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<ul><li> 1. TRUSSES ANALYSIS ! ! ! ! !!!Fundamentals of the Stiffness Method Member Local Stiffness Matrix Displacement and Force Transformation Matrices Member Global Stiffness Matrix Application of the Stiffness Method for Truss Analysis Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors Space-Truss Analysis1</li></ul><p> 2. 2-Dimension Trusses2 3. Fundamentals of the Stiffness Method y x 4 32(x2, y2)1123(x3, y3)1 (x1, y1) Node and Member Identification36 528 47 (x4, y4) Global and Member Coordinates Degrees of Freedom Known degrees of freedom D3, D4, D5, D6, D7 and D8 Unknown degrees of freedom D1 and D23 4. Member Local Stiffness Matrix x yqjjqiq 'i =iq' j = x y d i=1AE/Lx di i dAE/L d j =yAE/LAE AE d 'i d'j L L1AE/LAE AE d 'i + d'j L Lq 'i AE 1 1 d 'i q ' = 1 1 d ' L j j [q] = [k][d] ----------(1)x x dj j dAE 1 1 [k ' ] = L 1 1 4 5. Displacement and Force Transformation Matricesyxyymxj(xj,yj) xi(xi,yi) x = cos x = y = cos y =x j xi L y j yi L= =x j xi ( x j xi ) 2 + ( y j yi ) 2 y j yi ( x j xi ) 2 + ( y j yi ) 25 6. Displacement Transformation Matrices y djyx djy jmdiy idiiLocaly m xjdjx xdixGlobalxyd 'i = d ix cos x + d iy cos y d ' j = d jx cos x + d jy cos y d 'i x d ' = 0 j y00x[d] = [T][d] d ix 0 d iy d y jx d jy ----------(2) x [T ] = 0y00x0 y 6 7. Force Transformation Matrices qjyy qiyymxiqixjyqjxyqiy = q 'i cos y q jx = q ' j cos x q jy = q ' j cos yjx qixiLocalx ymxGlobal qix = q 'i cos xx qjy q ix x q iy = y q jx 0 q jy 0 0 0 q 'i x q' j y [q] = [T]T[q]where x y T [T ] = 0 0 0 0 x y ----------(3)7 8. Member Global Stiffness Matrix ----------(3)[q] = [T]T[q]Substitute ( [q] = [k][d] + [qF] ) into Eq. 3, yields the result, [ q ] = [ T ]T ([k][d] + [qF] ) = [ T ]T [ k ][T][d] + [ T ]T [qF] = [k][d] + [qF] [ k ] = [ T ]T[ k ][T][k][qF] = [ T ]T [qF][ k] =x y 0 00 0x y1-1xy00-1AE L100xyUVUVxxxyxxxyAE V yx L U x xyyyxyyxyxxxyV yxyyyxyyU [k] =8 9. Application of the Stiffness Method for Truss Analysis Equilibrium Equation: [Qa] = [K][D] + [QF] Partitioned Form: Joint Load Qk=QuUnknown Displacement K11K12DuK21K22DkReaction+QFk QFuBoundary Condition[Qk] = [K11][Du] + [K12][Dk] + [QF] [Du] = [Ku]-1(([Qk] - [QF]) -[K12][Dk])9 10. Member Forcesx qjy yymjx qixiq 'i AE 1 1 d 'i q ' F i q ' = 1 1 d ' + F L j q ' j j x 0 q 'i AE 1 1 x q ' = 1 1 0 L jy00xy00x d ix 0 d iy d y jx d jy