1.Analized the given structure by the slope deflection method, without using the modified equations for the hinged and overhanging end. 20k 3k/ft
E A
6k
2k/ft
3I
10I
2I
B 12ft θA
18k
2I
C 12ft
12ft
D
4ft
8ft
3ft
= 0 , θB =? , θC =? , θD = ?
Relative Stiffness Member AB BC CD
I
L
I/L
Krel
3I 10I 2I
12 24 12
3I/12 x 12/I 10I/24 x 12/I 2I/12 x 12/I
3 5 2
Fixed and moment 2
2
FEMAB = + WL = +3 X 12 = + 36 Kft 12 12 2 2 FEMBA = - WL = - 3 X 12 = -36 Kft 12 12 2 2 FEMBC = + WL = + 2 X 24 = + 96 Kft 12 12 2 2 = + WL + P ab 2 12 L 2 2 = 2 x 24 + 20 x12 x12 x 12 = +156 Kft 12 2 24 FEMBC = - 156 Kft 2 FEMCD = Pab 2 L 2 = + 18 x 4 x 8 = 32 Kft 2 12 2 FEMDC = - Pa b 2 L = - 18 x 42 x 8 2 12 = -16 Kft MDE = 6 x 3 = 18 Kft MAB = FEMAB + 2EI ( -2 θA - θB ) MAB = FEMBA + [ Krel] (-2 θ A– θB ) = 36 + 3 (2 θA - θB ) MAB = 36 - 3 θB MBC = FEMBC + [Krel] (-2 θB – θC)
= + 156 + 5(-2 θB – θC) = + 156 – 10 θB -5 θC MCB =FEMCB + [Krel ] (-2 θC - θB) = -156 + 5 (-2 θC - θB ) = -156 – 10 θC – 5 θB MCD = FEMCD + [Krel] (-2 θC – θD) = +32 + 2 (-2 θC – θD ) = 32 – 4 θC – 2 θD Joint equilibriun equations; At joint B , MBA + MBC = 0 -36 – 6 θB + 156 – 10 θB – 5 θC = 0 -16 θB -5 θC = 120 1 At joint C , MCB + MCD = 0 -156-10 θC - 5 θB +32-4 θC - 2 θD = 0 -5 θB -14 θC - 2 θD =124 2 At joint D , MDC +MDE =0 -16-4 θD -2 θC +18 =0 -2 θC -4 θD = -2
3
Solving the equation, θB = +11.9399 θC = -14.2076 θD = +7.6038
MAB = +36 -3 x (11.9399) = 0.18 Kft MBA = -36-6 (11.9399) = -107.64 Kft MBC = + 156-10(11.9399)-5(-14.2076) = 107.64 Kft MCB = -156-10(-14.2076)-5(11.9399) = -73.62 Kft MCD = +73.62 MDC = -18 Kft MDE = +18 Kft
2.
Use the slope deflection equations without modification E = 3000 Ksi , I = 1000 in 3I
10I
2I
2I
A
E B 12'
0.5"
C
D
24'
12'
3'
E = 30000 ksi, All FEM =0 ( no load) 4 I = 1000 in displacement at pt ; B = 0.5 " θB = ? , θC = ? , θD = ? Span AB, = / L -3 = 0.5" = + 3.47 x 10 rad 12 x 12 MAB = FEMAB + 2EI / L ( -2 θA -2 θB + 3) 3 = 0 + 2 x 30 x 10 x (3x1000) = 104 , 167 Kft 12 x (12x12) -3 MBA = 0 + 104167 (-2 θA – θB + [-2 θB – θA + 3(+3.4722 x 10 )] = -208,333 θB +1085.06 Span BC, = / L -3 = 0.5 = -1.736 x 10 rad 24 x 12 2EI = 2 x 30x103x10x1000 = 173611 Kft L 24' x 144 MBC = FEMBC + 2EI / L (2 θB – 2 θC + 3) = 0 + 173611 (-23 θB – θC +(3 x -1.736x10-3) = -347222 θB -173611 θC -904.16 MCB = FEMCB + 2EI / L (-2 θC - θB 3+3)
-3
= 0+ 17361 (-2 θC – θB +(3x -1.736x10 ) = -347222 θC -173611 θB – 904.16 Span CD, = 0 (no settlement) 3 2EI / L = 2x30x10 x2x1000 = 69444 Kft 12x144 MCD = FEMCD + 2EI / L [ -2 θC – θD -3) = 0 + 69444 (-2 θC – θD ) = -13888 θC - 69444 θD MDC = FEMDC + 2EI / L (-2 θD - θC ) = 0+ 69444 (2 θD - θC ) = -13888 θD - 694444 θC Joint Equilibrium equation, At Joint B, MBA + MBC -208333 θB +1085.06- 347222 θB -173611 θC – 904.16 = 0 -555555 θB - 173611 θC = -180.84 1
4
Joint C , MCB + MCD =0 -347222 θC - 173611 θB -904.16 - 138889 θC - 69444 -173611 θB - 486111 θC - 69444 θD = 904.16 Joint D, MDC + MDE =0 -09444 θC - 138889 θD =0 3
θD =0
Solving eq;n 1 , 2 ,and 3. -3 θB = +1.08150 x 10 rad -3 θC = -2.41916 x 10 -3 θD = +1.20958 x 10 End Moments, -3 MAB = -104167 x 10.18150 x 10 + 1085.06 = + 972.42 Kft MBA = +859.76 Kft MBC = - 859.76 Kft MCB = -252 Kft MCD = +252 Kft MDC = 0 3.
