164 6 Deflections of Elastic Structures
Actual state. Reaction of supportsRADRBDP. As usual, the axial forces in
the members are taken positive in tension. Equilibrium equations lead to the follow-
ing internal forces:
O 35!X
M 6 leftD 0 WRA2dCPdO 35 hD 0 !O 35 D4
3PU 24!X
M 3 leftD 0 WRAdCU 24 hD 0 !U 24 D4
3P (b)D 23!X
YleftD 0 W D 23 sin ̨CRAD 0 !D 23 D5
3PD 36!X
YleftD 0 W RAPD 36 sin ̨D 0 !D 36 D 0Unit state. Unit vertical loadPD 1 is applied at the point where vertical displace-
ment is to be determined. Reaction of supportsRNADRNBD0:5. Internal forces in
unit state are:
ON 35!X
M 6 leftD 0 WRNA2dON 35 hD 0 !ON 35 D
4
3
UN 24!X
M 3 leftD 0 WRNAdCUN 24 hD 0 !UN 24 D2
3
DN 23!X
YleftD 0 W DN 23 sin ̨CRNAD 0 !DN 23 D5
6(c)DN 36!X
YleftD 0 W RNADN 36 sin ̨D 0 !DN 36 D5
6Ta b l e6.3contains the data which is necessaryfor computation of displacement ac-
cording to (a). They are the length of the elements, their axial rigidity, and internal
forces in all members in actual and unit states. The last column contains application
of (a) for each member separately.
Summation of the last column of this table leads to the required displacement of
joint 6
y 6 DXNpNlN
EAD506P
18 EA; (d)Note.For calculation of displacement ofall jointsof the bottom (upper) chord of
the truss, the different unit states should be considered. For each unit state, the pro-
cedure presented in Table6.3should be repeated. Thus, the Maxwell–Mohr integral
requires calculation of the internal forces inallmembers of the truss foreachunit
state. This procedure is cumbersome and will be extremely simplified using the
elastic load method.