Dix 0 Diy q 'F i D + F y jx q' j D jy 10 11. q 'i AE x q ' = L x jy y xx Dix y Diy q ' F i D + F y jx q' j D jy Dxiqj = AE Lx yx qjy yymx qixyDyi Dxj+qj FDyjjxiMember Forces11 12. Member Forces Dxi qm = AE Lx yxy x qmy yymxDyi Dxj+qjFDyjjxiMember Forces12 13. Example 1 For the truss shown, use the stiffness method to: (a) Determine the deflections of the loaded joint. (b) Determine the end forces of each member and reactions at supports. Assume EA to be the same for each member. 5m3m50 kN 3m5m5m80 kN 4m4m13 14. 65m3m(-4,3)50 kN 3m535m5m180 kN 4m22(-4,-3) 4m4 31(0,0)1 38 42 = ( x j xi )i + ( y j yi ) j ij L LMemberx7 (4,-3)yUi Ui [ k ]m =Vicosy = y UjVjxx xy xx xy#1-4/5 = -0.8 -3/5 = -0.6#2cosx = x-4/5 = -0.83/5 = 0.6#34/5 = 0.8-3/5 = -0.6AE Vi yx yy yx yy L U j x x x y x x x y Vj yx yyyx yy14 15. 6Member x53221230.640.480.36- 0.8 0.60.64-0.480.360.80.64-0.480.3610.64 0.48 -0.64 -0.48AE 2 5 30.48 0.36 -0.48 -0.36 -0.64 -0.48 0.64 0.484-0.48 -0.36 1 11 [ k ]2 =AE 2 -0.48 5 5 -0.64 60.48 0.36 7650.36 0.48 -0.36 0.48 0.64 -0.480.48 -0.36 -0.480.368 1 10.36 0.48 -0.36 0.48 0.64 -0.480.48 -0.36 -0.4820.64 -0.48 -0.64 0.480.64 -0.48 -0.64 0.48AE 2 -0.48 [ k ]3 = 5 7 -0.64 82-0.641 [ k ]1 =-0.8 -0.6741y2#383 2x y#2 34x2#1 11y0.36[K] =AE 5 221.92-0.48-0.481.0815 16. 5m3mGlobal5 250 kN 3m65m5m113480 kN 4m2734m 1 1Q1 = -50 Q2 = -808=AE 5 2D1 D221.92-0.48D1-0.481.08D2+0102-250.65/AE=-481.77/AE16 17. 6Local532211[ q ' F ]m =14 33[AE x L y8 42xy#1-0.8-0.6#2-0.8 0.8-0.6[ ]]7 [qF]1 = AE 50.80.6-0.8-0.60.6#3x Dxi D yi x + q'F Dxj D yj MemberD4= 0.0= -97.9 kN (C)[qF]2 = AE 50.8-0.6-0.80.617.7 kN 50 kN36.87o97.9 kN 80 kNAE 5-0.8 +0.6= -17.7 kN (C)D1= -250.65/AE D2= -481.77/AE D5= 0.0 D6= 0.0= +17.7 kN (T)17.7 kN [q ] = F 3D1= -250.65/AE D2= -481.77/AE D3= 0.0+0.8 -0.6D1= -250.65/AE D2= -481.77/AE D7= 0.0 D8= 0.017 18. 6 5322117.7 kN14 350 kN1 336.89o8 42Memberxy#1-0.8-0.6#2-0.80.6#30.8-0.67 Check :17.7 kN97.9 kN 80 kN+ F = 0: 17.7 + 17.7 +50cos 36.89 x - 97.9cos73.78 - 80cos53.11 = 0, O.K 17.7(0.6)=10.62 kN 17.7(0.8)=14.16 kN 50 kN 97.9(0.6)=58.74 kN 80 kN 97.9(0.8)=78.32 kN17.7(0.6)=10.62 kN 17.7(0.8)=14.16 kN18 19. Example 2 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the deflections of the loaded joint. The support B settles downward 2.5 mm. Temperature in member BD increase 20 oC. Take = 12x10-6 /oC, AE = 8(103) kN. 8 kN D4 kNA oC3m0 +2B B = 2.5 mmC 4m19 20. 