אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא ١٠٣ א ١٠٣ W،،אא،א אאאאאאאאאא א א ،א א א א א א אאאאאא Wאאאאאא K אאא אאאאאאא א ،א אא א א א א א א א ، א،אאאא אאאאאאאאאאאאא א א ،א Kאא،אא א ? ? ? ? א א Kאאאאאא אאאאאאא ،א،אאאא Kאאאאא W א א א א Kא אאאא ١٠٣ אאאאStatics אא Kאאאאאא אEStaticsFא אא אאאאא ،אאא א،אאא،EאאFא Kאאאא Wאאא W Kאא J EאFאא J KFree Body Diagramאא J K J אאאאאאאאא J Kא KEאFאאאאאאא J Truss Systemאאאאאא J אאאאא Wאאאאאא א• א אא• א אאאאא • אא • אאא• א אא• א א• א אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] א אאא אאא אאא ١ אא אאא ١٠٣ Wאא אאאאאא Kאאאא Wאא Wאא אאא • אא • אאא • K٪100אאאWאאא ٧Wאא Wאא • • • Wאא אאאאאאא KVectorsא -١- אא אאא ١٠٣ Vectorא K אאאאא،אאא א ،אא אא Kא،א،א،אWאא K١ א אא٢{١ Coplanar VectorsאאE١ אאא F3 F1 F2 ١ ﺷﻜﻞ Colinear Vectors אE٢ K٢אא F2 F1 ٢ Concurrent Vectors אE3 K٣אאא F3 F2 F1 ٣ -٢- אא אאא ١٠٣ ٣{١ K٤אאאא F2 F1 F1 + F2 F1 F2 ٤ Wאא A+B=B+A K א٤{١ א (٥ )أﻧﻈﺮ ﺷﻜﻞ رﻗﻢαאBA A.B = A.B.Cos(α) W Kvector scalarא A α O B ٥ -٣- אא אאא ١٠٣ Kא٥{١ אא J ١ אא E٦FC, B,A W A = Acos(α1) + ASin(α1) B = BCos(α2) + BSin(α2) C = CCos(α3) + CSin(α3) KאאC, B, A אאאα3،α2،α1 y Cy C D Dy By Ay O B A Ax Bx Cx x Dx ٦ א אyאxאDא ٦א Wאא Dx = A.Cos(α1) + B.Cos(α 2) + C.Cos(α3) Dy = A.Sin(α1) + B.Sin(α 2) + C.Sin(α3) -٤- אא אאא ١٠٣ WDאא D = D x2 + D y2 Tan (α ) = Dy α = Tan ( −1 Dx Dy Dx) KאאDאאאאα א٦{١ K אאאאאאאאא א vectorאאא אאKNאkN א، Kאא، אאא،א KEאFEFאWאא W אאFא FxWW א،٧א KאאFאא هﻲ אα،٧אFy -٥- אא אאא ١٠٣ F Fy α Fx ٧ Fx = FCos (α ) Wאאא Fy = FSin(α ) WF אא F = 2 F x + F y2 α = Tan −1 ( Fy Fx ) W١ W٨אאאא F1 = 25 kN, F2 = 50 kN, F3 = 100 kN Wא אxאא ٨א KEFKy -٦- אא ١٠٣ אאא y 50 kN F2y 30 F3 100 kN F2x F2 25 kN F1y F1 60 45 F1x x F3y F3x ٨ F1x = F1cos 45 = 25cos 45 = 17.68 kN F1y = F1sin 45 = 25sin 45 = 25 kN F2x = J 50cos (60 ) = -25 kN F2y = 50sin (60 ) = 43.30 kN F3x = 100cos( 30 ) = -86.60 F3y = 100 sin( 30 ) = - 50 kN א א אF2x, F3x, F3y א K Rx = F1x + F2x + F3x = 17.68 – 25 -86.60 = - 93.92 kN Ry = F1y + F2y + F3y = 17.68 + 43.30 – 50 = 10.98 kN WRא -٧- אא אאא R = ١٠٣ 2 R x + R y2 α = Tan −1 ( Ry Rx ) Wא R = (−93.92)2 + (10.98)2 = 94.56 kN α = Tan−1(10.98/ 93.92) = 6.670 Graphical Method אאא٧{١ ٦אאאא אאאאאאא Kאאאאאא W F2 = 30 kN אאF1= 20 kNאא Kאאא٩ F1 = 20 kN F2 = 30 kN ٩ 10 kN -٨- Wא ١ אא אאא ١٠٣ אא 20 kNOא RאK١٠א30 kNאא K١٠אאאאאאא א אRאאK אאא Kאאאאאאא،א R F1 = 20 kN O F1 = 20 kN O OF1 = 20 kN ١٠ α F1 = 20 kN F1 = 20 kN W ،אאא אא אאאא Kאא Particleאא٨{١ אאאאאאאא KאאאאאאאKא -٩- אא אאא ١٠٣ אאאא٩{١ אא،EאאFאאאא אאאKEאFאאאא אz, y, xא(Fx, Fy, Fz) WE١١אF F = Fx + Fy + Fz Fx = F Cos(αx ) Fy = F Cos(αy) Fz = F Cos(αz) W xאאאאαx yאאאאαy zאאאא αz Wא F = Fx2 + Fy2 + Fz2 y Fy A F αy αx αz Fz Fx O B z x ١١ -١٠- אא ١٠٣ אאא Kאא١٢אאא K١ ١٢٠ kN ١٢٠ kN 110 ٤٥ E F ١٥ kN ٨0 kN ١٠٠ kN 20 kN EF E F ١٢ אאאK אא١٣אאא K٢ KאאאאKאאא ٢٠٠ kN ١٥٠ kN ٣٥٠ ٩٠ ٠ ١٣ ١٨٠ kN