Use the slope deflection method , for the frame shown in fig; . Frames without sidesway 8K
3.6 kft
D
B
2I
C
I
20'
A 6' θA = θC =
20'
0 , + = 0 , θB = ?
Relative stiffness Member
I
L
I/L
AB I 20 2I/20 x 20/I BC 2I 20 2I/20 x 20/I FEMAB = FEMBA = 0 2 2 FMBC = + WL = 3.6 x 20 = +120Kft 12 12 2 2 FEMCB = -WL = - 3.6 x 20 = -120 Kft 12 12 MBD = (8 x 6) = -48 Kft Slope deflection equations, MAB = FEMAB + [KAB] rel (-2 θA – θB ) = 0 +1(-θB ) = - θB MBA = FEMBA + [KBC]rel (-2 θB - θA )
Krel 1 2
2
= 0 + 1(-2 θB ) = -3 θB MBC = FEMBC + [ KBC] rel (-2 θB – θC ) = 120+ 2 (-2 θB ) = 120 - 4 θB MCB = FEMCB + [ KCB] rel (-2 θC – θB ) = -120 + 2 (-θB ) = -120 -2 θB Joint Equilibrium equations, Joint B, MBA + MBC – 48 = 0 -2 θB + 120 – 4 θB -48 = 0 θB = +12 End Moment, MAB = - θB = -12Kft MBA = -2 θB = -2 x 12 = -24 Kft MBC = 120-4( 12) = +72 Kft MCB = -120 -2(12) = -144 Kft 4. 60KN 10KN/m
A B
3m
θA = θC =
3m
0,
θB =
C
6m
EI Const;
?
For Span AB, FEMAB =
2
Pa b 2 L 2 = 60x3x3 = 45KN-m 2 6 FEMBA =-45 KN-m
For Span BC, FEMBC =
FEMCB
2
WL 12 2 = 10x6 = 30KN-m 12 = -30 KN-m
Member
I/L
Krel
AB
I/6 x 6/I
1
BC
I/6 x 6/I
1
MAB = FEMAB + Krel (-2 θA – θB ) = 45+ (-θB ) = 45 - θB MBA = -45- θB MBC = 30 -2 θB MCB = 30 - θB Joint equilibrium eq; Joint B, MBA+ MBC = 0 -4 θB =15 θB = -15/4 End moments, MAB = 41.25 KN-m MBA = -37.5 KN-m MBC = +37.5 KN-m MCB = -26.25 KN-m
5. 20K
2Kft
3I
6'
θA =
0,
θB =
?,
4'
θC =
FEMAB =
3I
3I
15'
5'
? AB = BC = 0 2
20x6x4 2 10 FEMBA = 28.8K-ft FEMBC = 37.5 K-ft FEMCB = -37.5 K-ft
= 19.2 K-ft
Slope deflection eq; MAB = FEMAB + 2EI/L (-2 θA – θB + 3) = 19.2 – 0.2 EI θB MBA = -28.8 – 0.4 EI θB MBC = 43.75 – 0.6 EI θB Joint equilibrium eq; Joint B,
MBA + MBC =0 FEMAB = 16.21 K-ft FEMBA = -34.78 K-ft FEMBC = +34.78K-ft Joint C, MCB + MC =0 MCB = -25 K-ft
6.