8 kN4D324 kN11(-4,0)A2oC3m0 +234mB = 2.5 mm1 (0,0) 36 CB2854(-4,-3)7 (0,-3)L+cosy = yUi [ k ]m =ViUjy-4/4 = -10#2Lx#1( y j yi ) jcosx = x UiMember-4/5 = -0.8-3/5 = -0.6#3 ij = ( x j xi )i0-3/3 = -1Vjxx xy xx xyAE Vi yx yy yx yy L U j x x x y x x x y Vj yx yyyx yy20 21. 4 32211(-4,0) 24(-4,-3)230.250-0.25-100.25000.1280.0960.072000.33320000- 0.8 -0.6 003 -0.2500.2500127[k]2 = 8x103562 0.096 0.072 -0.096 -0.072 5 -0.128 -0.096 0.128 0.096 6 -0.096 -0.072 0.096 0.0728100200.3337000080 -0.33300.333021 0.128 0.096 -0.128 -0.09600-1 104[k]3 =y2/L418x103x y/L#37 (0,-3)1[k]1 = 8x103x2/L#285y#136 3Member x1 (0,0)100 -0.333[K] =8x103210.378 0.09620.096 0.40521 22. Member 2: [q] = [k]m[d] + [qF] 4 2 1 3 1 2 1 (0,0) (-4,0) 23oC 20 +6 (-4,-3)1q5q1 q2 q1 q27 1.92 k =(T1)AE -6 3 (0,-3) = (12x10 )(20)(8x10 )25= 8x103 = 8x1031.152 kN60.128 0.096 -0.128 -0.09620.096 0.072 -0.096 -0.072d25-0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 1 2-1.536 1d16q6+20oC1.536 kN1= 8x1031.536 kN2+20oC34B = 2.5 mm q2285q11.152 kN1.92 kN10.128 0.096d120.096 0.072d210.128 0.096d120.096 0.072d2+ 8x103d5 = 0+d6 = -2.5x10-3 5 6 -0.128 -0.0960-1.152 2 1.536 5 1.152 6 5-3 -0.096 -0.072 -2.5x10 6+1.92 1.44++-1.536 1 -1.152 2 -1.536 1 -1.152222 23. 8 kN D4 4 kN32A1(-4,0) 2oC3m0 +2B = 2.5 mm134m1 (0,0) 36 CB2854(-4,-3)7 (0,-3)[Q] = [K][D] + [QF] Global:1Q1 = -4 Q2 = -8 D1 D2==8x103210.378 0.096D120.096 0.405D2-0.8514x10-31.44+-1.536 -1.152m-2.356x10-3+1.92m23 24. 4 3212112[ q ' F ]m =36[AE x L yx Dxi D yi x + q'F Dxj D yj 853Local4xy#1-10#2- 0.8 0-1[qF]1 =8x1031.0 0.0 -1.0 0.04-0.6#31.92 kN= -1.70 kN (C)1.92 kND2= -2.356x10-3 D3= 0.0 D4= 0.0 D1= -0.8514x10-3[qF]2 =8x1030.80.6 -0.8 -0.65= -2.87 kN (C)2+20oC[ ]D1= -0.8514x10-37Member][qF]3 = 8x103 0.0 1.0 0.0 -1.0 3 = -6.28 kN (C)D2= -2.356x10-3 + -1.92 D5= 0.0 D6= -0.0025 D1= -0.8514x10-3 D2= -2.356x10-3 D7= 0.0 D8= 0.024 25. 4 3212126 538 47cosxcosy[q]m-10-1.70#23Member #11- 0.8-0.6-2.87#30-1-6.288 kN 8 kN 1.70 kN2.87 kN4 kN14 kN1.70 kN 232.87(0.8) = 2.30 kN 6.28 kN 2.87(0.6) = 1.72 kN6.28 kN25 26. Example 3 For the truss shown, use the stiffness method to: (a) Determine the end forces of each member and reactions at supports. (b) Determine the displacement of the loaded joint. Take AE = 8(103) kN. 8 kN 4 kNA AD=+ = - 4 mm3mmm 3DCB 4m4m26 27. 