Determine the reactions of each support b Bingf (1) Facebookx B Module 13 Resources and Recorc Xi classroom.google.com/u/0/c/NDAWNTMOMDESMJEW/m/NDM1NjA3MjU4MjAy/details* A :POF Module 13 - Moment Distribution Method.pdfCEIT - 02 - 501PE Open with Google DocsJointsABDMembersАВBAВСCBCDDCDEMOMENT DISTRIBUTION METH3/74/7DF14/73/71FEM-7201080 |-941.667 758.333 -600.000600-400Example:1" Dist.720-59.286 -79.047 -90.476-67.857-200-29.643 360.000 -45.238 -39.524 -100.000 -33.929f ond Diet 29 643 134.898-179.864 79.728 59.796 | 33.929CO2. Determine the internal moments at each support ofthe 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 -29.898We can stop our computation in the 9th distributionrow since we already have a less than 1 moment. It 4 Dist.co-11.718 33.72520.848 -15.624 -14.949 15.63611.718 -23.388 -31.184 17.470| 13.103 -15.636may now be neglected.CO-15.592 -7.818co-11.6945.8598.7356.551• Determining the summation of moments.ΣΜΑΒ0EMBA = 1247.010EMBC = -1247.010EMCB = 659.438EMCD = -659.4385th Dist.11.694-6.255-8.34013.37710.033-6.551CO-3.1275.8476.689-4.170-3.2765.0166th Dist.3.127-5.372-7.1634.2553.191-5.016CO-2.6861.5642.127-3.582-2.5081.5957" Dist2.686-1.582-2.1093.4802.610-1.595co1.3431,3431.740-1.055 -0.7981.305EMDC = 400EMDE = -400Determining the reactions at each support. (Practice8* Dist. -1.343 | -1.321-1.7621.0580.794-1.305co-0.672 -0.6720.529-0.881-0.6520.397at Home)9th Dist.0.6720.0610.0810.8760.657-0.39727SumPage 27 I 30Q +HOUENIT DICTDIDUTION HETILODO 261682875_13163.jpgShow allNENG11:09 ama 4) OUS12 Dec 2021 150 ib300LL.400 bMOMENT DISTRIBUTION METHODExample:1212122. Determine the internal moments at each support ofthe beam shown in the figure. El is constant.• Stiffness Factors:4EI (3• Distribution Factors:• Fixed End Moments (FEMS)150(12)EIDFAR = 1.0FEMA-720lb · ftKAn =12 4EI3030DFBA = EI+ E/12wl?FEMBA =20150(12)2204EIKBc =124EI-= 1080lb - ft4EIPab?FEMBC = -(400)(3)(9)_ (300)(8)(4)²122/12El 4 +4EI/124EIKcp =EIDFnc == -941.667lb ft2/134/74EI1+ El 4!!12212 44Pa-bFEMC =-E2(400) (3) (9) (300) (8) (4)= 758.333lb · ft122122DFCB%3D%3DFEMCO50(12)²-600lb · ftElA1212wl2 50(12)?12FEMDE = -50(4)(2) = -400lb - ftDFCD%3DAEI/12 + El/A= 600lb ft12FEMDC =%3DDFDC= 1.0Joints

Answered: Image /qna-images/answer/a7387710-b1cb-431a-9b39-919aee3973a0.jpg

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

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

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Concept explainers

Equilibrium and Support Reactions

Civil engineers deal with structures that are composed of several members connected or with individual members such as beams and columns. To analyze such structures, the condition of statics must be applied. Under the static condition, the members are considered to be in equilibrium under the action of forces, that is, the members are assumed to be having negligible velocity.

Question

Determine the reactions of each support

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

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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