10K
A
1 K/ft
I
3I B
6'
θA =
0,
C
4'
θB =
?,
θC =
15'
?
FEMAB = 9.6K-ft FEMBA = -14.4 K-ft FEMBC = +18.75 K-ft FEMBC = - 18.75K-ft Slope deflection eq; MAB = FEMAB + 2EI/L (-2 θA – θB + 3) =9.6– 0.2 EI θB MBA = -14.4– 0.4 EI θB MBC = 18.75 +9.375 – 0.6 EI θB Joint B, MBA + MBC =0 EI θB =13.725 FEMAB = 6.855 K-ft FEMBA = -19.89 K-ft FEMBC = +19.89 K-ft 7. 6K
18K 3 Kft
2Kft
A 3I
3' θA =
0,
θB =
B
10I
9'
?,
C
3' θC =
FEMAB = 10.125K-ft FEMBA = -3.375 K-ft
?,
θD =
21'
?
2I
D
12'
2I
3'
FEMBC = 41.343 K-ft FEMCB = -5.906 K-ft FEMCD = 36 K-ft FEMDC = -36 K-ft MC = 6 x 1.5 = 9 K-ft Slope deflection eq; MAB = FEMAB + 2EI/L (-2 θA – θB + 3+) MAB =10.125– 0.5 EI θB MBA = -3.375 – EI θB MBC = 41.343 +1.66 EI θB – 0.83 EI θC MCB = -5.906 - 1.66 EI θC – 0.83 EI θB MCD = 36 – 13.5 – 0.5 EI θB Joint equilibrium eq; Joint B, MBA + MBC =0 37.968 -2.66 EI θB – 0.83 EI θC = 0 Joint C, MCB + MCD = 0 43.594 -1.33 EI θB –1.66 EI θC = 0 Joint D, MDC + MC = 0 MDC =-9 K-ft solving eq; 1 and 2, EI θC = 19.767 EI θB = 8.105 MAB = 6.072 KN-m MBA = -11.48 KN-m MBC = +11.48 KN-m MCB = -45.446 KN-m MCD = -45.446 KN-m 8.
1
2
3
Compute the end moment for the beam due to the following support moment E= 30x10 Ksi Support 'a' vertically 0.01 ft down, Rotate 0.001 radian clockwise, Support 'b' vertically 0.04 ft down. Support 'c' vertically 0.0175 ft down. 4 I = 1000 in
a
I
b
3I
c
3I d
0.01'
10'
15'
5'
0.04'
0.0175'
ab
There is no load, all FEM=0 θA = -0.001 (clockwise direction)
ab = / L = +0.003 ab = -0.0015 θB = ? , θC = ? Slope deflction eq; MAB = FEMAB + 2EI/L (-2 θA – θB + 3) MAB = – 0.2 EI θB + 0.0014 EI MBA = –0.4 EI θB+ 0.0016 EI -4 MBC = -0.6EI θB – 9x10 EI Joint equilibrium eq; Mba + Mbc = 0 -4 -EI θb =-7 x 10 EI -4 θb =7x 10 MAB = 262.5 K-ft MBA = 275 k-ft MBC = -275 K-ft Joint 'C' MBC + MC = 0 MCB =0 9.
The closed or box frame is used for culvents . It is always symmetrical and usually symmetrically loaded. There will be no sidesway. A 0.8 K/ft B 0.24K/ft EI const; 0.624K/ft
D 0.54K/ft
From symmetry
10'
C 1.42K/ft 10'
θB =
-
,
θA
θC = θD ,
θA = ? , θD = ? 2
= 0 2
FEMAB = WL = 0.8x 10 = 6.67 K-ft 12 12 2 2 FEMAD = - 0.24 x 10 + 0.324x 10 = +0.92 k-ft 12 30 2 FEMDC = -0.8x 10 = - 6.67 K-ft 12 Slope deflection eq; MAB = FE MAB + Krel MAB (-2 θA – θB ) = 6.67 - KθA MAD = -0.92+ K (-2 θA – θB ) MDA =+0.38 +K (-2 θA – θB ) MDC = -6.67 +K(-2 θA – θB ) = -6.67 – K θD Joint equilibrium eq; MAB + MAD = 0 -3 K θA -K θD + 5.75 = 0 1 Joint D, MDA + MDC = 0 -3 K θD -K θA + 6.29 = 0 2 eq; 1 and 2 solving, K θD = -3.08 K θA = +2.94 End moments, MAB = +3.37 K-ft MAD = -3.37 K-ft MDA = +3.60 K-ft MDC = -3.60 K-ft
10.