8 kN 4 kNA AD=+Dmm 31 = - 4 mm3m24 CB 4m = ( x j xi )i + ( y j yi ) j ij L L[ k ]m =3453548(0,-3)7 (4,-3)ViMemberxy#1-4/5 =-0.8-3/5 = -0.6cosy = ycosx = x Ui36(-4,-3)4mUi212(0,0) 1#20-3/3 = -1Vj#34/5 = 0.8-3/5 = -0.6#44/4 = 10#54/4 = 10Ujxx xy xx xyAE Vi yx yy yx yy L U j x x x y x x x y Vj yx yyyx yy27 28. 2 1 25m4 23 (-4,-3)3m14354m(0,-3) 4 m 2480-10.8-0.6x2/L xy/L y2/L 0.128 0.096 000.072 0.3330.128 -0.096 0.0727 (4,-3) 430.128 0.096 -0.128 -0.09620.096 0.072 -0.096 -0.0723-0.128 -0.096 0.128 0.0964-0.096 -0.072 0.096 0.072[k]3 = 8x10361251[k]2 = 8x103-0.6#31[k]1 = 8x103-0.8#251y#15 m36xMember(0,0) 100020 0.3330 -0.33350000 -0.33300.33327810.128 -0.096 -0.128 0.0962-0.096 0.072 0.096 -0.0727-0.128 0.096 0.128 -0.09680.096 -0.072 -0.096 0.072061028 29. 2 1 25m4 23 (-4,-3)3m14354mx2/L xy/L y2/L100.2500#55100.25004(0,-3) 4 m56750.250-0.250600007-0.2500.250800008 7 (4,-3) 634530.250-0.250400005-0.2500.2506[k]4= 8x103y#45 m36xMember(0,0) 10000 1Global Stiffness Matrix[k]5= 8x10325871[K] = 8x1032 5 729 30. Global Stiffness Matrix 121 20.096 0.072 -0.096 -0.0723-0.128 -0.096 0.128 0.096 5100020 0.3330 -0.3335000060 -0.33301[k]3 = 8x103[k]4= 8x1030.33320-0.250400005-0.2500.2500000[k]5= 8x10370.250-0.250600007-0.2500.2508000080.128 -0.096 -0.128 0.0967 -0.128 0.096 0.1286507852 -0.096 0.072 0.096 -0.072 80.2562156-0.096 -0.072 0.096 0.072 1[k]2 = 8x103430.128 0.096 -0.128 -0.0964[k]1 = 8x1036343[K] = 8x103 -0.0960.096 -0.072 -0.096 0.0721 1 2 520.2560.0570.0 -0.1280.0 0.477 0.0 0.096 0.0 0.0 0.50 -0.257 -0.128 0.096 -0.25 0.37830 31. 8 kN 4 kNA AD=+1 14 AD4.8 kN=+4mm 3AE/L = 10.67 kNAE/L = 4.8 kN 3.84 kN = -4 mm3.84 kN3558 74m2.88 kN 126 3CB 4m2D = - 4 mm3mmm 3Global Fixed end forces2Fixed End10.67 kN-3.841-2.88 + 10.67 = 7.79 2 5 0.0 7 0.02.88 kN31 32. 8 kN 4 kNA AD=+D1 = -4 mm3mmm 3244m36 3CBGlobal:215458 74m[Q] = [K][D] + [QF] 71 Q1 = 4 Q2 = -8 Q5 = 0150.2560.00.0 -0.128=8x1030.0 0.477 0.0 0.0 0.50 0.00.096 -0.25D2 D50.378D76.4426x10-3 -5.1902x10-3 2.6144x10-3 5.2288x10-3m m m m-3.84D17 -0.128 0.096 -0.25=Q7 = 0 D1 D2 D5 D72 52+7.79 0.0 0.032 33. Member forces2 Dxi D yi x + q'F Dxj D yj 214548576.4426x10-3D1 D2 D5 D7= AD 4.8 kN=+m m m m-5.1902x10-3 2.6144x10-3 5.2288x10-3 4.8 kN1mm 3 yix -0.8-0.6#20-110.67 kN]D1 [qF]1 = 8x103 0.8 0.6 -0.8 -0.6 5 = -1.54 kN (C)10.67 kN 2[ ]iyx36 3[Member[ q ' F ]m =AE x L#11D2 0+ -4.80 D1[qF]2 = 8x103 0.0 1.0 0.0 -1.0 3 = -3.17 kN (C)D2 D5+ 10.67033 34. 