Advanced Methods of Structural Analysis

Appendix 573

Influence lines for shear forceatsection6areshownbelow.

024681012

lll

14 16 18

0.09870.12340.00001.00000.69130.3087

• 0.1234-0.0987
  ⎠

⎞

⎝

IL⎛Q 6 right

Example.ForcePis applied at point 8. Construct the bending moment diagram.

Solution.

1.Bending moment at point 6 (supportB)isM 6 D0:0789P l.
2.Ordinate of influence lineQ 0 D0:0789, so reaction of supportAisRAD
0:0789Pand directed downward.
3.SinceQright 6 DRACRBD0:6913P, then reaction of supportBis

RBDRAC0:6913PD0:0789PC0:6913PD0:7702P:

4.Reaction of supportD:RD!

P

MCDRDlCRA2lRBlCP

2

3

lD 0!
RDD0:0543P
5.Bending moment at point 12 (supportC)isM 12 D0:0543P l:Thesameresult
may be taken immediately from TableA.15for section 6, if loadPis located at
section 10.

Final bending moment diagram is presented below.

M

RA=0.0789P

A

0.1515Pl

C

D

B

RB=0.7702PRC=0.363PRD=0.0543P

0.0789Pl P 0.0543Pl