אאK אא١٤אאא K٣ KאאאאKאאאא ٢٠٠ kN ٩٠ kN ٦٥ ٠ ٤٥ ١٨٠ kN ١٤ -١١- אא אאא ١٠٣ K١٥אAא KאאאαאאEF KאאאEF 40 kN ٣0 kN α A α ٦0 kN ١٥ -١٢- אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא אאא אאא ٢ אא ١٠٣ אאא Wאא א K Wאא Wאא • א • אא • אא • K٪100אאאWאאא ٥Wאא Wאא • • • Wאא Kאאאאאא - ١٣ - אא ١٠٣ אאא אא١{٢ KOאאאא E١אF O Wאא EאאאאאאאFxEאFZא Moment = Force x Perpendicular Distance M=Fxd W kN.mN.mOא F אM KmOאF אאאd אא،אאאאא א Kאאאא O - + d F ١ - ١٤ - אא ١٠٣ אאא W١ K٢אOאF= 400 Nאאא :א F = 400 N 2m O ٢ Mo = + 400 x 2 = + 800 N.m WOאא Kאאאאאאא W٢ K٣אאF= 250 kNOאא 10 m 250 kN O ٣ WOאא Mo = - 250 x 10 = - 2500 N.m Kאאאאאאא - ١٥ - אא ١٠٣ אאא W٣ K٤אOאאא 500 kN O ٤ Wא ،dא Oאא٤א WKOאא Mo = 500 x 0 = 0 kN.m ٢{٢ א ،אאא Kאא W٤ K٥אאאאA אא 50 kN 100 kN 100 kN 50 kN 1m A 1m 3m 3m ٥ - ١٦ - 50 kN 2m 2m אא ١٠٣ אאא Wא ،Aאא אאאAאא Kאאאאאא KAא MA = J 50x0 H100x2 J 100x3 H50x6 J 50x1 = = 0 H200 J 300 H300 J 50 = H150 kN.m אאאא50 kNאא אאAאאאאAא KA א Varignon’s Theorem ??٣{٢ OFE אF?? אאא،O אא Kאא W٥ K٦אOאF = 100 kNא 100 kN ٣٠٠ 10 m O 40 m ٦ - ١٧ - אא ١٠٣ אאא 30٠ 10 m O ٨٧ kN 40 m ٧ Wא ٣٠א אF = 100 kNא ٦א אא Oאא،אא KOא WE٧FFאאא Fx = Fcos(30) = 100xcos(30) = 87 kN Wאא Fy = Fsin(30) = 100xsin(30) = 50 kNWאא ZOאאא H87 x 10 = H870 kN.m אאאא WOאאא - 50x40 = -2000 kN.m Kאאאאא WאאאOאF אאא Mo = -2000 +870 = -1130 kN.m - ١٨ - אא ١٠٣ אאא W٦ KEF٨אAאF = 1000 Nא 3m 3m Fx = 1000Cos(50) 50 500 ٠ ١٠٠٠ N ١٠٠٠ N Fy = 1000Sin(50) 10 m 10 m A A EF E F ٨ Wא EF٨א Fy FxFא KAא Fx = Fcos(50) =1000xCos(50) = 642.78 N Fy = Fsin(50) = 1000xSin(50) = 766 N Zאא Zאא WAאאא Mx = +642.78x10 = +6427.8 N.m WAאאא My = +766x3 = +2298 N.m WאאאאAאFאאא MA = +6427.8 +2298= +8725.8 N.m Kאאאאא - ١٩ - אא ١٠٣ אאא Coupleאא٤{٢ א אאאאאא אאא،אא Moאאאאאאאא K٩א A F h F B ٩ אאאא אאא WK٩א h/2 א M1 = Fxh/2 M2 = Fxh/2 Zאא M1 + M2ZF(h/2 + h/2)ZFxh/2 + Fxh/2 ZFxh אאאאא אאא אאאK אאאא K١٠אאאא - ٢٠ - אא ١٠٣ אאא F + d M = Fd F ١٠ א،אא אאאdאFאא א K١١א F O h d F ١١ KhאE١١FאאאO WOאאאאא Mo = Fx(d + h) – Fxh = Fxd + Fxh –Fxh = Fxd אאאאא Kאאא - ٢١ - אא ١٠٣ אאא W١ K١٢אאאאאאאאא 50 kN 50 kN 10 m ١٢ Wא אאאאא ،١٢א אאאאא،אא W C = J 50 x 10 = -500 kN.m אאאאאא אאא אK אאאאא K١٣אאא500 kNאאא 500 kN.m ١٣ - ٢٢ - אא ١٠٣ אאא אא٥K٢ FBAFא אא Bא–F F אEEF١٤ KEF١٤אAאאFאאא אBאא–FאAאאFא אאאא אאא WEF١٤אאא C = -Fxd אא KEF١٤אאא F -F B B d A EF F d A EF ١٤ F B d F A C =F.d EF W٢ AאאאBאאF = 20 kNאאא KEF١٥א 1` - ٢٣ - אא ١٠٣ אאא B 20 kN B 20 kN B d A EF A 450 20 kN ٢0 m ٢0 m ٢0 m A 450 20 kN EF ١٥ C = 282.8 kN m 450 20 kN EF Wא A א אFאא אBאאKEF١٥א20 kNא אא אאAא WC C = +20xd kN.m Wאאאd 0 d = 20xSin(45 ) = 20x0.707 = 14.14 m Wאא C = +20x14.14 = +282.8 kN.m K אאאאאאא אא AאBאF א א،EF١٥א+282.8 kN.mאA א KEF١٥אEF١٥א - ٢٤ - אא ١٠٣ אאא ٦{٢ F = 50 kNאאOאא K١٦א E١ ٥٠ kN 30٠ O 5m ١٦ K١٧אAאFאE٢ ١٠ ٤٥٠ F = 100 kN ٢٥ A ١٧ Oאא Fא K١٨א E٣ - ٢٥ - אא ١٠٣ אאא 50 kN 20 kN 5m 5m O 15 m F ١٨ K١٩אAאאאאא E٤ 20 kN A ٥m ٥m ٥m ٥m B 10 kN 25 kN ١٩ K٢٠אאאאBאא E٥ 100 200 ٥٠ kN A ٣٠٠ ٥٠ kN B 20 m ٢٠ - ٢٦ - ١٠m אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא אאאא אאאא ٣ אא אאאא ١٠٣ Wאא free אא،אאאEאFאא Kאאbody digram Wאא Wאא אאאאאאא • אאאא • אאאאאא • K٪100אאאWאאא ١١Wאא Wאא • • • Wאא אאאאאא Kאא - ٢٧ - אא אאאא ١٠٣ ١{٣ אאK אאאE אFאא אא אאאאא אאא K אא Kאאאאאא Roller Support אא٢{٣ אאא אאא אאא אאK١ אא אאאא אKאאא (Ax = 0Fאאא EAאאאFA אא K١א y x A Ay A A Ay ١ Ay Hinge Support אא٣{٣ אא אא אאא אKEF٢אAא KEF٢אאא - ٢٨ - אא אאאא ١٠٣ Ax A A Ay EF E F ٢ Fixed or Cantiliver Support אאא٤{٣ A אאאאאאאא W אKEF3א KEF٣אAאא MA Ax A Ay ()أ ٣ ()ب Equilibrium Conditionsאא٥{٣ אאא אאKאאא אאאK אאא א אאאאאא אא אKאאא אאאאאאא Kא - ٢٩ - אא אאאא ١٠٣ אאא٦{٣ ،4אא F2y F1 F2 y F2x F4 F٣ R1 x R2 א –٤ yאxאאא Wאאאא WאאאxאאאE١ ∑F x =0 F4 – F2x – F3 =0 WאאאyאאאE2 ∑ Fy = 0 R1 + R2 – F1 –F2y = 0 אE אאאFאE3 ∑M = 0 Wאא Kא??אאאאא - ٣٠ - אא אאאא ١٠٣ Wאאא٧{٣ Colinear Forces אאאE Kא٥אאאא א אא א אא F2אF1אKאאאא KF1אאא אאאאא Wאאא ∑F x =0 F2 -F1= 0 F1 = F2 F2 F1 א W٥ Concurrent Forces אאאE K אא אאא אא אאK٦א אOאאאאW אא Kyxא - ٣١ - אא אאאא ١٠٣ F2 F3 O F1 W٦ Parallel Forces אאאE אאאאא Wאא،K٧אא ∑F x =0 ∑M = 0 K ∑Fy =0אא א : ٧ - ٣٢ - אא אאאא ١٠٣ Opposite Couples אאE א אאאאאא K٨אאאא +M -M א אW ٨ Free-Body Diagramאא٨{٣ אאאאאאא א אאאאאאאא אאאאאFree-Body Diagramאא KEאFאא Wאא١ KEF٩אאABאאא Wא אאאאAB אא hinge AאאKאאאא Bא،AyAxsupport אFByroller support KEF٩אABאא،E - ٣٣ - אא אאאא ١٠٣ BאA אאאאא Wאאא ∑F x Wאא = 0 WאאE Ax = 0 WAאאE ∑MA = 0 +100x2 +100x4 –Byx6 = 0 +200 +400 – 6By = 0 +6By = 600 By = + 600/6 = + 100 kN אאאא+100 kN BאByאאאאא Kyxאא אאAאE٩FA אא KAy א W،אאE ∑ Fy =0 Ay + By -100 -100 =0 Ay = 200 -By WAyאאEFאBy = +100 kN Ay = 200 – 100 = +100 kN 100 kN 100 kN D C A 2m 2m – EF٩ - ٣٤ - B 2m אא אאאא ١٠٣ 100 kN 100 kN y Ax A Ay x B D C By אאא א אWEF٩ W٢ AאאEF١٠אאאאא KאאBא 100 100 kN 450 B A 2m 2m 2m WEF ١٠ Wא אאאאAB אא AאK אאאאא Bא،AyAxhinge support Byroller support ABאא،EאF - ٣٥ - אא אאאא ١٠٣ ٤٥אאאאKEF١٠א W +70.7 kN = +100Cos(45) Wאא +70.7 kN = +100Sin(45) Wאא ∑F x Ax +70.7 = 0 Wאאא = 0 WxאאאEF Ax = - 70.7 kN א א א،Aאא Kxאא אא ∑ M A = 0 WAאאE Kאא +100x2 + 70.7x4 – Byx6 = 0 +200 + 282.8 = 6By By = +482.8/6 = + 80.47 kN KAאAאא ∑F =0W،אאE y Ay + By -100 -70.7 = 0 Ay + By = +170.7 Ay = +170.7 – By = +170.7 – 80.47 = +90.23 kN Ay = +90.23 kNWAאAyאאאאא Kאאא - ٣٦ - אא אאאא ١٠٣ 100 100 kN 50cos(45) Ax 450 B A Ay 2m 2m 2m y 50sin(45) By x א אWEF ١٠ W٣ 50 אאא،EF١١אאאאא KCאkN 10 m 2m A C 1000 5m B א WEF١١ Wא EF١١אEF١١אאאא ∑F WאאE١ x = 0 Wxאאא א Bx – Ax =0 - ٣٧ - Wא אא אאאא ١٠٣ אא אאא א Wy ∑F =0 y Ay – 50 -1000 = 0 Ay = + 1050 kN EאאאF ∑ M A = 0 WAאאאא +1000x10 + 50x2 – Bxx5 = 0 +10000 +100 -5Bx = 0 +10100 = 5Bx Bx = +10100/5 = +2020 kN E٢ WAxאאBx Bx – Ax = 0 Ax = Bx = +2020 kN אאאאAxאאAxאא KאxאAx א،אא 10 m 2m Ax A ١ C Ay 5m Bx 50 kN 1000 B א א: EF١١ - ٣٨ - אא אאאא ١٠٣ ٩{٣ EF١٢EF١٢אאאאא E١ Kאא 500 kN A C 5m A 5m 500 kN ٤٥٠ B 3m C B 3m 3m EF 3m EF ١٢ E،،،F١٣אאאאא Kאאאא E٢ 100 N 1٥0 N A B 6m 6m EF 6m - ٣٩ - אא אאאא ١٠٣ 1٢0 N 240 N 100 N A B ١٠ m ١٠ m ١٠ m EF 240 N M = +100 kN.m A ١٠ m ١٠ m EF 300 kN 250 kN 600 A B ٥m 5m ٥m EF ١٣ B ١٠ m - ٤٠ - אא אאאא ١٠٣ אאK١٤١א אאאאE٣ Kאא 800 kN 3m 3m 500 kN A B 2m ٢m 1m ١٤ - ٤١ - אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא אאא אאא ٤ אאא ١٠٣ אאא Wאא אא،אאאאאאאא אאאאא אא Kאאא Wאא Wאא אאאאא • אאאא • אאאאאאאא • אאאא • K٪100אאאWאאא ١٦Wאא Wאא • • • Wאא אאאאאא Kאא - ٤٣ - אאא ١٠٣ אאא א١{٤ EF אא KאאאEאאFאא אאא٢{٤ אKאEאFאאא W١אאאא EF١אSimple Beam • אKE٣F Kאא EF١אContinuous Beam • אאא Kא،אא EF١אCantiliver Beam • אKE٣F Kאא EF١אOverhanging Beam • אKE٣F Kאא EF١אBuilt-in Beamא • א٦א Kא،אאא End-supported Cantiliver Beam • EF١א א٤א Kא،אאא - ٤٤ - אאא ١٠٣ אאא Compound Beam • EF١EF١א hinge אאאאא،roller א אאא Kאא Simple BeamWEF١ Continuous Beam WEF١ Cantiliver BeamWEF١ - ٤٥ - אאא ١٠٣ אאא אWEF١ Built-in Beam WEF١ End-supported cantiliver Beam א WEF١ Compound Beam composed of cantiliver and simple beam connected by internal hinge WEF١ Compound Beam composed of cantiliver and simple beam connected by internal roller - ٤٦ - אאא ١٠٣ אאא אאאא٣{٤ Wא אE١ ،EF٢א K Wאא אE٢ אאאאאא אאאאאKEF٢אא Kאאא Wא E٣ אא KEF٢א Wא אE٤ א KEF٢א W E٥ KEF٢א KEאFWאאא Kא،א،אא،א،אאאאאאאא אאאאאאאאאאא אאKא،א،אא Kאאאאאא - ٤٧ - אאא ١٠٣ אאא F kN y x WEF w kN/m אWEF w2 kN/m w1 kN/m אWEF w (kN/m) WEF - ٤٨ - אאא ١٠٣ אאא M (kN.m) WEF אאאאW٢ אאא٤{٤ אKxyאאאאאא אאאKyxאאאאא Kא אאאא٥{٤ א אE١ אאאאאאא Wא ∑F =0 ∑Fy =0 ∑M = 0 Kאא x Kאא BAאא - ٤٩ - אאא ١٠٣ אאא Wא אE٢ אאאאאEאאFאאא אOאFאא Kxאא אאEEF٣אA אx KEF٣אOB אAOאFא WVxא Aאxאאאא Vx אOאVxא K٣א KEF٣אאאאאא KEF٣אאא WMx א Cאא CאMxא KOאאOB אאAOא Wא אE٣ Kאאא - ٥٠ - אאא ١٠٣ אאא y P x A w C B O C x WEF٣ P Vx Mx w Mx B A RA O Vx EF٣ +V RB +M א אWEF٣ - ٥١ - אאא ١٠٣ אאא W١ KEF٤אאCאאא 10 kN 5m A 5m B C 3m WEF4 10 kN 5m 5m A B C RB = 5 kN RA = 5 kN 10 kN MC A MC 5m 3m B C VC VC RB = 5 kN RA = 5 kN אאWEF4 - ٥٢ - אאא ١٠٣ אאא Wא אאאאCאא E١ WAC ∑F =0 y RA – Vc =0 Vc = RA = + 5 kN WACאCאאCאאE٢ ∑M = 0 RA.3 – Mc = 0 Mc = 5x3 = + 15 kN.m אאאא٦{٤ אאאE١ אאאאא ، א אאאE אאFאאא Kאאאאא W٢ KE١FEF٤אאאאאא K٥אRA = RB = +5kN Wא١א 10 kN 5m A x a C a b 5m B b x RB = 5 kN RA = 5 kN W٥ - ٥٣ - אאא ١٠٣ אאא אAאxCAאa-a אא K٦אאאאא٥א אא א אK אאא Wאא ∑F =0 y 5–V=0 V = 5 kN K+5kN٥xאא ∑M a = 0 Wa-aאא 5.x – M = 0 M = 5x אאא Kx אא WE٥m٠xFx ٥ ٤ ٣ ٢ ١ ٠ EmFx 25 20 15 10 ٥ ٠ Mא kN.m x M V 5 kN a-aאאW٦ - ٥٤ - אאא ١٠٣ אאא אAאxBCאb-b אא K7אאאאא ٥ א b-אאא אא אK אאא KE١٠٥xאFWb ∑F =0 y 5 – 10 -V = 0 V = -5 kN K-5 kN١٠٥xאא ∑M =0 b Wb-bאא 5.x – ١٠.