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Slope-Deflection

THEORY
Course

Civil Engineering (BSCE 01)

136 Documents
Students shared 136 documents in this course
Academic year: 2023/2024
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Klistel Mainopaz
Ateneo de Davao University
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SLOPF · DICTION IOUATIONS COMPLETE CASES OF PROBLEMS ( SAMPLE : BENDING : RIGHT LEFT "A & B & C & D

⑱

itrary loading i P B = 0 because there is no settlements

  • ⑦ Span AB Formula applicable twice , MAB & MBA On ! # ② Span BC formula applicable twice , MDC & McB D ③ Span CD Formula applicable only once , McD . · x - Ma : (210) + -B - 3 (0)) + FEMAB > In slope deflection assumes that clockwise rotation clockwise moments are positive . Min = (201 + 0 - 3 (0)) + FEMBA CONJUC ATE BEAM & LOADING DUE TO INTERNAL MOMENTS AT N & F Fi MBc = -[201 + + - x(0)) + FEMBC / M - Mc = (20 + &B - 302] + FEMCB
  • It is downward " McD = (0-0) + FEMcD - EMe because the shear " rce rotation I I * It is upward because OF EN is clockwise ON 13/3 the shear forcecw on right side On LEFT /IDE I M is positive . RESTRAINED BEAMS & FEM's
  • the rotation of bending > if the beam is continuous. IMAGINE EACH SPAN ARE FIXED REGARDLESS moment is clockwise positive because it is downward at THE SUPPORT IS FIXED OR NOT ARE CONSIDERED FIXED. right side . -Assume the bending rotation · THE moments produced or contributed by the loads are called IEMs · DUE TO CONCENTRATED POINT SLOPE DEFLECTION 18 : When there are moments in Near end and Far end then - MN = (20N + 0+- 3(1)) : can be used For My P apply it twice - * EMN FEMF - it is concave down it means it is negative I C / b
  • moments at the ends · my = (N + 20 - 3(4)] L We only expect positive moments here because it is concave up - ADDING THE FEMS ① Apply area moment
  • The tangent of N and I are horizontal MN (20N + 0+ - 3)) , Note : R B = ① settlement my = (N + 20 - 3(4)] -E NOTE: Y the sign conventions in applying the SDE CW moments , (Wrotation (8) , CWO: ARE POSTIVE IQ. TO SOLVE RF & FEMF : When the other end of the beam FIONF = 0 FINIT = 8 , say I has Mr= FFMF = - Pa2b but not all the time Far end is a 12 with moment arm > There are case that Near end is Mix=0 When When it is ROLLER OR PIN SUPPORT . FEMN = - Pab 12 CAN BE USED ON Far end [MF] when one end - MN= applicable For moment of the beams 3 (tw - ()) + + EMN - EME area has O moment.

SAMPLE : ⑦ Calculate FEM For Span AB and for Span BC. El : constant W & calculate the moments at A and B For the beam shown. El : constant T = 45 solve in two ways. PdX = 15 x - 20dX &P = 24dX 50kN as aim! N = 15x - 20 Nima -x 45KI ↓ 120kN-M - B - C X 24kN/M 13 I - x x + 4 am " I 5 - x ~ B It x G c FEMAB=) - " (20dx) (x) /9 - x) 2m - 4m A Sm 3m 3m - zu 3m 2m 2m 92 FIRST WAY : Area - Moment F EMAB = RC(9) 10 RB 123 MB Mc 2 : 3- · 47. 3 300 358202. 5 * SPAN AB BY AREA MOMENT FEM BA = MB ! DORB) (10) + 10 MB - = (200)(5) - -(50)(7) = 0 50RB +IOMB = 1725 -

  • /B = 0 = 50RB(-) + MB(5) = (n00(3))"!) + (250)(x)(E) 8 RB + 50MB =
  1. 1 & 2 : RB = 41. 78 kN FEMBA = -134 kN-M FEMAB = MA = 10 (G1 . 70) +( - 124) - 300 - 250 FEMAB = MA= - 149 KN-M SPAN AB BY DIRECT FORMULA : FEMAB=( - (Wax) (x) (L-22 - 12 abe = (8 -(24dx)(x)(10 132 - x)2 - 50(7)(7) 102 FEMAB = - 147. 5-31-
    • EMAB = -149 KN-M FEMBA= (3-(24) (x)1 -x) - 50 (n) 102 FEMBA = -176 KN-M SPAN BC : FEMB = - ( - (15xdx))8 - x)(x + 4)2 / 3120)(2) 92 q2(2(a)
  • x(2)) FEMBC=-121. 972 kN-M SPAN CB :
  • <Mc : -R0(x) (2(a) - n(x)) -(3(1xdx)(
  • x)2(x + 4) 92 92 FEMCB = - 0 - en C FEMCB=- 75. 94 IN
  • EMcB= 75. 94kN((W)

