F. E. Davydov.

Graphical Method for Selecting Cross-Sections Under Combined Loading

The proposed graphical method for selecting cross-sections under combined loading makes it possible quickly, accurately, without special effort, by means of a straight line and a circle, to construct diagrams and, from them, to select and check cross-sections for beams of any individual and built-up shapes.

Oblique Bending

Oblique bending occurs when the plane of the bending moment does not coincide with either of the principal axes of the section.

As a consequence of oblique bending, the following is obtained: a) the neutral line is located at an acute angle to the plane of the moment; b) because the plane of the deflection arrows is always perpendicular to the neutral line, it will not coincide with the plane of the bending moment; c) the plane of the deflection arrows lies between the plane of the moment and the plane of easiest bending.

The determination of stresses in oblique bending is carried out by the formula:

\[ \sigma_A=\frac{M_y y}{J_{x_0}}+\frac{M_x x}{J_{y_0}}, \tag{1} \]

where: \(\sigma_A\)—the stress at point \(A\);

\[ M_y=M\cos\alpha;\qquad M_x=M\sin\alpha; \]

\(x\) and \(y\)—the coordinates of point \(A\) with respect to the principal axes of inertia,
\(J_{x_0}\) and \(J_{y_0}\)—the principal moments of inertia.

The selection of cross-sections in oblique bending is carried out by the formula:

\[ W_x=\frac{M_y+C M_x}{R}, \tag{2} \]

where: \(W_x\)—the section modulus of the section with respect to the principal central axis \(X_0\);

\(M_x\) and \(M_y\)—the components of the bending moment along the principal axes \(X_0\) and \(Y_0\) from the bending moment in the plane of the acting forces;

\(R\)—the allowable stress in bending;

\[ C=\frac{W_x}{W_y} \]

—the coefficient, adopted for the preliminary selection of the section and not constant for all shapes; it is taken as the mean value for an individual shape.

Thus:

1. For an I-beam section \(C=8.7\)
2. For channel iron \(C=6.5\)
3. For wooden beams \(C=1.3—1.4\).

Having determined \(W_x\) by formula (2), one selects from the table of shapes a suitable number. The selected number is checked by the formula:

\[ \sigma=\frac{M_y}{W_x}+\frac{M_x}{W_y}\le R \]

If this condition is not satisfied, a larger or smaller section number is taken and the check is repeated.

For other sections, such as Z-sections, angles, and built-up sections, the selection of cross-sections is carried out by trial, even without a preliminary coefficient \(C\).

Consequently, the selection of sections under oblique bending is indeterminate. We decided to eliminate this indeterminacy by resorting to a graphical method, as the most visual and rapid way of solving particular problems.

The basic formula for determining stresses (1) can be transformed as follows:

\[ \sigma_A=\frac{M_y y}{J_{x_0}}+\frac{M_x x}{J_{y_0}} = M\left[ \frac{y\cos\alpha}{J_{x_0}}+\frac{x\sin\alpha}{J_{y_0}} \right] \]

\[ \frac{y}{J_{x_0}}\cos\alpha+\frac{x}{J_{y_0}}\sin\alpha-\frac{\sigma_A}{M}=0 \tag{3} \]

The obtained equation (3) is nothing other than the normal form of the equation of a straight line in rectangular coordinate axes \(\dfrac{y}{J_{x_0}}\) and \(\dfrac{x}{J_{y_0}}\).

For constant values of \(\dfrac{y}{J_{x_0}}\) and \(\dfrac{x}{J_{y_0}}\), but variable \(\dfrac{\sigma_A}{M}\) and \(\alpha\), we obtain a pencil of straight lines \((1—1,\ 2—2,\ 3—3,\ A—A_1,\ A—A_2,\ \text{etc.})\), passing through the point \(A\left[\dfrac{y}{J_{x_0}},\ \dfrac{x}{J_{y_0}}\right]\) (see Drawing 1).

At the angle \(\alpha=0\), i.e., when the acting moment passes along the axis \(Y_0\), equation (3) takes the form:

\[ \frac{y}{J_{x_0}}-\frac{\sigma_A}{M}=0 \quad (\text{equation of the line } A—A_1) \]

or

\[ \sigma_A=\frac{M_y}{J_{x_0}} \]

—the formula for determining normal stresses under plane bending in the plane \(YOZ\).