(x – 5) -M = 0 M = 5x -10.x +50 = 50 – 5.x = 5(10 – x) 10 kN 5m M x 5 kN V b-bאאW7 אאא Kx אא WE10 m٥xFx - ٥٥ - אאא ١٠٣ אאא ١٠ ٩ ٨ ٧ ٦ ٥ EmFx ٠ ٥ ١٠ ١٥ ٢٠ ٢٥ Mא kN.m x אאאאאא 10 kN A 5m K٨א C 5 kN 5m B 5 kN +5 kN א + - Shear Force V - 5 kN א Bending Moment M + +25 kN.m אאW٨ - ٥٦ - אאא ١٠٣ אאא W٢ K٩אאאאאא Wא א E١ Wאאאאא P = 5x10 = 50 kN א،אאא W،EאאF אא RA = RB = 50/2 = +25 kN a-aאאא E٢ א9אאAאxa-a KאאאK10אאאא Wאאאאאא xאא a-aאx אאא P = 5.x WE١٠F KAאx/2א ∑F =0 y 25 – 5.x - V = 0 V = 25 – 5x xFxאא ∑M - ٥٧ - KE 10 m ٠ a = 0 Wa-aאא אאא ١٠٣ אאא 25 . x − 5 x . x − M = 0 2 x2 − M = 0 2 5 M = 25 x − x 2 2 25 x − 5 אא Kx אאא E١٠٠xFxאאא Wאאא 10 9 -25 0 8 7 EmFx 6 ٥ ٤ ٣ ٢ ١ ٠ -5 5 15 25 א 0 Mא -20 -15 -10 0 10 20 22.5 40 52.5 60 62.5 60 52.5 40 22.5 kN kN.m x אאאא א אא١١אא K١١א אK אx=5 mא אK אאא אאFאאא KEאאאאא - ٥٨ - אאא ١٠٣ אאא a 5 kN/m A B 10 m a x 25 kN 25 kN W9 5 kN/m M x V 25 kN 5.x kN x/2 x M V 25 kN a-aאאW10 - ٥٩ - אאא ١٠٣ אאא 5 kN/m A 5m C B 5m 25 kN 25 kN +25 kN א + - 25 kN א 0 0 ++ Mmax =+62.5 kN.m אאW١١ - ٦٠ - אאא ١٠٣ אאא אאאאאא ٧{٤ אאאאא אאאאK אאא אK אאאא אאא א אאאאאאאא אאK אאאאאא אאאאא א אאאאאא K אאא אאאאאDr. BeamPro K١٢א אאאאא אאאא Dr. BeamPro א :١٢ - ٦١ - אאא ١٠٣ אאא W٨{٤ ،١٦،١٥،١٤،١٣אאאאאאא K٢٣،٢٢،،٢١،٢٠،١٩،١٨،١٧ 10 kN/m A 3m ١٣ 20 kN 3m 5m 3m A 15 kN B D C ١٤ 10 kN 6m 5 kN 6m A 12 m ١٥ 10 kN/m A 3m C D 5m B 3m ١٦ 10 kN/m A 5m 5m ١٧ - ٦٢ - אאא ١٠٣ אאא ١٠ kN ٢0 kN/m A 5m 5m ١٨ 20 kN 20 kN 10 kN/m 2m A 2m 2m C 2m B D 5m ١٩ 20 kN 10 kN/m A C B 5m 3 m ٢٠ 5 kN/m A B 10 m ٢١ 10 kN 5 kN C B A 2m 2m D 2m ٢٢ - ٦٣ - 2m E אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא א א ٥ אא ١٠٣ א Wאא אא،אאאאאאאא אאאאאא Kאאא Wאא Wאא אאאאא • אאאא • EאאFאאא • אאאאאאאא • K٪100אאאWאאא ١١Wאא Wאא • • • • Wאא אאאאאאא Kאא - ٦٥ - אא ١٠٣ א א١{٥ אאE Fאאא אאאאאא אאאjointאאאאK אא K١אאא אאאK אאא אאK٢אאאאא א אאאאא אאא K אא אאאאא אאאאאK٥،٤، ٣א Kאאstatically determinate trusses אW١ - ٦٦ - אא ١٠٣ א אאא٢ ٣ PRATT ENGLISH HOWE אאאאW٤ - ٦٧ - אא ١٠٣ א PRATT WARREN אאאאWEF٤ PRATT BAILEY אאאאW٥ - ٦٨ - אא ١٠٣ א WARREN אאאאWEF٥ Determinacyאאאא٣{٥ אאEFא K١אאjointsא א،אא Wא אאHEאאאFאZא אEFbWאא אאr אjointsEאFאJ b + r Zאאא ∑F =0 ∑F =0Wאא y x Kb + rZא2JZאאא Kאאאאאא - ٦٩ - אא ١٠٣ א א אא unstableEFא b + r 〈 2J b + r = 2J statically determinate א EF F statically indeterminateא אאא b + r〉 2J Eאא KEאאFאאא b + r ≤ 2J Wא אאא EF٦אאאKrigid frameEF٦א א nonrigid frameEFא אEF٦אBא FאEFאא، KDאBאCאAEא B B C A C D A EF E F אאW٦ - ٧٠ - אא ١٠٣ א W١ K٧אאאא ٧ Wא b א٦Jאא٧א W3 rא٩ b + r = 9 + 3 = 12 2J = 2x6 = 12 b + r = 2JWאא KstableEFא W٢ K٨אאאא Wא J = 5 EאFא b = 6א r = 3א b+r=6+3=9 2J = 2x5 = 10 - ٧١ - W b + r 〈 2J Wאא אא ١٠٣ א KunstableEFא ٣ ٤ ٢ ٥ ١ ٨ אאJ ٤{٥ Section MethodאJoints MethodאWאא א ٩אא אאאאEFאאאא Kא Joints Method EאFאאE١ אאאאאאאא אאאאאאFא א אאא،E אאאאאא אאא אאFאאK אא א אKE אא KE٢אF ٢א Wאאאאא - ٧٢ - אא ١٠٣ א W١ אאE אאFאאא Kjoints methodאא٩א 3 kN E F 15 m C B A 20 m RA 20 m 9 kN D 20 m 12 kN RB ٩ Wא Kאאאא א ∑ M D = 0 WDאאE١ א RA.60 – 9x40 – 3x40 – 12x20 = 0 Rax60 = 720 RA = 720/60 = 12 kN ∑F =0Wאא y RA + RB – 9 – 3 -12 = 0 RA + RB = +24 RB = 24 - RA = 24 – 12 = 12 kN - ٧٣ - אא ١٠٣ א WאאאאE٢ EFA אE אAEאאאאאא K FAE ٥ A FAB ٣ ٤ RA = 12 kN FAEyFAExFAE א FAEy = 3/5 FAE 4/5 FAE ZFAEx WAאאא RA + FAEy = 0 12 + (3/5)xFAE = 0 FAE = - (5/3)x12 = - 20 kN ∑F =0 y אAEאאאאא K WAאאא FAB + FAEx = 0 FAB + (4/5)xFAE =0 FAB = - (4/5)xFAE = -(4/5)x (-20) = +16 kN - ٧٤ - ∑Fx=0 אא ١٠٣ א אABאאאאא K BאE FBE B FBC FBA = 16 kN ٩ kN FBC – 16 = 0 FBC = + 16 kN ∑ Fx = 0 Tension FBE J 9 = 0 FBE = + 9 kN Tension ∑F =0 y EאE 3 kN ٥ 3 FEF E 20 kN ٤ 9 kN FEC ∑F =0 y - ٧٥ - אא ١٠٣ א -9 -3 – (3/5)xFEC + (3/5)x20 = 0 (3/5)xFEC = 12 -12 = 0 kN FEC = 0 ∑Fx=0 FEF + (4/5) FEC + (4/5)x20 = 0 = -(4/5)x20 – (4/5)xFEC = - 16 kN Compression FEF CאE FCF FCE = 0 ٥ C FBC FCD FCD ٤ 12 kN 3 FCD - FBC = 0 = FBC = + 16 kN Tension FCF - 12 = 0 FCF = +12 Tension ∑Fx=0 ∑F =0 y FאE F FEF 12 kN - ٧٦ - FFD אא ١٠٣ א FFD ∑Fx=0 (4/5) xFFD + FEF = 0 (4/5)xFFD = -FEF = - 16 kN = - (5/4)x16 = - 20 kN Compression אאאא١٠א K–אHא 3 kN E 12 kN 0 kN B + 16 kN + 16 kN 9 kN H F + 12 kN A - 16 kN + 9 kN - 20 kN Wא J - 20 kN C ١٠ D + 16 kN 12 kN 12 kN Section Method אאE٢ אאאאאא אאאאKא א א אK Kא אאאא א א אאאא K٣אאא Wאאאאא - ٧٧ - אא ١٠٣ א W٢ אאCEEF BCאE אאFאאא Ksection methodאא٩א Wא אאא אאאא א CEEFBCאa-a١١אK K١١א 3 kN a E F A RA C B 9 kN a ١١ 12 kN D RB אאa-a א KCEEFBCאאאאאא،١٢א א אאאאא KEאאאFא - ٧٨ - אא ١٠٣ א 3 kN FEF E FEF F 15 m FEC FEC A FBC B C 20 m D FBC 12 kN 9 kN 12 kN 12 kN 12 Cאאa-aאאאא WFEFאאא WCאא E١ ∑MC =0 -FEF x 15 -12x20 =0 FEF = -240/15 = - 16 kN Compression ﻹﻳﺠﺎدEאאa-aאאא אE٢ ∑ME =0 W FBC ﻗﻴﻤﺔ اﻟﻘﻮة +12x20 – FBCx15 = 0 FBC = +240/15 = + 16 kN Tension אאa-aאאא אE٣ ∑Fy=0 W FEC اﻟﻘﻮة 12 - 9 -3 - (3/5) FEC = 0 12 - 12 - (3/5) FEC= 0 FEC= 0 אECאא K - ٧٩ - אא ١٠٣ א ٥{٥ אאא،١٣אאאE١ Kjointsא A B C 5m D E 5m F 5m G 5m 10 kN 13 K١٤אאאאE٢ A B 14 אא،١٥אאא E٣ Kאאjointsאא - ٨٠ - אא ١٠٣ א C 20 kN 5m A 10 m B 10 m 20 kN ١٥ אא،١٦אאא E٤ Kאאjointsאא E F 10 m A 10 m 10 m B D C 10 m 10 Kn 20 Kn ١٦ אא،١٧אאא E٥ Kאאjointsאא - ٨١ - אא ١٠٣ א 15 kN 25 kN A 45 45 45 5m 5m 45 B 10 kN ١٧ Kjointsאא١٨אאאאE٦ 150 Kn E 2m D 4m 150 Kn C B A 120 Kn 4m 8m ١٨ אאcd, Cd, CD, BC, cCאא E٧ אא section methodאא١٩א K - ٨٢ - אא ١٠٣ א d c b g e 40 m A C B E D H G 30 Kn 40 Kn 40 Kn ٢٠אאא CH, CI, BJאא E٨ Kאאsection methodאא 10 kN 10 kN 10 kN I 10 m 10 kN H 10 kN G A B C D 5@5 = 25 m 20 - ٨٣ - ١٩ J 6x30 = 180 m K E F אא ١٠٣ א K٢١אאא AD DBאאE٩ A 6m B 50 kN 5m D 50 kN C E 5m F G 2 3 3 3 ٢١ אאא٢٢אאאאE١٠ K 5 kN 20 kN 5 kN C B A 1.6 m D 3m 3m E 12 kN ٢٢ - ٨٤ - F אאא ã¹]<gè…‚jÖ]æ<ËÖ]<Üé×Ãj×Ö<íÚ^ÃÖ]<퉉ö¹] אאאא אא א א ٦ C,B,A ١٠٣ Wאא אאאאאאאאאאא אאאא אאאKא Kאאאא Wאא Wאא אאאאאאא • אאאאא • אאאא • אא • K٪100אאאWאאא ٨Wאא Wאא • • אאאא • Wאא אאאאא אאא Kאאא - ٨٥ - C,B,A ١٠٣ אאאאא١{٦ אאאאא א א אאאא אאאKא FאאאKאא אא FאK١א