③ for the restrained beam shown , compute Fixed end moments two ways. Ea 1 and 2 : RB = 40. 59 dp = 3ndX MB = - 98. 7kN - M

####### ~l ne im M A = 10(49 . 59)-98 . 7 -150 - 270

In 2M 3m 4 - X 1m I MA = - 112. 8 kN-M THE OTHER WAY : FORMULA BASED RA(4)

  • EMx = - )-(7-2xdx)(
  • x))x + 2) 102 MA FEMA = - 112. 8 KN - M 2- += MB = - ( - (t - 2xdx)(8 - x)2(x + 2) 182
  1. 5
  2. 5 3n()(1) = 99 FEMB= -98 .7 kN - M F10AB = (4) (Ma) + (4Ra)(4) - (148 . 5)(3) - 148. 35) = 199)() 18RA + GMA = 444. 5 - ① ...... --- [I , BIA = O Ma(()(412) + (4RA)(4)(41)
  • (18. 5)(2)(- + 1) - (99)()30. 5) - (148. 5)(1)(0. 5) "Thee 34RA +18MA = 250. 425 - ⑫ FEMA = MA = 57. 0425 kN - M FIRST WAY : AREA MOMENT Rn = Dat kit MB = ( )(4) + ( - 57. 0425)
    1. 5 - 99 MB = FEMB = - 74 - 9475 KNM 10 RA MA Another Method : ↑ 2- FEMA = -(3(3dx)(
  • x)(4 - x 42 120 - 120(2) = 240 MA = FEMA = - 57. 0425 kN-M L FEMis = Mi = -(3(32dx)(
  • x)234 - 40 30(4) = 120 1n - 514. 5 FEMB = MB = - 74. 9475 kN-M 2
  • 2- ④ For the restrained beam snown , compute the fixed moments in two ways (n0) (2) (1. 5) = 175 90 34 KNI & P = 7. 2x dX ↓ * = Y 8 - X X W = 7. 2X /
  • 3m i 5m 12m B 10RB [18AB = = 10Rn)(10) + 10MA - (120)(4) - 12034) - :(24)(4) - (40)34) - (4)(44 - /(120)(4) + i((5)(3) + (1 - 5)()
  • (13. 5)(1) - ((45)(3) - m5() - 1987(1) MB 50RA +10 MA = 1844. 425 - 1
  • 800 - 948 - 440 - 192 2 EIBI = - 50R(10(3) + (10MA) (1012) - (120)(4)(4(4 + 4) - (120)(4)34(2) - (240)(4)(4(n) - i(401(4)( + 4)
  • 320 - 220 12 - 2 4-75 2. 25 - 2n . 25 - 47. 5 (40)24)34() - (120)(4)(412) + j(1. 5)(2))415 + 1) + (5)()(42) + (1 . 5)(1)(1) - (inx)()(y + 1) - ((n5)(1)(055) 150 - - 22 - 5 -12(90)(1)(8. 5) I (n)(b)(n) : 50 RA + 50MA= 2534. 05 -> 270 2 Eq 1 & 2 : FIRST METHOD : ARE A MOMENT RA = 48. 2809 kN xB = MA = - 154 - 942 kN . M = (PRA) (10) + MA(18) - =(150)(5) - 150(2) - (270) (a) 50RB + 18 MB = 1042. 5 - 79. 1 MB = 10(48. 2809) - 154. 942 -120 - 240 - 48 - 120 + 1. 5 + 12. 5 - 175 - 90 MB = - 201. 13 kN - M (1 + x/B = 50RB(1(2) + (18MB)(5) - (150)(5)(5/ + 2) - 150(n)(212) - (270)(n)(n(z) 50 RA + 50 MB = 1820

(1 = 200x10-4/200x10") [1 = 43880(I) 20dX ↓ dp = 8x = X - dp = 20dX # - x * 1 x ↓ 1" # ↑ MA : - 20(2) = -90kN - m MAB : - 40 = 4/20n + 0is - 2(4))

