On the Calculation of Frame Systems
Abstract Generated abstract
This paper examines a limitation of conventional frame-system calculation schemes that represent members only by their geometric axes and supports. It argues that when member height is taken into account, the usual equilibrium and deformation-continuity assumptions at frame joints may be violated, and proposes a practical approximation in which the joint region is treated as an absolutely rigid body. Using the displacement method, the author derives modified canonical equations and bending-moment expressions that include additional moments caused by linear displacements of section centroids. A numerical example indicates that even for a relatively small joint dimension, the resulting bending moments may differ from those obtained by the usual method by about 20 percent, with larger discrepancies expected as the joint-size-to-span ratio increases.
Full Text
On the Calculation of Frame Systems
B. P. Mitrofanov
(Presented by the scientific seminar of the Department of Strength of Materials)
Calculation schemes of frame systems are usually represented in the form of networks of the geometric axes of the frame elements and support connections.
The subsequent calculation provides for satisfying the conditions of equilibrium and continuity of deformations for the frame elements.
However, these conditions are violated if the calculation scheme is considered with allowance for the height of the elements.
Let us consider this question proceeding from the general principles adopted in structural mechanics for the calculation of frame systems made of members of solid cross-section.
Suppose that the frame joint shown in Fig. 1 possesses only angular mobility in the plane of the drawing.
Fig. 1. Fig. 2.
Then the element \(M\), considered in the section of the girder, will have a displacement \(\Delta x\), while when this same element is considered in the section of the column, it will have a displacement \(\Delta y\).
Consequently, the condition of continuity of deformations and the law of plane sections are not satisfied.
The resulting contradiction may be eliminated in various ways.
Let us consider one of the simple variants that is practically acceptable in a number of cases.
If the deformations of the element \(ABCD\), which, incidentally, are usually very small, are neglected, i.e., if the element \(ABCD\) is assumed to be an absolutely rigid body, then the design scheme of this joint can be represented in the form shown in Fig. 2.
It is obvious that the centers of gravity of the cross sections \(BC\) and \(DC\) will have linear displacements \(\Delta_1\) and \(\Delta_2\), which will lead to the appearance of additional bending moments.
The quantities \(\Delta_1\) and \(\Delta_2\) may be represented in the form:
\[ \Delta_1 = a z_1, \]
\[ \Delta_2 = b z_1. \]
The corresponding bending moments are equal to
\[ M_1 = 6 i_1 \frac{\Delta_1}{l_1}; \]
\[ M_2 = 6 i_2 \frac{\Delta_2}{l_2}, \]
where \(i_1, i_2\) are stiffness coefficients; \(l_1, l_2\) are the lengths of the members.
The canonical equation of the displacement method for a frame joint can be represented in the form
\[ r_{11} z_1 + \sum R_{1p} + \sum M_{\Delta 1i} = 0, \]
where the term \(\sum M_{\Delta 1i}\) represents the total bending moment arising as a result of the linear displacements \(\Delta_1\) and \(\Delta_2\).
The influence of this quantity may be substantial.
Let us consider an example.
Let us estimate the significance of the indicated circumstance from the change in the bending moments of the frame shown in Fig. 3.
Fig. 3. a) design scheme of the frame; б) diagrams \(M\) for the primary system; г) diagrams \(M\), constructed by the usual method (dashed line) and for design scheme a) (solid line).
Let us write the canonical equation of the displacement method, taking into account the bending moments from the displacement \(\Delta\) in the reaction of the constraint \(r_{11}\):
\[ r_{11} z_1 + R_{1p} = 0, \]
where the reactive moment in the imposed constraint from the external load is
\[ R_{1p} = - \frac{Pl}{8} - \frac{Pa}{2} = -0.125\,Pl - 0.5\,\frac{a}{l}\,Pl, \]
and the reactive moment in the imposed constraint caused by a unit rotation of joint \(B\) is
\[ r_{11} = i \left[ 8 + 24\,\frac{a}{l} + 24\left(\frac{a}{l}\right)^2 \right]. \]
The angle of rotation of joint \(B\) is
\[ z_1 = \frac{0.125\,Pl + 0.5\,\dfrac{a}{l}\,Pl} {i \left[ 8 + 24\,\dfrac{a}{l} + 24\left(\dfrac{a}{l}\right)^2 \right]} . \]
The bending moments in sections \(A, B, C\) are equal to
\[ M_A = \frac{\left(2 + 6\,\dfrac{a}{l}\right) \left(0.125\,Pl + 0.5\,\dfrac{a}{l}\,Pl\right)} {\left[ 8 + 24\,\dfrac{a}{l} + 24\left(\dfrac{a}{l}\right)^2 \right]} ; \]
\[ M_B^{\mathrm{st}} = 0.125\,Pl - \frac{\left(4 + 6\,\dfrac{a}{l}\right) \left(0.125\,Pl + 0.5\,\dfrac{a}{l}\,Pl\right)} {\left[ 8 + 24\,\dfrac{a}{l} + 24\left(\dfrac{a}{l}\right)^2 \right]} ; \]
\[ M_B^{\mathrm{rig}} = 0.125\,Pl + \frac{\left(4 + 6\,\dfrac{a}{l}\right) \left(0.125\,Pl + 0.5\,\dfrac{a}{l}\,Pl\right)} {\left[ 8 + 24\,\dfrac{a}{l} + 24\left(\dfrac{a}{l}\right)^2 \right]} ; \]
\[ M_C = 0.125\,Pl + \frac{\left(2 + 6\,\dfrac{a}{l}\right) \left(0.125\,Pl + 0.5\,\dfrac{a}{l}\,Pl\right)} {\left[ 8 + 24\,\dfrac{a}{l} + 24\left(\dfrac{a}{l}\right)^2 \right]} . \]
Figure 3 gives the bending-moment diagram according to the formulas obtained. For \(a = 0.05l\), the dotted line shows the diagram constructed by the usual method.
The difference in the bending moments reaches \(20\%\) and will increase as the ratio \(\dfrac{a}{l}\) increases.