At \(\alpha=90^\circ\), i.e., when the acting moment passes along the axis \(X_0\), equation (3) takes the form:

\[ \frac{x}{J_{y_0}}-\frac{\sigma_A}{M}=0 \quad (\text{equation of the line } A—A_2) \]

or

\[ \sigma_A=\frac{M_x}{J_{y_0}} \]

—the formula for determining normal stresses under plane bending in the plane \(XOZ\).

The angle \(\alpha\) in the graph is measured counterclockwise from the positive direction of the axis \(\dfrac{y}{J_{x_0}}\), and in the section clockwise from the negative direction of the axis \(Y_0\).

The segment \(OK_i\), where the index \(i=1,2,3\), etc., multiplied by the scale of the drawing, is equal to the ratio \(\dfrac{\sigma_A}{M}\). Suppose that \(OK_i\) is equal to some

the quantity “\(a\),” i.e. \(\dfrac{\sigma_A}{M}=a\), then, for a known bending moment \(M\), we can determine the stress \(\sigma_A=aM\), and, for a known stress \(\sigma_A\), determine \(M\), which will produce this stress at point \(A\), i.e.

\[ M=\frac{\sigma_A}{a}. \]

When selecting cross-sections we choose the dangerous points for which the condition is imposed that the normal stress not exceed the allowable value, i.e. \(\sigma \le R\). Let us plot, in a coordinate field with axes \(\dfrac{y}{J_{x_0}}\) and \(\dfrac{x}{J_{y_0}}\), the dangerous points of the individual numbers of some profile with their coordinates \(\dfrac{y}{J_{x_0}}, \dfrac{x}{J_{y_0}}\) or \(\dfrac{1}{W_x}, \dfrac{1}{W_y}\). For definite values of \(\dfrac{R}{M}\) and \(\alpha\) the straight line

\[ \frac{y}{J_{x_0}}\cos\alpha+\frac{x}{J_{y_0}}\sin\alpha-\frac{R}{M}=0 \]

will occupy a definite position in the field. Then one of the plotted points may lie on this straight line or be closer to it than the others. The profile number fixed by this point should be regarded as the most suitable for satisfying the values \(R, M\), and \(\alpha\).

All right triangles \(OK_iA\) have one and the same hypotenuse \(OA\), and this means that the locus of the points \(K_i\) is the circle constructed on the diameter \(OA\). Having constructed the circle on the diameter \(OA\) and drawn straight lines through the origin at various angles \(\alpha\), we obtain segments of straight lines between the origin and the point of intersection of these lines with the circle, which give us the values \(\dfrac{R}{M}\).

Let us now turn directly to the construction of graphs for characteristic individual profiles.

Construction of a graph for wooden beams.

In drawing 2(a) a wooden beam is shown with dimensions \(b=16\ \mathrm{cm}\), \(h=22\ \mathrm{cm}\). \(A, B, C\), and \(D\) are the dangerous points in bending. \(X_0\) and \(Y_0\) are the principal central axes of inertia. \(M\) is the maximum bending moment from the acting forces; \(M_x\) and \(M_y\) are the components of the moment \(M\) along the axes \(X_0\) and \(Y_0\). \(\alpha\) is the angle measured from the negative direction of the \(Y_0\) axis to the plane of the acting bending moment \(M\).

In drawing 2(b), in the coordinate field with axes \(\dfrac{y}{J_{x_0}}\) and \(\dfrac{x}{J_{y_0}}\), the dangerous points of the profile are plotted:
\(A(0.000774;\ 0.001064)\), \(B(-0.000774;\ 0.001064)\), \(C(-0.000774;\ -0.001064)\), and \(D(0.000774;\ -0.001064)\), and circles are constructed on the diameters \(OA, OB, OC\), and \(OD\). The quantities \(0\!-\!1\), \(0\!-\!2\), \(OK_i\), multiplied by the scale of the drawing, as we have already seen, are equal to \(\dfrac{R}{M}\), i.e.

