Advanced Methods of Structural Analysis

232 7 The Force Method

 Statical verification of bending moment diagram. The free-body diagram of

jointD(closed sectiona/is shown in Fig.7.11g. Equilibrium condition of this

joint is X

MD11:268C1:81713:085D0:

 Kinematical verification of bending moment diagram. Displacement in the

direction of any primary unknown in the given system must be equal to zero.

This condition is verified by multiplying bending moment diagramMPin the

actual state by bending moment diagramMNiinanyprimary system.

Displacement in the direction of the first primary unknown is

 1 D

MN 1 MP

EI

D

1

1 EI



5

6

.8:036 10 1:817 10 C 4 1:32410/

C

1

2 EI



4

6

. 2 13:085 10 C 2 11:349 6 13:085 6 C 10 11:349/

C

1

2 EI



1

2

11:349 6 

2

3

 6 D

1

EI

.195:453195:511/0:

The relative error is 0.029%. Similarly it is easy to check that displacement in the
direction of the second primary unknown is zero, i.e., 2 DMN 2 MP=EID 0.
Shear forces may be calculated using differential relationshipsQDdM=dx.This
formula leads to the following results:

Q 6  7 D

13:085.11:349/

4

D6:1085kN

Q 7  8 D

11:349

6

D1:8915kNI Q 4  5 D

11:268

3

D3:756kN:

The portionA-3 subjected to loadqand couples at pointsAand3isshownin
Fig.7.11h.

HA!

X

M 3 D 0 W HAD6:2438kN!QADC6:2438kN;

H 3!

X

MAD 0 W H 3 D3:756kN!Q 3 D3:756kN:

The final shear force diagram is shown in Fig.7.11i.
Axial forces may be derived from the equilibrium of rigid jointD; a correspond-
ing free-body diagram is presented in Fig.7.11i. The shear at point 3 is negative, so
this force, according to the sign law, should rotate the body counterclockwise. It is
assumed that unknown forcesN 6  7 andN 3  4 are tensile.

N 6  7!

X

XD 0 W N 6  7 D0;

N 3 A!

X

YD 0 W N 3 AD6:108kN: