Tcos HyDNy QMT Case 3 cos HXDFX cos 2 HXDFX cos 0x cos HyDFX cos 2 HXDFX cos 0x cos HyDFX Case 4 AE Tcos HyDFy cos 0x cos HyDFy AE 2 Tcos HyDFy gcos 0x cos HyDFy AE 2 Tcos HyDFy CEE 379 2 DOF Spring Example D3 D1 D2 D4 gt gt gt gt Q3 Q4 gt Q11OkN Q225kN Sprinq ForceDeformation Relationships 01 k39 d Spring 1 3 1 k1 500 qN 500 500 dN 3 kNm qF 500 500 dF 1 Spring 2 1 2 k2 1000 qN 1000 1000 dN 1 kNm qF 1000 1000 dF 2 Spring 3 2 4 k3 250 qN 250 250 dN 2 kNm qF 250 250 dF 4 Compatibility Boundary dN1 D3 0 Nodal dF1 dN2 D1 Conditions dF3 D4 0 Connectivity dF2 dN3 D2 System Stiffness Eguations K D Q 1 2 3 4 1500 1000 500 o 1 D1 Q1 1000 1250 o 250 2 D2 Q2 500 o 500 o 3 D3 Q3 0 250 o 250 4 D4 Q4 CEE 379 2 DOF Spring Example Solve for Nodal Displacements K11 Du Qk 1 2 1500 1000 1 D1 Q1 1000 1250 2 D2 Q2 K1141 OR Du 00014 00011 0014 m 00011 00017 25 0031 m Spring End Forces positive to right negative to left k39 d q Spring 1 500 500 0000 714 kN 500 500 0014 714 kN Spring 2 1000 1000 0014 1714 kN 1000 1000 0031 1714 kN Spring 3 250 250 0031 786 kN 250 250 0000 786 kN Spring Internal Forces tension positive compression negative Spring 1 T1 714 kN Spring 2 T2 1714 kN Spring 3 T3 786 kN

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