\[ OK_i=a=\frac{R}{M}, \]

whence \(M=\dfrac{R}{a}\). For known allowable stresses \(R\) and angle \(\alpha\), the bending moment \(M\) depends on the segment \(OK_i=a\). The larger the segment \(OK_i\), the smaller the \(M\) that brings the stress at the point under consideration to

$R$, and conversely. In view of the symmetry of the profile, the points $A$, $C$ and $B$, $D$, for all values of the angle $\alpha$, are brought to $R$ in pairs by identical $M$. Then we may consider not all the points, but only one of each pair, say, $A$ and $B$, and vary the angle $\alpha$ not from $0^\circ$ to $360^\circ$, but from $0^\circ$ to $180^\circ$. For $0<\alpha<90^\circ$ the moment $M$ required to bring the stress to $R$ at point $A$ will be less than for point $B$; consequently, we do not consider the part of the circumference for point $B$, $6'$, $0$, $3'$ and $4'$. For $90^\circ<\alpha<180^\circ$ the moment required to bring the stress to $R$ at point $A$ will be greater than for point $B$; then we do not consider the part of the circumference for point $A$, i.e. $4, 5, 0, 1$. But when the angle $\alpha$ changes from $90^\circ$ to $180^\circ$, the moment $M$ changes in magnitude in the same way as for point $A$ when the angle $\alpha$ changes from $90^\circ$ to $0^\circ$. Consequently, we may consider the change of the angle $\alpha$ not from $0^\circ$ to $360^\circ$, but from $0^\circ$ to $90^\circ$, i.e. consider only the part of the circumference of point $A$ lying in the first quadrant.

Thus, in order to construct the graph for selecting wooden beams of rectangular cross-section, it is necessary for each beam number to construct the parts of circumferences located in the first quadrant. Through the origin, draw straight lines for measuring the angles $\alpha$, and describe from the origin a series of circumferences with radii $OK_i$ for measuring

$$ \frac{R}{M}. $$

This graph is presented in drawings 3 and 4.

In an entirely analogous manner, graphs (drawings 5, 6, 7) are constructed for an I-beam profile and channel iron. Only because the graph is indistinct in the lower zone can this place be drawn on a large scale, for convenience in using it, which we have done for the I-beam profile.

Construction of the graph for the Z-profile OST 29.

When constructing the graph for the Z-profile by the accepted method, we encounter the difficulty of measuring the angle $a$, since the principal axes for the individual numbers are not equally inclined to the vertical wall of the profile. To eliminate this inconvenience, we construct the graph in the following way.

In the coordinate field $\dfrac{y}{J_{x_0}}$ and $\dfrac{x}{J_{y_0}}$ we plot the dangerous points $A$ and $B$ and connect these points with the origin—$O$. Taking into account that $\alpha=\varphi+\beta$, the values of which are clear from the profile shown on the graph (see drawing 8), and the possibility of rotating the axes, we rotate the lines $OA$ and $OB$ to the right by the angle $\varphi$. On them, as on diameters, we draw circumferences. The further construction of the graph is carried out by the accepted method. Consequently, in such a construction of the graph, we measure not the angles $\alpha$, but $\beta$, the determination of which is extremely simple in each particular case.

Example 1. A vertical uniformly distributed load $q=0.4\ t/m$ acts on a purlin of an I-beam profile, serving to support the roofing and laid in the joints of the upper chord of the truss parallel to the ridge of the roof, of length $l=2\ m$. The angle of inclination of the roof slope to the horizon is $\alpha=30^\circ$. The allowable stress is $R=1200\ kg/cm^2$.

Solution. The greatest bending moment:

$$ M_{\max}=\frac{q l^2}{8}=\frac{40\cdot 200^2}{8}=20000\ kg\cdot cm. $$

Ratio:

$$ \frac{R}{M}=\frac{1200}{20000}=0.06. $$

From the found value \(\dfrac{R}{M}\) and the angle \(a = 30^\circ\), we find from the graph that the most suitable number of the I-beam section is No. 14.

Check of stress: \(\sigma = OK_i \cdot M = 0.057 \cdot 20000 = 1140\ \text{kg}/\text{cm}^2.\)

Example 2. Under the conditions of Example 1, select the number of the Z-section profile.

