Advanced Methods of Structural Analysis

4.3 Influence Lines for Reactions and Internal Forces 93

Internal Forces in Sectionk

The internal forces can be found in a similar way, using the relevant influence lines.
They are the following:

MkDP 1 1:96q

2:0788C1:0394

2

 8 P 2 0:5194D9:500kN m;

QkDP 1 0:3883Cq

0:177C0:0885

2

 8 CP 2 0:04425D1:405kN;

NkDP 1 0:40194q

1:1039C0:5519

2

 8 P 2 0:2759D19:473kN:

The magnitudes of just found internal forcesMk,Qk,andNkcoincide with those
computed in Sect. 342 and presented in Table4.1.

Example 4.1.Let us consider design diagram of the arch in Fig.4.7. It is necessary
to find bending moment in section 3 due to the forcePD 10 kN, applied at point 7.

Solution.The feature of this problem is as follows: we will compute the bending
moment at the section 3withoutinfluence line forM 3 but using influence line for
reaction. As the first step, we obtain the vertical reaction and thrust, which are nec-
essary for calculation of internal forces.

Step 1.FindHandRAfrom previously constructed influence lines presented in
Fig.4.7

RADPyRD 10 0:125D1:25kNI

HDPyHD 10 0:25D2:5kN;

whereyRandyHare ordinates of influence line forRAandH, respectively,
under concentrated forceP.
Step 2.The bending moment in section 3, considering left forces, becomes

M 3 DHy 3 CRAx 3 D2:57:596C1:25 12 D3:99kN m;

wherex 3 andy 3 are presented in Table4.1.

This example shows that one of advantages of influence line is that the influence
lines for reactions and thrust constructedoncemay be used for their computation
for different cases of arbitrary loads. Then, by knowing reactions and thrust, the
internal forces at any point of the arch may be calculated by definition without using
influence line for that particular internal force.
This idea is the basis of complex usage of influence lines together with fixed
load approach, which will be effectively applied for tedious analysis of complicated
structures, in particularly for statically indeterminate ones.