2 1[ q ' F ]m =21436 3D1 D2 D5 D754=7 m m m mMemberxy#3 #40.8 1-0.6 0#510 yx[ ]]856.4426x10-3 -5.1902x10-3 2.6144x10-3 5.2288x10-3[AE x L Dxi D yi x + q'F Dxj D yj D1 [qF]3 = 8x103 -0.8 0.6 0.8 -0.6 5 = -6.54 kN (C)D2 D7 0 0[qF]4 = 8x103 -1.0 0.0 1.0 0.0 4 = 5.23 kN (T)0 D5 0 D5[qF]5 = 8x103 -1.0 0.0 1.0 0.0 4 = 5.23 kN (T)0 D7 034 35. Memberxy[q]#1-0.8-0.6-1.54#220-1-3.17#3 #40.8 1-0.6 0-6.54 5.23#5105.231 21436 3544 kN1.54 kN 1.54 kN 5.23 kN8578 kN 4 kN3.17 kN 3.17 kN8 kN6.54 kN 4 kN6.54 kN 5.23 kN0.92 kN3.17 kN3.92 kN35 36. Special Trusses (Inclined roller supports)36 37. Transformation Matrices 2 1[ q* ] = [ T ]T[ q ]31464*528573* y y*ii 3*2 j14*jx = cos j 1jq1=q2ix iy0 0jx jy0 0jy = sin j x [ T ] = [[ix = cos iT ]T]T=qi qj [T]Tixiy0000jxjyiy = sin i x*qj 1qiq*3 q*4ij37 38. [ k ] = [ T ]T[ k ][T][ k ]m =ix iy 0 00 0jx jy UiUi [ k ]m =ixix1-1ixiy00-1AE L100jxjyViUjVjixiy ixjx ixjyAE Vi iyix iyiy iyjx iyjy L U jxjx jxjy j jx ix jx iy Vj jyix jyiyjyjxjyjy38 39. Example 5 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant.30 kN 3m45o 4m39 40. 2Member 1: 13m34*6 5i = 0, 5 ix = cos 0 = 1, iy = sin 0 = 0213* 45o4m4*6qi3* o 45 o o [q*] ij = -45 = 135 , ix = cos (-45o) = 0.707, iy = sin(- 45o) = -0.707 1qj1ij[q*] = [T*]T[q] + [T*]T[qF]q5 q6 q3* q4*=1 0 0 00 0 0.707 -0.707qi qj [T*]T40 41. 2Member 1: 13m34*6 15i = 0 ; 5 ix = cos 0 = 1, iy = sin 0 = 023* 45o4m[k*]1==qj1i0 0 0.707 -0.707 5[k*]13* o 45 o o [q*] ij = -45 = 135 , ix = cos (-45o) = 0.707, iy = sin(- 45o) = -0.707 1qiAE 1 1 [k *] = [T T ] 1 1 [T ] L 1 0 0 04*65 0.25 6 0 AE 3* -0.1768 4* 0.1768AE 41 -1-1 1j1000000.707-0.70763*4*0 0 0 0-0.1768 0 0.125 -0.1250.1768 0 -0.125 0.12541 42. 2Member 2: 21 3m34*6[k *] = [T T ]3* 4*45o4m AE 1 1 1 1 [T ] L [k*]2=0 -1 0 00 0 -0.707 -0.707AE 33* j 90o+45o j = -135o = 215o , =135o ix = cos (-135o) = -0.707, iy = sin(- 135o) = -0.7071 -1-1 10-1001[k*]2=1 2 AE 3* 4*i 25o o 1 i = -90 = 270 , 90o ix = cos(-90o) = 0, 2 iy = sin(-90o) = -121qi23*0 0.3333 -0.2357 -0.23570 -0.2357 0.1667 0.16670 -0.2357 0.1667 0.16670-0.707 -0.7074*0 0 0 00qj42 43. 22Member 3: 13m34*6[k ] = [T T ]63* 45o4m AE 1 1 1 1 [T ] L [k]3==qj336.87oi5 qi i = j = 36.87o ; ix = jx = cos (36.87o) = 0.8, iy = jy = sin(36.87o) = 0.60 0 0.8 0.6AE 55[k]30.8 0.6 0 0j3215136.87o6120.096 0.072 -0.096 -0.072-0.128 -0.096 0.128 0.096-0.096 -0.072 0.096 0.0725 0.128 6 0.096 AE 1 -0.128 2 -0.0961 -1-1 10.80.600000.80.643 44. 