300 LL. 150 ib 400 b MOMENT DISTRIBUTION METHOD 50 n.com Example: 12 12 2. Determine the internal moments at each support of the beam shown in the figure. El is constant. • Stiffness Factors: 4EI (3 • Distribution Factors: • Fixed End Moments (FEMS) 150(12) EI DFAR = 1.0 FEMA -720lb · ft KAn = 12 4 EI 30 30 3/7 /12 El4 wl? FEMBA = 20 150(12)? 20 4EI DFBA = 4EI 1080lb - ft KBc = 12 4EI Pab? FEMBC = - (400)(3)(9) (300)(8)(4)? 122 (400) (3) (9) (300) (8) (4) 4EI Kcp = 12 EI DFnc = 4/7 = -941.667lb ft 21/13 4/7 4EI/12 !! 122 12 4 El 4 + 4 Pa-b FEMca =-I-12 = 758.333lb · ft 122 122 DFCB %3D %3D 50(12)? 12 FEMCD = = -600lb - ft ElA 12 12 wl2 50(12)2 12 FEMDE = -50(4)(2) = -400lb ft DFCD 3/7 %3D 4El/12+ El/4 FEMDC = = 600lb - ft %3D 12 %3D DFDC = 1.0
300 LL. 150 ib 400 b MOMENT DISTRIBUTION METHOD 50 n.com Example: 12 12 2. Determine the internal moments at each support of the beam shown in the figure. El is constant. • Stiffness Factors: 4EI (3 • Distribution Factors: • Fixed End Moments (FEMS) 150(12) EI DFAR = 1.0 FEMA -720lb · ft KAn = 12 4 EI 30 30 3/7 /12 El4 wl? FEMBA = 20 150(12)? 20 4EI DFBA = 4EI 1080lb - ft KBc = 12 4EI Pab? FEMBC = - (400)(3)(9) (300)(8)(4)? 122 (400) (3) (9) (300) (8) (4) 4EI Kcp = 12 EI DFnc = 4/7 = -941.667lb ft 21/13 4/7 4EI/12 !! 122 12 4 El 4 + 4 Pa-b FEMca =-I-12 = 758.333lb · ft 122 122 DFCB %3D %3D 50(12)? 12 FEMCD = = -600lb - ft ElA 12 12 wl2 50(12)2 12 FEMDE = -50(4)(2) = -400lb ft DFCD 3/7 %3D 4El/12+ El/4 FEMDC = = 600lb - ft %3D 12 %3D DFDC = 1.0
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Determine the reactions of each support
Transcribed Image Text:b Bing
f (1) Facebook
x B Module 13 Resources and Recorc X
i classroom.google.com/u/0/c/NDAWNTMOMDESMJEW/m/NDM1NjA3MjU4MjAy/details
* A :
POF Module 13 - Moment Distribution Method.pdf
CEIT - 02 - 501P
E Open with Google Docs
Joints
A
B
D
Members
АВ
BA
ВС
CB
CD
DC
DE
MOMENT DISTRIBUTION METH
3/7
4/7
DF
1
4/7
3/7
1
FEM
-720
1080 |-941.667 758.333 -600.000
600
-400
Example:
1" Dist.
720
-59.286 -79.047 -90.476-67.857
-200
-29.643 360.000 -45.238 -39.524 -100.000 -33.929
f ond Diet 29 643 134.898-179.864 79.728 59.796 | 33.929
CO
2. Determine the internal moments at each support of
the beam shown in the figure. El is constant.
co
-67.449 14.821 39.864 -89.932 16.964 29.898
• For our tabular calculations:
3rd Dist.
67.449 -23.437 -31.249 41.696 31.272 -29.898
We can stop our computation in the 9th distribution
row since we already have a less than 1 moment. It 4 Dist.
co
-11.718 33.725
20.848 -15.624 -14.949 15.636
11.718 -23.388 -31.184 17.470| 13.103 -15.636
may now be neglected.
CO
-15.592 -7.818
co
-11.694
5.859
8.735
6.551
• Determining the summation of moments.
ΣΜΑΒ0
EMBA = 1247.010
EMBC = -1247.010
EMCB = 659.438
EMCD = -659.438
5th Dist.
11.694
-6.255
-8.340
13.377
10.033
-6.551
CO
-3.127
5.847
6.689
-4.170
-3.276
5.016
6th Dist.
3.127
-5.372
-7.163
4.255
3.191
-5.016
CO
-2.686
1.564
2.127
-3.582
-2.508
1.595
7" Dist
2.686
-1.582
-2.109
3.480
2.610
-1.595
co
1.343
1,343
1.740
-1.055 -0.798
1.305
EMDC = 400
EMDE = -400
Determining the reactions at each support. (Practice
8* Dist. -1.343 | -1.321
-1.762
1.058
0.794
-1.305
co
-0.672 -0.672
0.529
-0.881
-0.652
0.397
at Home)
9th Dist.
0.672
0.061
0.081
0.876
0.657
-0.397
27
Sum
Page 27 I 30
Q +
HOUENIT DICTDIDUTION HETILOD
O 261682875_13163.jpg
Show all
N
ENG
11:09 am
a 4) O
US
12 Dec 2021
Transcribed Image Text:150 ib
300
LL.
400 b
MOMENT DISTRIBUTION METHOD
Example:
12
12
12
2. Determine the internal moments at each support of
the beam shown in the figure. El is constant.
• Stiffness Factors:
4EI (3
• Distribution Factors:
• Fixed End Moments (FEMS)
150(12)
EI
DFAR = 1.0
FEMA
-720lb · ft
KAn =
12 4
EI
30
30
DFBA = EI+ E/12
wl?
FEMBA =
20
150(12)2
20
4EI
KBc =
12
4EI
-= 1080lb - ft
4EI
Pab?
FEMBC = -
(400)(3)(9)_ (300)(8)(4)²
122
/12
El 4 +
4EI/12
4EI
Kcp =
EI
DFnc =
= -941.667lb ft
2/13
4/7
4EI1+ El 4
!!
122
12 4
4
Pa-b
FEMC =-E2
(400) (3) (9) (300) (8) (4)
= 758.333lb · ft
122
122
DFCB
%3D
%3D
FEMCO
50(12)²
-600lb · ft
ElA
12
12
wl2 50(12)?
12
FEMDE = -50(4)(2) = -400lb - ft
DFCD
%3D
AEI/12 + El/A
= 600lb ft
12
FEMDC =
%3D
DFDC
= 1.0
Joints
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