Question 18.2: Determine the reactions and the member end forces for the th...
Determine the reactions and the member end forces for the three-span continuous beam shown in Fig. 18.12(a) by using the matrix stiffness method.
Degrees of Freedom From the analytical model of the beam shown in Fig. 18.12(b), we observe that the structure has two degrees of freedom, d_1 and d_2, which are the unknown rotations of joints 2 and 3, respectively. Note that the member local coordinate systems are chosen so that the positive directions of the local and global axes are the same.
Therefore, no coordinate transformations are needed; that is, the member stiffness relations in the local and global coordinates are the same.
Structure Stiffness Matrix
Member 1 By substituting L = 10 m into Eq. (18.6), we obtain
k =\frac{E I}{L^{3}}\left[\begin{array}{cccc}12 & 6 L & -12 & 6 L \\6 L & 4 L^{2} & -6 L & 2 L^{2} \\-12 & -6 L & 12 & -6 L \\6 L & 2 L^{2} & -6 L & 4 L^{2}\end{array}\right] (18.6)
By using the fixed-end moment expressions given inside the back cover of the book, we evaluate the fixed-end moments due to the 80-kN load as
Q_{f 2} = \frac{80(6)(4)^2}{(10)^2} = 76.8 kN-mQ_{f 4} = -\frac{80(6)^2(4)}{(10)^2} = -115.2 kN-m
The fixed-end shears Q_{f 1} and Q_{f 3} can now be determined by considering the equilibrium of the free body of member 1, shown in Fig. 18.12(c):
+\circlearrowleft ∑M_② = 0 76.8 – Q_{f 1}(10) + 80(4) – 115.2 = 0
Q_{f 1} = 28.16 kN
+↑∑F_y = 0 28.16 – 80 + Q_{f 3} = 0
Q_{f 3} = 51.84 kN
Thus, the fixed-end force vector for member 1 is
F_{f 1} = Q_{f 1} = \begin{bmatrix} 28.16 \\ 76.8 \\ 51.84 \\ -115.2 \end{bmatrix}\begin{matrix} 0 \\ 0 \\ 0\\1 \end{matrix}From Fig. 18.12(b), we observe that the structure degree of freedom numbers for this member are 0, 0, 0, 1. By using these numbers, the pertinent elements of K_1 and F_{f 1} are stored in their proper positions in the structure stiffness matrix S and the fixed-joint force vector P_f , respectively, as shown in Fig. 18.12(d).
Member 2 By substituting L = 10 m into Eq. (18.6), we obtain
The fixed-end moments due to the 24-kN/m load are
Q_{f 2} = -Q_{f 4} = \frac{24(10)^2}{12} = 200 kN-mApplication of the equations of equilibrium to the free body of member 2 yields (Fig. 18.12(c))
Q_{f 1} = -Q_{f 3} = 120 kNThus,
F_{f 2} = Q_{f 2} = \begin{bmatrix} 120\\ 200 \\ 120 \\ -200 \end{bmatrix}\begin{matrix} 0 \\ 1 \\ 0\\2 \end{matrix}By using the structure degree of freedom numbers, 0, 1, 0, 2, for this member, we store the relevant elements of K_2 and F_{f 2} into S and P_f , respectively, as shown in Fig. 18.12(d).
Member 3 L = 5 m:
The elements of K_3 are stored in S using the structure degree of freedom numbers 0, 2, 0, 0. Note that since member 3 is not subjected to any external loads,
F_{f 3} = Q_{f 3} = 0Joint Load Vector Since no external moments are applied to the beam at joints 2 and 3, the joint load vector is zero; that is,
P = 0
Joint Displacements The stiffness relations for the entire continuous beam, P – P_f = Sd, are written in expanded form as
\begin{bmatrix} -84.8 \\ 200 \end{bmatrix} = EI \begin{bmatrix} 0.8 & 0.2 \\ 0.2 & 1.2 \end{bmatrix}\begin{bmatrix} d_1 \\ d_2 \end{bmatrix}By solving these equations simultaneously, we determine the joint displacements to be
EId_1 = -154.09 kN·m^2 EId_2 = 192.35 kN-m^2
or
d = \frac{1}{EI}\begin{bmatrix} -154.09 \\ 192.35 \end{bmatrix} kN-m^2Member End Displacements and End Forces