2Global Stiffness: 133m4*623*15545o4m[k*]1=5 0.25 6 0 AE 3* -0.1768 4* 0.17681[k*]2=1 2 AE 3* 4*23*0 0 0 00 0.3333 -0.2357 -0.23570 -0.2357 0.1667 0.16670 -0.2357 0.1667 0.1667=3*4*0 0 0 0-0.1768 0 0.125 -0.1250.1768 0 -0.125 0.1254*5[k]365 0.128 6 0.096 AE 1 -0.128 2 -0.0966120.096 0.072 -0.096 -0.072-0.128 -0.096 0.128 0.096-0.096 -0.072 0.096 0.0721[K]=23*1 0.128 0.096 0 2 0.096 0.4053 -0.2357 AE 3* 0 -0.2357 0.291744 45. 2Global :130 kN 3m3m34*6 1545o 4m23* 45o4m[Q] = [K][D] + [QF] 1Q1 = 30 Q2 = 0 Q3*= 0D1 D2 D3*==23*1 0.128 0.096 0 AE 2 0.096 0.4053 -0.2357 3* 0 -0.2357 0.29171 AED1 D2 D3*352.5 -157.5 -127.345 46. 2Member Forces : Dxi D yi x + q'F Dxj D yj 1 3m34*6 15[ q ' F ]m =2= y[ ]]x3*045o4mD1 D2 D3*[AE x L1 AE[qF]1 = AE -1 0 0.707 -0.707 4 = -22.50 kN, (C)352.5 -157.5 -127.3iy#110#20-1-0.707-0.707 -0.7070.80.60.8jy[qF]2 = AE 0 1 -0.707 3 = -22.50 kN, (C)0.707 -0.707#3jx0 D1ixMember0 D3*0.6D2 D3* 0 0[qF]3 = AE -0.8 -0.6 5 = 37.50 kN, (T)0.80.60 D1 D246 47. 2Reactions :130 kN 3m33m4*6 1545o 4m Member[q]1Member Force (kN) -22.50[q]23* 45o[q]3-22.504m237.5022.50 kN 37.50 kN 45o36.87o 22.50 kN7.50 kN45o31.82 kN 22.50 kN47 48. Example 6 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant. 30 kN3m45o 4m4m48 49. 2Member 1:3m8 136742i = 0o, 5 ix = cos 0o = 1, iy = sin 0o = 054* 3* 154m4*64mqio [q*] ij = -45 ix = cos (-45o) = 0.707, iy = sin(- 45o) = -0.707qj1i3* o 451j[q*] = [T*]T[q] + [T*]T[qF]q5 q6 q3* q4*=1 0 0 00 0 0.707 -0.707qi qj [T*]T49 50. 2Member 1:3m8 1236 14m4m[k*]1==qj1i0 0 0.707 -0.707 5[k*]13* o 45 o o [q*] ij = -45 = 315 , ix = cos (-45o) = 0.707, iy = sin(- 45o) = -0.707 1qiAE 1 1 [k *] = [T T ] 1 1 [T ] L 1 0 0 04*6i = 0o, 5 ix = cos 0o = 1, iy = sin 0o = 054* 3* 5745 0.25 6 0 AE 3* -0.1768 4* 0.1768AE 41 -1-1 1j1000000.707-0.70763*4*0 0 0 0-0.1768 0 0.125 -0.1250.1768 0 -0.125 0.12550 51. 2Member 2:3m8 123674o 1 i = -90 , 90o ix = cos(-90o) = 0, 2 iy = sin(-90o) = -154* 3* 4m[k *] = [T T ]4*4mAE 1 1 1 1 [T ] L [k*]2=0 -1 0 00 0 -0.707 -0.707AE 33* j 90o+45o j = -135o = 215o, qj o) = -0.707, =135o ix = cos (-135 iy = sin(- 135o) = -0.7071 -1-1 10-1001[k*]2=1 2 AE 3* 4*i 215qi223*0 0.3333 -0.2357 -0.23570 -0.2357 0.1667 0.16670 -0.2357 0.1667 0.16670-0.707 -0.7074*0 0 0 0051 52. 2Member 3:3m8 12362 7415364* 3* 4m[k *] = [T T ]AE 1 1 1 1 [T ] L i[k]3==0 0 0.8 0.6AE 55[k]3qj5 qi i = j = 36.87o ; ix = jx = cos...</p>