A simply supported girder of 20 m span is traversed by a moving udl of 6 m Tength with an intensity of 20 kN/m, from left to right. Find the maximum bending moment and maximum positive and negative shear forces at sections 4 m from left support. Also find the absolute maximum bending moment that may occur any where in the girder.

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Hello please do it by ild method like I'm doing mine may not be correct so please do full question by the method i have mentioned 

(Plan)
A simply supported girder of 20 m span is traversed by a moving udl of 6 m
Tength with an intensity of 20 kN/m, from left to right. Find the maximum bending
moment and maximum positive and negative shear forces at sections 4 m from
left support. Also find the absolute maximum bending moment that may occur
any where in the girder.
Draw influence lines for forces inzreznoiteLze re.reanZEre aZZA72
tevátion)
Transcribed Image Text:(Plan) A simply supported girder of 20 m span is traversed by a moving udl of 6 m Tength with an intensity of 20 kN/m, from left to right. Find the maximum bending moment and maximum positive and negative shear forces at sections 4 m from left support. Also find the absolute maximum bending moment that may occur any where in the girder. Draw influence lines for forces inzreznoiteLze re.reanZEre aZZA72 tevátion)
PAGE No
20 Sumbly Aubboted gordex o ao m Apam w traverced
ude of
smangth unth intoNsity 20 KNIM from loft to sught
manumum BM and Man. tue $-ve shear Lau at secton
4m ko uy20 KNIm
sught Fund
from
m Lokt supporl Also und absol man BM that ma
c 20 KN/m
anywithure
Occur
ID gorder
ANS
下 个 C
Now
有ム
20m
16m
Nau SFe -
20
- वर = पन
41
tue
FOT Nan SFc place UD e
/2
415
16
6 M
1om
Hence Hax tue SFc= Ix[+/x6.
8十5
13x3
3.90
ot 1e等
Transcribed Image Text:PAGE No 20 Sumbly Aubboted gordex o ao m Apam w traverced ude of smangth unth intoNsity 20 KNIM from loft to sught manumum BM and Man. tue $-ve shear Lau at secton 4m ko uy20 KNIm sught Fund from m Lokt supporl Also und absol man BM that ma c 20 KN/m anywithure Occur ID gorder ANS 下 个 C Now 有ム 20m 16m Nau SFe - 20 - वर = पन 41 tue FOT Nan SFc place UD e /2 415 16 6 M 1om Hence Hax tue SFc= Ix[+/x6. 8十5 13x3 3.90 ot 1e等
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