  • FEM MB = - 20(3)(1 - 5) = - 90kN . M -90 = 2) -000(y+x + 8B-3) 5)] - )! /B * -Y MB = - 4 * (0-07 - EM, I FEMF - 98 = 40 , 0000x + 150000B + - -
  • 1375/12 - 25. 2 - 425/24 - 24. 4 30 , 0008 A + 150000B = - 39. 475 =
  • 90 = - * (40) [0B -0) - (P (0dx)(x)(10 - x)2 - (40)(4)(92 - -)5(20dx) (x10-x) - ) MBA = 150008A + 30 , 0008B3 + + ( 182 182 10 2 -x 82
  1. 30B = 185. 225 MBA = 150000A + 30 , 0008B+ 129. 275 + ⑪ 8 B = 75 ⑱ , 000 (201 + 0-)-(3)) -(3(84)(2 +x)(n-x)
  • El MBC = 52 SOLVING REACTION : MBc = 440000B + 420008c : (- 94) - SMB = 0 MBC = 440000B + 320000c + 89. 088 - 2
  • RD(10) + 40(7) + 20(8))812 - 4) = 0 Mc = 320008B + 440008 - ( - 94) + (3(8x)(e + 52 x)(3 - x) RD = 44kN / MCB = 320008B + 440000 +94 + 18. 288 SMc = 0 Right McB = 32 , 000 &B + 44000Oc+ ( . 288 - ⑬ Mc - 44(5) + 40(2) = 0 Mc - 00 (20 + ) - - ))- / Mc = 148kN-M MBC = - 90 = (2)-) + x - (7) - Mix: soos or 10 to a re
  • 90 = 280. 4 + 0. 20 - 0. 08EDC) -

####### (170-48 - 0. 244) = - 4 = ⑪ Mic 1 800080(-2s s

Mee Mc = - 140 = -(2+ 5 + 20 - 3))) + 20c Mpc : 4008 + % - ⑤

  • 10 = 140. 3 + 0. 80cE -0 . 24 DC+ (l(0. 80c - 0. 24b) = - 9 - ② MBA + MBC = 0 Eq 1 & 2 : 150008A + 30 , 000 B+ 129. 275 + 44000B +320008c+ 89. 088 : 0 Oc= 150000A +94000tB+ 32 , 000 Oc +218 , 443 = 0 E , an Ac = 120 245t 150000A + 94 , 0000B +32 , 000tc = - 218 , 443 -> 49. B McB + McD = 0 8c = 1 , El 2 32 , 000 &B + 440008c+( - 288+08 , +gp+ 2 q : Ac : - 5832 , 000 &B + 440008c+( - 288+008 , + 2 32 , 0008B + 2006- - 145. 399 = tq MA = 0 o = 2 ⑰ = (20x
  • (2) - my) - EG. A , B , C : 0 : ** + My - D - 15 0A = - 3 - 421x10 - 4
  • (0x - DX) = - & B = - 1. 9407x1y - 3 Oc = - 9. 1144 x 10 - 4 90 : M / 25 , 1)-2 + I -A = 3 - 321x18 - 4 A A = 849/E & = 1 - 9407x10- AA = 849/E) - Mic = Mb = - (w0 , -9 . 144x1 -4) + a MD = - 42. 29 kN M

FEMAB = (40) 2 (1) (216) - 3(1)1 = 15 aw FEMBA= (40)) 5) ne [2(4) - x(517 : -25 kNm can Min = (0 C -r( + ( - 25) - (15) MBA = 0. 58B - - ① MAR = -(20x + 0B-3( * )) - ** M * EMBc= -(2(8) - 3(4)) = 14 - 825kNMCW) Ma - 100) (210. 002) + 0is-(2)))) FEMLB = -2)(2(8) - x(z) = - 28. 125 kNm(e(w) MAB = - + 4008B - 200 - 45 Mic= 34(0B -0) + 12 - 875-12) - 28. 125) Ma = goo - a. A MBc = + 4 - ② MBA + MBC = 0

iwowrowilayam

a 0. 57B - + 0B+ 5 = 0 Mie : -00 (201 + 0

  • ()) - -)) - 81 ⑤ B - MBC = 20 , 000tB + 10 , & B = 1. 7857/E) 0008c/-- 10 MBC = 20 , 0000B +10 , 0008 - - a. C MBc = MB = (1. 7857)+ Mc = 30(2)(1) = 40kN - m MB = 31 - 487 KN-M MBc = 20 = 1 ,00/os+exc- (n)) 8)) s 40 = 10 , 000 & B +20 , 000 + + lu 40 = 10 , 000 &B + 20 , 000tc 10 , 000 &B + 20 , 0 c= - 177. 5 - ① Equilibrium at B : MBA + MBC = 0 800 , - + 20 , 0008B + 16 , 0008- = 1505 ⑱000B + 10 , 00002 = - [4 1 & 2 : O B = 7. 47 x10- 02 = -0. 01871 MAB = MA = /r1000) Mr = -89. 4 kN-M Mic = 20 , 0000B + 10 , 000tc-122. 5 MB = 20 , 000(7 - 47x10 -3) + 10 , 000(-0 . 01071) - 124. 5 MB= - 74. 2 kN-m