F אא א אFאא N/mm2 N/m2σא א Wא F σ = A W kN/mm2, N/m2, N/mm2 ﺑـאאא :σ kg, kN , N EאFאאאאWF m2 , cm2, mm2אWA a F F a F σ Section a-aW١ - ٨٦ - C,B,A ١٠٣ W١ אאא،P = 30 kNא K3cmx2cmאא Wא WAא A = 3x 2 = 6 cm2= 6x10 m2 -4 W30 kN אאאא σ= 3 P 30 x10 = = 5 x10 7 N = 50 N − 4 m2 mm2 A 6 x10 אאאאא٢{٦ אFL0א Fא،٢א∆L EאFא WאεאאאEא ε= ∆L L0 W m , cm, mmאא∆L mm, cm, m אאL0 KEFאε - ٨٧ - C,B,A ١٠٣ F F ∆L L0 אW٢ W٢ 30 L0 אא،K٣אאא אא70 kNא F10 cm cm KאאאאK0.02 cm Wא 70 kN 10 cm 70 kN 30 cm ٣ A= π D2 = π AאE (0.1) 2 = 7.85 x10−3 m 2 4 4 70 x103 F σ= = = 8.92 x106 N 2 = 8.92 MN 2 −3 m m A 7.85 x10 אσאאE - ٨٨ - C,B,A ١٠٣ 70 x103 F σ= = = 8.92 x106 N 2 = 8.92 MN 2 −3 m m A 7.85 x10 אεאE ∆L 0 . 02 x 10 − 2 = = 0 . 67 x 10 ε = L0 30 x 10 − 2 −3 אאא٣{٦ Kאאאאאאא high-strength אאאא א ٤א אאאא אKsteel א אא200 N/mm2 א٠ אא אאא אHooke אאEElastic modulus (Young’s modulus)א KאאאEא١אא (Ν/mm2) 300 σ 200 ﻣﻨﻄﻘﺔ اﻟﻠﺪوﻧﺔ ﻧﻘﻄﺔ اﻻﻧﻜﺴﺎر 250 150 100 50 0 ﻣﻨﻄﻘﺔ اﻟﻤﺮوﻧﺔ Stress اﻹﺟﻬﺎد 350 0 .0 5 S tra in ε اﻻﻧﻔﻌﺎل 0 .1 אאאאW4 - ٨٩ - C,B,A ١٠٣ E= WEא σ ε σ = E.ε kN/m2 , N/mm2 Wאא E Fא אאא ٤א Kאאאאא KאאאאאEא Wא אאאאWF אWA אאWL0 : F אאא א: ∆L E= σ ε Wεσ F σ A = F .L 0 E = = ∆L ε ∆ L.A L0 - ٩٠ - C,B,A ١٠٣ L0אאAאEאF אא W ∆L = F .L0 E. A אאאאא٤אא אאא א K٤אא - ٩١ - C,B,A ١٠٣ אאאאW١ EGpaFEא א ٢١٠–١٩٠ א ٧٩J ٧٠ א ١١٠J ٩٦ א ١٤–٠{٧ א ٣١–١٧ EאFא ١٣J ١١ א W٣ Fy yield loadאא2 cmאא א Fmax = 150 kNא= 80 kN K٥אFb = 70 kNא Fmax Load (κΝ) 150 80 Fy 70 Fb 0 Elongation אאW٥ - ٩٢ - C,B,A ١٠٣ Wאא אא E١ אא אE٢ אאאאא E٣ K1 cm Wא אאE١ A0 = A0אא π 4 (0.02) 2 = 0.314 x10 −3 m 2 Wאא σy = Fy A0 = 80 x103 = 254 x10 6 N m2 0.314 x10 −3 WאאE٢ σ max = Fmax 150 x103 = = 477 x10 6 N m2 0.314 x10 −3 A0 אאE٣ A= π 4 אא (0.01) 2 = 0.0785 x10 −3 m 2 Wאא σb = Fb 70 x103 = = 892 x10 6 N m2 A 0.0785 x10 −3 - ٩٣ - C,B,A ١٠٣ אאא٤{٦ א E Fאאאא ،٧٦אshear stressא Wאא τ= V A W N/m2, kg/cm2, N/mm2 אאאאτ kG, kN, N אאאV m2, cm2, mm2אA WE٧Fγאא ∆L γ = L W m, cm, mmאאאאW∆L m, cm, mm אאL Kאאγ K אאאאא Wאא τ = Gγ - ٩٤ - C,B,A ١٠٣ W ،אkN/m2, N/mm2 אG אK אאאGא 2א W E EאFאאG E G = 2(1 + υ ) W Poisson’s ratio ?א?ν אאאאW2 EGpaFGא א 80–75 א 30J 26 א 41J 36 א 10 kN 10 kN 10 kN אW٦ - ٩٥ - 10 kN C,B,A ١٠٣ א אאW٧ א٥{٦ 8א Wא،אאאאא σ= M y I W kG.m , N.m , kN.m אאאאאWM mm4 , cm4, m4 moment of inertia אWI Kאאאאneutal axisאאאWy - ٩٦ - C,B,A ١٠٣ KEאFאאאאאA א٣א אאאאאW٣ א א bh3 Ix = 12 hb3 Iy = 12 y h x b y Ix = Iy = πR 4 4 = 4 πD 64 x R - ٩٧ - D C,B,A ١٠٣ W K٩אאאאא 10 kN A B ١٠ cm ١٠ m ٥ cm א ٩ Wא RA = RB = +٥kNZאא M = 5 x5 = 25kN .m W b.h3 5 x103 I= = = 0.42 x103 cm 4 12 12 Wא Wאאאא 10 y = h = = 5cm 2 2 - ٩٨ - Wא C,B,A ١٠٣ 25 x10 3 x10 N N M σ = x 5 2976 29 . 76 y= = = I 0 .42 x10 3 cm 2 mm 2 ٦{٦ ،F = 20 kNא 2.5 cmאא KאאE = 70 GN/m2א 6 cm 40 cmאא E١ E٢ א אא،1 MN/m2אא אKאאאאאא E = 200 GN/m2. 