applied solely at its ends. The sign convention is as follows: an end moment is positive if
it is clockwise; an angular displacement is positive if the rotation is clockwise; the trans-
verse displacement A is positive if it rotates the member in a clockwise direction.
Computing the end rotations gives Oa = (L/6EI)(2Ma, - Mb) + A/L; Sb = (LI6EI)(-Ma +
2Mb) + A/L. These results may be obtained by applying the moment-area method or unit-
load method given in Sec. 1.- Solve the foregoing equations for the end moments
Thus,
M¥)(
2
^-¥) M^ffV^-f) (4)
\ L I \ L I \ L /\ L IThese are the basic slope-deflection equations.- Compute the value of I/L for each member of the bent
Let K denote this value, which represents the relative stiffness of the member. Thus Kab =
100/20 = 5; Kcd = 144/24 = 6; Kbe = 300/30 = 10; Kce = 60/15 = 4. These values are
recorded in circles in Fig. 18. - Apply Eq. 4 to each joint in turn
When the wind load is applied, the bent will deform until the horizontal reactions at the
supports total 10 kips (44.5 kN). It is evident, therefore, that the end moments of a mem-
ber are functions of the relative rather than the absolute stiffness of that member. There-
fore, in writing the moment equations, the coefficient 2EIIL may be replaced with I/L; to
view this in another manner, E =
1
A.
Disregard the deformation associated with axial forces in the members, and assume
that joints B and C remain in a horizontal line. The symbol Mab denotes the moment ex-
erted on member AB at joint A. Thus Mab = 5(6b - 3A/20) = (^56) b - 0.75A; Mdc =
6(0C - 3A/24) = (^60) C - 0.75A; Mec = 4(0C + 3A/15) = (^40) C + 0.8OA; Mba = 5(20b - 3A/20) =
1006 - 0.75A; Mcd = 6(2BC - 3A/24) = (^120) C - 0.75A; Mce = 4(20C + 3A/15) = (^80) C + 0.8OA;
Mcd = W(20b + O 0 ) = 2Q9b + (^100) C; Mcb = 10(0 6 + (^20) C) = lQOb + (^200) C.
- Write the equations of equilibrium for the joints
and for the bent
Thus, joint B, Mba + Mbc = O, Eq. a; joint C, Mcb + Mcd + Mce = O, Eq. b. Let H denote the
horizontal reaction at a given support. Consider a horizontal force positive if directed to-
ward the right. Then Ha, + Hd + He + 10 = O, Eq. c. - Express the horizontal reactions in terms of the end moments
Rewrite Eq. c. Or, (Mab + Mba)/2Q + (Mdc + Mcd)/24 - (Mec + Mce)/l5 + 10 = O, or 6Mab +
6Mba + 5Mdc + 5Mcd - %Mec - %Mce = -1200, Eq. c'. - Rewrite Eqs. a, b, and c' by replacing the end moments
with the expressions obtained in step 4
Thus, 3006 + (^100) C - 0.75A = O, Eq. A- IQ^ + (^400) C + 0.05A = O, Eq. £; 9QOb - (^60) C -
29.3OA =-1200, Eq. C.
- Solve the simultaneous equations in step 7 to obtain the
relative values of 0& O 0 , and A
Thus Ob = 1.244; Oc = -0.367; A = 44.85. - Apply the results in step 8 to evaluate the end moments
The values, in foot-kips, are: Mab = -27.42 (-37.18 kN-m); Mdc = -35.84 (-48.6 kN-m);
Mec = 34.41 (46.66 kN-m); Mba = -21.20 (-28.75 kN-m); Mcd = -38.04 (-51.58 kN-m);
Mce = 32.94 (44.67 kN-m); Mbc = 21.21 (28.76 kN-m); Mcb = 5.10 (6.92 kN-m).