Member 1 By using the member’s structure degree of freedom numbers, we obtain the member end displacements:
u_1 = v_1 = \begin{bmatrix} v_1 \\ v_2 \\v_3 \\v_4\end{bmatrix} \begin{matrix} 0 \\ 0 \\ 0\\1 \end{matrix} = \begin{bmatrix} 0 \\ 0 \\0\\d_1\end{bmatrix} = \frac{1}{EI}\begin{bmatrix} 0 \\ 0 \\0\\-154.09\end{bmatrix}By using the member stiffness relations Q = ku + Q_f (Eq. (18.4)), we compute member end forces as
Q = ku + Q_f (18.4)
F_1 = Q_1 = EI\begin{bmatrix}0.012 & 0.06 & -0.012 &0.06 \\ 0.06 & 0.4 & -0.06 &0.2 \\ -0.012 & -0.06 &0.012 &-0.06\\0.06 & 0.2 &-0.06 &0.4 \end{bmatrix} \frac{1}{EI}\begin{bmatrix} 0 \\ 0 \\0\\-154.09\end{bmatrix} + \begin{bmatrix} 28.16 \\ 76.8 \\ 51.84 \\ -115.2 \end{bmatrix}= \begin{bmatrix}18.91 kN \\ 45.98 kN-m\\61.09 kN \\ -176.84 kN-m \end{bmatrix}
Member 2
u_2 = v_2 = \begin{bmatrix} v_1 \\ v_2 \\v_3 \\v_4\end{bmatrix} \begin{matrix} 0 \\ 1 \\ 0\\2 \end{matrix} = \begin{bmatrix} 0 \\ d_1 \\0\\d_2\end{bmatrix} = \frac{1}{EI}\begin{bmatrix} 0 \\ -154.09 \\0\\192.35\end{bmatrix}Q = ku + Q_f
F_2 = Q_2 = \begin{bmatrix}0.012 & 0.06 & -0.012 &0.06 \\ 0.06 & 0.4 & -0.06 &0.2 \\ -0.012 & -0.06 &0.012 &-0.06\\0.06 & 0.2 &-0.06 &0.4 \end{bmatrix}\begin{bmatrix} 0 \\ -154.09 \\0\\192.35\end{bmatrix} + \begin{bmatrix} 120 \\ 200 \\120\\-200\end{bmatrix}
= \begin{bmatrix}122.3 kN \\ 176.83 kN-m\\117.7 kN \\ -153.88 kN-m \end{bmatrix}
Member 3
u_3 = v_3 = \begin{bmatrix} v_1 \\ v_2 \\v_3 \\v_4\end{bmatrix} \begin{matrix} 0 \\ 2 \\ 0\\0 \end{matrix} = \begin{bmatrix} 0 \\ d_2 \\0\\0\end{bmatrix} = \frac{1}{EI}\begin{bmatrix} 0 \\ 192.35 \\0\\0\end{bmatrix}Q = ku + Q_f
F_3 = Q_3 = \begin{bmatrix}0.096 & 0.24 & -0.096 &0.24 \\ 0.24 & 0.8 & -0.24 &0.4 \\ -0.096 & -0.24 &0.096 &-0.24\\0.24 & 0.4 &-0.24 &0.8 \end{bmatrix}\begin{bmatrix} 0 \\ 192.35 \\0\\0\end{bmatrix} = \begin{bmatrix} 46.16 kN \\ 153.88 kN-m \\-46.16 kN\\76.94 kN-m\end{bmatrix}
The end forces for the three members of the continuous beam are shown in Fig. 18.12(e).
Support Reactions Since support joint 1 is the beginning joint for member 1, equilibrium considerations require that the reactions at joint 1, R_①, be equal to the upper half of F_1 (i.e., the forces at end 1 of member 1).
R_① = \begin{bmatrix} 18.91 kN\\45.98 kN-m\end{bmatrix}in which the first element of R_① represents the vertical force and the second element represents the moment, as shown in Fig. 18.12(f ). In a similar manner, since support joint 2 is the end joint for member 1 but the beginning joint for member 2, the reaction vector at joint 2, R_②, must be equal to the algebraic sum of the lower half of F_1 and the upper half of F_2.
R_② =\begin{bmatrix} 61.09\\-176.84\end{bmatrix} + \begin{bmatrix} 122.3 \\176.83 \end{bmatrix} = \begin{bmatrix} 183.39 kN\\-0.01≈0\end{bmatrix}Similarly, at support joint 3, R_③ can be determined by algebraically summing the lower half of F_2 and the upper half of F_3.
R_③ =\begin{bmatrix}117.7\\-153.88\end{bmatrix} + \begin{bmatrix} 46.16\\153.88\end{bmatrix} = \begin{bmatrix} 163.86 kN\\0\end{bmatrix}Finally, the reaction vector at joint 4 must be equal to the lower half of F_3:
R_④ = \begin{bmatrix} -46.16 kN\\76.94 kN-m\end{bmatrix}The support reactions are shown in Fig. 18.12(f ).
Equilibrium Check Applying the equations of equilibrium to the entire structure (Fig. 18.12(f )), we obtain
+\uparrow \sum F_{Y}=018.91 – 80 + 183.39 – 24(10) + 163.86 – 46.16 = 0 Checks
+\circlearrowleft \sum M_{④}=045.98 – 18.91(25) + 80(19) – 183.39(15) + 24(10)(10) – 163.86(5) + 76.94 = 0.02 ≈ 0 Checks