6 cm P 40 cm 6 mm19 mm אאאא،א K116 kN Kאאאאאא - ٩٩ - E٣ C,B,A ١٠٣ P = 116 kN 19 m m 6 mm ABC BCא50 mmABא،Pא E٤ א40 MpaABאאאאK38 mm .BC ﻓﻲ اﻟﻤﻨﻄﻘﺔσא A C B P 13.3 א12 mm1mא E٥ אאאאא،אkN Kאאאאאא،0.5 mmא 12 mm 13.3 kN 1m - ١٠٠ - 13.3 kN C,B,A ١٠٣ q = 6.4 kN/m 3.5 m E٦ א א אאאK א (b = 140 mm, h = 240 mmFא Kא q kN/m 240 mm 3.5 m 140 mm אq = 60 kN/m 2 m Kא(b = 50 mm, h = 150 mmFא E٧ Kאאא אE١ c = 500 mmאאBאאE٢ Kאאd = 25 mm אאא q kN/m B 150 mm d 50 mm c L=2m - ١٠١ - C,B,A ١٠٣ P א א E٨ אאאKא= 16 kN Kא(b = 100 mm, h = 250 mmF Wאאאאאא אאאא E١ Kאאא E٢ אאא E٣ 5.5 kN/m3אאא Kאאא P P B A C D 250 mm 100 mm 2.5 m 0.5 m 0.5 m - ١٠٢ - C,B,A ١٠٣ אאאJ א Area Moments of Inertia - ١٠٣ - C,B,A ١٠٣ אאאאJ B Areas and Volume אאא - ١٠٤ - C,B,A ١٠٣ 1. Right Angles B c A b a C a = cos B c b cos A = = sin B c a tan A = = cot B b c sec A = = cosec B b c cosec A = = sec B a b cot A = = tan B a sin A = 2. Derived Relationships a = c sin A = c cos B = b tan A = b cot B = c 2 - b 2 b = c cos A = c sin B = a cot A = a tan B = c 2 - a 2 c= a a b b = = = = sin A cos B sin B cos A - ١٠٥ - a 2 + b2 C,B,A ١٠٣ 3. Oblique Triangles B c a A C b a- Sine Law a b c = = sin A sin B sin C b- Cosine Law a 2 = b 2 + c 2 - 2 bc cos A b 2 = a 2 + c 2 - 2 ac cos B c 2 = a 2 + b 2 - 2 ab cos C 4. General Trigonometric Formulas sin A = 2 sin 1 2 1 A cos A = 2 1 - cos 2 A = tan A cos A 1 1 2 2 cos A = 2 cos 2 A - 1 = 1 - sin 2 A 1 1 2 2 cos A = cos 2 A - sin 2 A = 1 - sin 2 A tan A = sin 2A sin A = = cos A 1 + cos 2A a- Addition and Subtraction Identities - ١٠٦ - sec 2 A - 1 C,B,A ١٠٣ sin (A ± B) = sin A . cos B ± sin B . cos A cos (A ± B) = cos A . cos B ± sin A . sin B tan (A ± B) = tan A ± tan B 1 ± tan A . tan B 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 sin A + sin B = 2 sin ( A + B) . sin (A - B) sin A − sin B = 2 cos ( A + B) . sin (A - B) cos A + cos B = 2 cos ( A + B) . cos (A - B) cos A − cos B = 2 sin ( A + B) . sin (A - B) b- double-Angle Identities sin 2A = 2 sin A . cos A cos 2A = cos 2 A - sin 2 A = 1 - 2 sin 2 A = 2 cos 2 A - 1 tan 2A = 2 tan A 1 − tan 2 A c- Half-Angle Identities sin A = 2 1 - cos A 2 cos A = 2 1 + cos A 2 tan A sin A = 2 1 + cos A - ١٠٧ - אא ١٠٣ K١٩٩٠،אאאא،אW١אא،אאKK١ א–אאאK٢ א،،אא KE٠٠٤٠J ٤J ١٢F١٩٩٩،א א،،אאK٣ KE٠٠٤٠J ٤J ١٢F١٩٩٩،א . Engineering Mechanics – Statics, 2nd Edition.٤ By Anthony Bedford and Wallace Fowler, Addison-Wesley, 1999. Chapman & Hall, New York, 1986.. Jafar Vossoughi, Statics for Architects,٥ for Engineers,. Ferdinand P. Beer and E. Russell Johnston, Vector Mechanics٦ McGraw-Hill Book Company, Singapore, 1990. - ١٠٨ - א ١٠٣ א ١ ٢ ٢ ٢ ٢ ٤ ٥ ٨ ٩ ١٠ ١١ ١٣ ١٤ ١٦ ١٧ ٢٠ ٢٣ ٢٥ ٢٧ ٢٨ ٢٨ KKKKKKKאאאWאא Vectorא ١{١ אא٢{١ ٣{١ א٤{١ א٥{١ א٦{١ Graphical אאא٧{١ Method Particleאא٨{١ אאאא٩{١ ١٠{١ ………………….………………….. אאאWאא א א١{٢ ٢{٢ Varignon’s Theorem ??٣{٢ Coupleאא٤{٢ אא٥K٢ ٦{٢ .………………….. 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