Advanced Methods of Structural Analysis

2.2 Application of Influence Lines for Fixed and Moving Loads 29

The value of reactionRAdue to given fixed loadsP,q,andMequals

RADPyCq!CMtan ̨DP

1

„ƒ‚...^2

y

Cq

1

2

 1 l

„ƒ‚...

!

CM

1

„ƒ‚...l

tan ̨

D

P

2

C

ql

2

C

M

l

:

The change of values of loads and/or their position does not change the procedure of
calculation of any factor using corresponding influence line. From this example we
observe an advantage of influence line: once constructed influence line may be used
for calculation of relevant function due to arbitrary loads.

Example 2.1.A design diagram of the simply supported beam with overhang is
presented in Fig.2.12. Calculate shear at sectionnusing influence line. The loads
areP 1 D 12 kN;P 2 D 8 kN;q 1 D 3 kN=m;q 2 D 2 kN=m.

Fig. 2.12 Design diagram of
the beam. (a) Influence line
for shearQn.(b) Shear force
diagram

RA

q 1

P 1

RB

6m 4m 3m

n

P 2

q 2

+

1

Inf. line Qn

0.4

0.6

(^1) 0.3
−
a

• 14.1
  Shear diagram Q
  14
  15.9
  8
  3.9 −
  b
  Solution.First of all, the influence line for required shear at sectionnshould be
  constructed (Fig.2.12a).
  The shear force at sectionncaused by the fixed loads isQnD
  P
  PiyiC
  P
  qi!i.
  If loadP 1 is located infinitely close to the left of the sectionn,thenordinate
  y 1 D0:6.IfloadP 1 is located infinitely close to the right of sectionn,then
  ordinatey 1 D 0:4.Ordinatey 2 D0:3. Areas of the influence line within the
  distribution loadsq 1 andq 2 are
  ! 1 D
  1
  2
   6 0:6D1:8mI! 2 D
  1
  2
   3 0:3D0:45m:
  The peculiarity of this problem is that forceP 1 is located at the section where it is
  required to find shear. Therefore, we have to consider two cases when this load is
  located infinitely close from the left and right sides of the sectionn.
  If loadP 1 is located to theleftof sectionn,then
  QnD 12 .0:6/C 8 .0:3/C 3 .1:8/C 2 .0:45/D15:9kN: