#### Read Microsoft PowerPoint - Structural Analysis Moment Distribution Lecture 9 [Compatibility Mode] text version

`!&quot;# ! %') \$ &amp;(&quot; * ) + * &quot; &quot; + . ) * &quot; #! \$ , &quot; + &quot; )/ &quot;&quot; 0 &quot; 1 0 ) 1 &quot; &quot; &quot; &quot; )2 &quot; &quot; &quot;3/4 &quot;* &quot;* &quot; )M AB1 = M BA 2/4 5 * ) 5 ) 0 * &quot; &quot;I M BA = 4 E tan  L I = 4E (for small values of  ) L/4 &quot;* 5 )) 5 * 5I (for small values of  ) L&quot; &quot; *&quot; 6 )M BA = 3E/) + &quot;) &quot; 7 ) )/4 &quot; &quot;* &quot; )M AB = M BA 6EId = 2 L&quot; &quot; &quot; 5/4 &quot; &quot;* ) &quot; &quot;M BA3EId 6EId = 2 = 2 2L L1! *7 )% 8* 71! *1 # \$ 727 #! \$ , )% 87 7 ) 7 ) *1 &quot;01! *, !FEM AB (being anti - clockwise) =-% 87 8WL - 65.0 × 3.0 × 3.0 = = -48.75 kNm 12 12FEM BA (being clockwise) = +48.75 kNm78FEM BC (being anti - clockwise) =- WL - 32.0 × 3.6 × 3.6 = = -34.56 kNm 12 12FEM CB (being clockwise) = +34.56 kNm1! *9 : 1 4 0 3 0 )3% 872 &quot; . ) ; )&gt; &lt;= ?. 7 ; )&gt; 0 ) &lt;= ? 3 5 71% &amp; 3 &lt;)% 0 ; )&gt;1' @ ; &lt;)% 0 ) &lt;= ? &lt;)?A % &amp; 31! *7 7 &quot;% 8&quot; ) , 7 727 )7 B &quot;)1% &amp; 3 &lt;)% 0 )720 )#, \$1! *3 A?C CC&lt;% 87A' 7 A' ) )C )@5000 × 10 4 K AB = = 16.67 ×103 3.0 ×103 5000 ×10 4 K BC = = 13.89 ×103 3.6 × 103 16.67 × 103 = 0.545 DFAB = 3 3 16.67 × 10 + 13.89 × 10 13.89 × 103 DFBC = = 0.455 3 3 16.67 ×10 + 13.89 × 101! *% 8&quot; )0.545 × -14.19 = -7.73 kNm being added to the end BA 0.455 × -14.19 = -6.46 kNm being added to the end BC1! *, 2 7 5 &quot;)% 8carry - over process 0.5 × -7.73 = -3.865 kNm transferred to the end A 0.5 × -6.46 = -3.23 kNm transferred to the end C1! *&quot; D ) &quot; 7 72 4G ) H =% 8&quot; &quot; &quot; ) 1?)@ 1&lt;% ( 1 ( ' )C 2')' 0 ) % '3 9 E F 71! *1 7 !( 8)B 7 =C A ?C&lt;8 B7 A ?C &lt; B ! ?C @C A ?C&lt;1! *2) * * ) ! .( 8&quot; &quot;, ! 72&quot;1! *72 5 &quot; ) )( 87A) &quot;3 8500 ×10 4 3 K AB = × = 14.17 × 103 4 4.5 ×103 4 6500 × 10 4 K BC = = 17.33 × 103 3.75 × 103 3 3 5500 ×10 4 K CD = × = 11.00 × 103 4 4 3.75 ×1031! *,DFBA = DFBC DFCB DFBC 3 K AB 4( 8,83 K AB + K BC 4 = 1 - 0.45 = 0.5514.17 × 103 = 0.45 = 3 3 17.73 × 10 + 14.17 × 10K BC 17.73 × 103 = = 0.612 = 3 3 3 17.73 × 10 + 11.00 × 10 K BC + K CD 4 = 1 - 0.612 = 0.3881! *3 72 *3PL 3 ×120 × 4.5 = 101.25 kNm = 16 16 2 PL 2 × 70 × 3.75 = = -58.33 kNm FEM BC = - 9 9 FEM CB = +58.33 kNm FEM BA = + FEM CD = - WL 220 × 3.75 = = -103.13 kNm 8 8( 888M DE = - M DC = -50 × 1.2 = -60 kNm1! *4 , 2 )I ) &quot;( 8) )1! *( 8; &lt;)= %A; C #= ' C)%\$1% 0 ) ';?)'1%' 'A &lt;)= 3 0 3 72 )Distribution BA = (+ 101.25 - 58.33)× 0.45 = 19.31 Distribution BC = (+ 101.25 - 58.33)× 0.55 = 23.61Distribution CB = -14.8 × 0.612 = 9.06 Distribution CD = -14.8 × 0.388 = 5.741! *B 1 5( 8), )1! *, 7 &quot; )( 8/8) 1 &quot; )+5 ) + *1 * ))/&quot; 5 1 ) ) ! !B ! )8&quot;*1( J&quot; , ) !`

#### Information

##### Microsoft PowerPoint - Structural Analysis Moment Distribution Lecture 9 [Compatibility Mode]

31 pages

Find more like this

#### Report File (DMCA)

Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:

Report this file as copyright or inappropriate

224732

### You might also be interested in

BETA
1748-7161-4-S1 1..33
0273.tif