= - 30 dp = 10x dX 3 12 - 5 6 - x d 8 X ↓ 4 20 + A 1 + x b 2 +x + y - x - *- ****, **Y W = 10 X - - X Il-I I By slope deflection: Ma = (210) + 8B - 0) - +xi) e 234 - x) - f) ! (4)(atx)cy- Mix = De 34( - 0)+)- 2 w MBA = 0. 50B + 4y - 1 - 2 ) MAB = OB - 80 - Ea . BA 0 . So I 20 I see Min = 2-(0 + 20B - r] + ) * (12 . 5x)(+x)2)7 - x) 52 MBC = 0. 80B + 0. 48c - 19. 2 = 19 - MBA = * 83 + 90 - Eq . McB = 0 .40 B + 0. 80 + Me McB = 0. 40B + 0. 88 + 28. 8 - 198 Mic = 2 (20B + 0

  • ( = (0xx20)(4 - x)(x + 4) 0 McD = y"(0 - 0) -((0x)(
  • x)( ,)2 - 4)-)3X)(
  • x)(a- x) MBL = 0. 80B + 0. 402-255 . 75 + E9 - - 2 42 McD = 0. 756c - - - ( - 403(un) McB = 8. 40B + 0. 80c + + eq - Mcp = 0. 150c+ a McD : -(20 + 0 - -) - (40)(2)(4)2 - (((40)(8 -x) (x) 82 82 McD : 0. 758c - 95 + Ea MiBA + MBC = 8
  1. 58B + 2 + 0. 88B +0 - 208 - 19. 2 = 0 Mac : 2, , (0+0 - 0) + 23)+(x)23x) 82 1 - 38B + 0. urtc = - * ** G MDc= =c + 15 + 110 Myc = - + 125 - Eq McB + McD = 0
  2. 40tB + 0. 88c+ 28. 8 + 0. 782 - 1197/44 = 0 MBA
  • MBC = 0
  1. 550
  2. 40B + 157c = - - + eq - * *B + 9 , 40
    1. 88B+0. 40 - 225. 75 = 0 E9. 1 & 2 : OB
    1. 400c = = Ea - &B = - 54. 0824 McB + McD = 0 OC = 7. 1844 0. 40B + 0. 88 + 24 + 0 - 758c - 95 = 0
  1. 48B + 1. 550tc = - - + Eq. B MB = MBC = 0. 8) -53. 082) + 0 - 40(7. 1844) - 19 - 2 MB = - 58. 792 /N-m 59 A & B : & B = 134. 807/E) CW & c = - 59. 04/E) Mn = MAB = 219(134 . 807) - MA = - 73. 25 kN. M MBC = MB = 0. 8) 134. 807) +0. 4)-59. 844) - 255. 75 MB = -171 - 519kN . M Mc = McD = 0. 75) -59. 044) - Mc= - 139. 277 kN-M Myc = (8) - 54 - -4) + 125] MDc = - 102. 842 KN - M

ap : 18dX 10X H e +

  • 4 - X - x t X dp= 25dX - - I I j- - x 2 + X I
    • X I- 4 - X Ma = 15)(0 + 8B - 320)] -(3(18)(X)(
  • x) 4 42 MBA = 2(5)(0 + 20B - x(0)) + (3(18)(x
  • x) 42 MBA = 3 1 000s [OB - (2)) +( - " (25ax) (e+xY4 - (
  • 42 2) MBC : MBA = 20000 +B - + 3 + 2 MBA = 20 , 000+B + 2 + Eq MBC

~ <naay side sway pay ang member are not equal maskin diff . Support >Wally sidesway pag equal side member ↓ 24dX

8 - X I

MAB = (210) + +1 - 3(0)] MAB = B = E9 - 1 Man = (

  • 20B - 307) MBA = -B = Eg- Mic = -(201 + 8 - x(0)) -((24)(x)(
  • x) 82 MBC = 0. 5 Op + 0. 250c-121. 5 - Ea - 3 (24) (x)2)8 -x) McB = 0 - 258B + 0 - 58c+ /o 82 McB = 0. 250B + 0. 50c +94 . 5 - E9. 4 Mc - My/20 + 0 - 0] McD = = C + Fa . 5 Mac : - - Eac MBA + MBC = 0 2B + 0. 50 8B + 0 - 2581-121 . 5 = 8 88B + 0. 250 = 121 - 5 + ⑪ McB + McD
  1. 250B + 0. 5 C +94. 5 + c 0
  2. 25 0B + / > = - 94. 5 - OB Eq. A & B :
P Slope deflection

1 Moment distribution

⑳

1 combination of JDE & Memor PLAMBWIM

ALL BEAMS
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Slope-Deflection

Course: Civil Engineering (BSCE 01)

136 Documents
Students shared 136 documents in this course

University: Ateneo de Davao University

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