Advanced Methods of Structural Analysis

8.2 Canonical Equations of Displacement Method 289

Ta b l e 8. 1 Calculation of unit and loaded reactions at introduced constraint 1
DiagramFree-body diagram of joint 1 Equilibrium equation Reaction

MN 1

r 11 0.6EI

1.0EI

0.8EI

X

MD 0

r 11 C.1:0C0:6C0:8/EID 0

r 11 D

2:4EI.kN m=rad/

MN 2

r 12

0.24EI

0.333EI XMD 0

r 12 0:24EIC0:333EID 0

r 12 D

0:093EI.kN m=m/

MP^0
R 1 P
4.1667

15.36

X

MD 0

R1P15:36C4:1667D 0

R1PD

11:1933kNm

Ta b l e 8. 2 Calculation of unit and loaded reactions at introduced constraint 2
DiagramPortion (crossbar) of the
structure

Equilibrium equation Reaction

MN 1 0.333EI

0.24EI r^21

X

XD 0

r 21 C0:24EI0:333EID 0

r 21 D

0:093EI.kN=rad/

MN 2 0.111EI

0.096EI r^22

X

XD 0

r 22 0:111EI0:096EID 0

r 22 D

0:207EI.kN=m/

MP^0

5 R 2 P

X

XD 0

R2PC 5 D 0

R2PD 5 kN

Let us consider in detail the procedure for the calculation of reactive forcesr 21.
When a section is passed infinitely close to joint 1 from above (Fig.8.8d), the mo-
ment3i 4 - 5 is applied. Since the extended fibers are located to the right of member
4-5, then moment3i 4 - 5 is directed clockwise considering the top part of the portion
4-5. Moment3i 4 - 5 is equilibrated by two forces,3i 4 - 5 =l 4 - 5 D0:333EI.(Theforce
at support 5 is not shown.) Force0:333EIis transmitted to the horizontal member
and has an opposite direction.
Member 1-3 should be considered in a similar way: pass a section infinitely close
to joint 1 from below, apply two bending moments4i 1 - 3 (the top of the member)
and2i 1 - 3 (the bottom of the member) to correspond to the location of the extended
fibers and then equilibrate them by two forces,6i 1 - 3 =l 1 - 3 D0:24EI.Asthelast
step, this force is transmitted to the horizontal member in the opposite direction.