Question 16.3: Determine the member end moments and reactions for the conti...

This beam was previously analyzed in Example 13.6 by the method of consistent deformations.

From Fig. 16.7(a), we can see that all three joints of the beam are free to rotate. Thus the beam can be considered to have three degrees of freedom, θ_A , θ_B, and θ_D, and it can be analyzed by using the usual slope-deflection equations (Eq. (16.9)) for members rigidly connected at both ends. However, this approach is quite time consuming, since it requires solving three simultaneous equations to determine the three unknown joint rotations.

M_{nf} = \frac{2EI}{L}(2θ_n + θ_f – 3ψ) + FEM_{nf}                         (16.9)

Since the end supports A and D of the beam are simple supports at which no external moment is applied, the moments at the end A of member AB and at the end D of member BD must be zero. (This can easily be verified by considering moment equilibrium of the free bodies of joints A and D shown in Fig. 16.7(b).) Thus the end A of member AB and the end D of member BD can be considered to be hinged ends, and the modified slope-deflection equations (Eqs. (16.15)) can be used for these members. Furthermore, since the modified slope-deflection equations do not contain the rotations of the hinged ends, by using these equations the rotations θ_A and θ_D of the simple supports can be eliminated from the analysis, which will then involve only one unknown joint rotation, θ_B. It should be noted that once θ_B has been evaluated, the values of the rotations θ_A and θ_D, if desired, can be computed by using Eq. (16.16). In the following, we use this simplified approach to analyze the continuous beam.

M_{rh} =\frac{3EI}{L}(θ_r – ψ) +(FEM_{rh} – \frac{FEM_{hr}}{2})                      (16.15a)

M_{hr} = 0                (16.15b)

θ_h = -\frac{θ_r}{2} + \frac{3}{2}ψ – \frac{L}{4EI}(FEM_{hr})                     (16.16)

Degrees of Freedom θ_B

Fixed-End Moments

FEM_{AB} = \frac{15(10)^2}{12}= 125 kN-m\circlearrowleft             or           +125 kN-m

FEM_{BA} = 125 kN-m \circlearrowright           or       -125 kN-m

FEM_{BD} = \frac{60(10)}{8} + \frac{15(10)^2}{12} = 200 kN-m \circlearrowleft         or      +200 kN-m

FEM_{DB} = 200 kN-m \circlearrowright        or      -200 kN-m

Slope-Deflection Equations Since both members of the beam have one end hinged, we use Eqs. (16.15) to obtain the slope-deflection equations for both members. Thus

M_{BA} = \frac{3EI}{10}(θ_B) + (-125 -\frac{125}{2}) = 0.3EIθ_B – 187.5                 (1)

M_{BD} = \frac{3E(2I)}{10}(θ_B) + (200 + \frac{200}{2}) = 0.6EIθ_B + 300                  (2)

Equilibrium Equation By considering the moment equilibrium of the free body of joint B (Fig. 16.7(b)), we obtain the equilibrium equation

M_{BA} + M_{BD} = 0                             (3)

Joint Rotation To determine the unknown joint rotation θ_B, we substitute the slope-deflection equations (Eqs. (1) and (2)) into the equilibrium equation (Eq. (3)) to obtain

or

from which

Member End Moments The member end moments can now be computed by substituting the numerical value of EIθ_B into the slope-deflection equations (Eqs. (1) and (2)). Thus

M_{BA} = 0.3(-125) – 187.5 = -225 kN-m        or       225 kN-m \circlearrowright

Member End Shears and Support Reactions See Fig. 16.7(c) and (d).

Equilibrium Check See Fig. 16.7(d).

+↑∑F_y = 0                52.5 – 15(20) + 225 – 60 + 82.5 = 0                        Checks

-52.5(20) + 15(20)(10) – 225(10) + 60(5) = 0                     Checks