10KN A
24KN/m B
I
C 1.5m
12KN
I 1.5m
D 1.5m
θD = θC
= 0,
4m
θB = ?
Member
I/L
Krel
DB
I/3 x 3/I
1
BC
I/4 x 3/I
0.75
2
2
FEMDB = Pab = 12 x 1.5 x 1.5 = 4.5 KN-m 2 2 L 3
FEMDB = -4.5 KN-m 2 2 FEMBC = WL = 24 x 4 = 32KN-m 12 12 FEMCB = -32 KN-m FEMBA = -(10 x 1.5 ) =- 15 KN-m Slope deflection eq; MDB = FEMDB + Krel (-2 θD – θB ) = 4.5 - θB MBD = -4.5 -2 θB MBC = 32- 1.5 θB MCB = -32-0.75 θB Joint equilibrium eq; At joint 'B' θB = 3.57 End moments, MBD + MBC + MBA =0 MDB =0.93 KN-m MBD = -11.64 KN-m MBC =26.645 KN-m MCB =-34.677 KN-m MBA =-15 KN-m 11.
A
Use the moment distribution method. 3K/ft 20K 2K/ft
3I
B
12'
10I 12'
2
18K
C 12'
FEMAB = 3 x 12 = +36 K-ft 12 FEMBA = -36 K-ft FEMBC = 2 x 242 + 20 x 12 x 12 2 = +156K-ft 2 12 24 FEMCB = -156 K-ft 2 FEMCD = 18 x 4 x 8 = +32 K-ft 2 12 2 FEMDC = -18 x 4 x 8 = -16 K-ft 2 12 MDE = 6 x 3 = 18 K-ft Member I/L Krel AB 3I/12 x 12/L 3 BC 10I/24 x 12/I 5 CD 2I/12 x 12 /I 2 DFAB =1 DFBA = 3/ (3+5)= 0.375
2I 4'
6K
D 8'
2I E 3'
DFBC =5/(3+5) = 0.625 DFCB =5/ (5+2) = 0.714 DFCD = 2/ (5+2) = 0.286 Joint Member Krel D.F FEM Cycle 1 Balance M C.O.M Cycle 2 Balance M C.O.M Cycle 3 Balance M Final M
A AB 3 1.0 +36 -36 -22.5 +22.5 -4.93 +4.93 0
B BA 3 0.375 -36 -45 -18 -9.85 +11.25 -9.38 -106.98
C BC 5 0.625 +156 -75 +44.27 -16.42 +13.75 -15.63 +106.97
CB 5 0.714 -156 +88.54 -37.5 +27.49 -8.21 +12.20 -73.48
D CD DC DE 2 2 0.286 1.0 +32 -16 +18 +35.46 -2.0 -1.0 +17.73 +11.01 -17.73 -8.87 +5.51 +4.88 -5.51 +73.48 -18.0 +18.0
1.2 K/ft 12. const: EI A
B
C
30ft
D
40ft
13.
30ft
3.6 K/ft 8K
D
B
2I
C
I
20ft
A 6ft
20ft
14. 10K
a 15.
1K/ft
I 6ft 3K/ft
b
3I 15ft
4ft 20K 2K/ft
c 18K
6K
A
3I
B
10I
12`
C
12`
2I
12`
D
4`
2I
8`
E
3`
16. 40KN
12KN/m
60KN
10KN
A I 2m
17.
` B
3I 6m
2m
10KN
C
2I 2m
D 4m
E 2m
24KN/m
A
B
I
C 1.5m
12KN
I 1.5m D
1.5m
4m
18. 6K
18K
3K/ft 2K/ft
A
3I
B
3ft` 9ft
10I 3ft
19.
C
21ft
2I
D
12ft
60KN
2I
E
3ft
10KN/m
A B 3m
3m
C 6m
EI const:
20.
3.6K/ft
A
I
8K
B
D
2I
9K
10ft
10ft C 20ft
6ft
BY TU (Monywar)
[email protected] 071-21476/21460 09-2132597,09-2133070