Solution.

\[ M_{\max}=20000\ \text{kg}\cdot\text{cm}; \qquad \frac{R}{M}=\frac{1200}{20000}=0.06, \]

\(\beta = 150^\circ\); suitable numbers of the Z-section profile: No. 6 or No. 8 (see Fig. 8).

Check:

\[ \sigma_{A_t}=0.073 \cdot 20000 = 1460\ \text{kg}/\text{cm}^2 \]

\[ \sigma_{A_b}=0.0423 \cdot 20000 = 846\ \text{kg}/\text{cm}^2. \]

We adopt Z-section profile No. 8.

Below we give the procedure for constructing and using the graphs.

Construction of the Graphs

1. In the coordinate field we plot the necessary dangerous points of the section with their coordinates \(\dfrac{y}{J_{x_0}}\) and \(\dfrac{x}{J_{y_0}}\), and connect them with the origin of coordinates.

2. On the obtained straight lines, as on diameters, taking into account the relationship and the period of variation of the stresses of the dangerous points of the section, we draw parts of circles.

3. In the case of different positions of the principal axes of inertia \(X_0\) and \(Y_0\) relative to the horizontal and vertical axes of individual profile numbers (Z-section iron), the straight line obtained in item (1) is rotated relative to the origin of coordinates clockwise or counterclockwise by the angle \(\varphi\)—the angle of rotation of the principal axes.

4. For measuring the angles \(a\) and \(\beta\), we draw straight lines through the origin of coordinates at various angles.

5. To determine the value \(OK_i = \dfrac{R}{M}\), we describe concentric circles from the origin of coordinates.

Use of the Graphs

1. Determine the maximum bending moment \(M_{\max}\).

2. Calculate the ratio

\[ \frac{R}{M_{\max}}. \]

1. From the \(a\)-angle between \(Y_0\) and the plane of action of the maximum bending moment \(M_{\max}\), or the \(\beta\)-angle between the vertical axis and the plane of action of the moment, and from the found ratio \(\dfrac{R}{M}\), the suitable profile number is found.

2. Check the stress: \(\sigma_{\max}=OK_i \cdot M_{\max} \ll R.\)

The force \(P\) is applied at any point of the end face of the beam and in an arbitrary direction (see drawing No. 10).

Projecting the force \(P\) onto the normal and onto the plane of the end face of the beam, we obtain:

\[ P_1 = P \sin \varphi; \qquad P_2 = P \cos \varphi . \]

Here \(\varphi\) is the angle between the force and the plane of the end face of the beam. The eccentric force \(P_1\) will cause bending, the moment of which will be constant along the entire length of the beam,

\[ M_1 = P_1 d_1 = P d_1 \sin \varphi \]

and tension from a central force equal to \(P_1\).

The force \(P_2\) will cause bending, the moment \(M_2\) of which at the place where the beam is fixed will be equal to:

\[ M_2 = P_2 l = Pl \cos \varphi, \]

and a twisting moment

\[ M_t = P_2 d_2 = P d_1 \sin(\beta_1-\beta_2)\cos\varphi = P d_1 \sin\beta \cos\varphi . \]

We add the two bending moments \(M_1\) and \(M_2\) geometrically in the following manner:

1) Lay off the distance \(d_1\), on the scale of lengths \(m_l\), on a line at an angle \(\varphi\) to the horizontal (see drawing No. 10). Let this be the segment \(OA\). Projecting it onto the vertical axis, we obtain the segment

\[ OB = d_1 \sin \varphi . \]

Multiplying \(OB\) by the scale of lengths \(m_l\) and by the magnitude of the force \(P\), we obtain the bending moment \(M_1\):

\[ M_1 = OB \cdot m_l P = d_1 \sin\varphi \, m_l P = P d_1 \sin\varphi . \]

To obtain the position of the moment \(M_1\), rotate the segment \(OB\) through the angle \(\beta_1\). After this we obtain the segment \(OC\), characterizing the magnitude and position of the moment \(M_1\).

2) Lay off the length of the beam \(l\), on the scale of lengths \(m_l\), on a line at an angle \(\varphi\) to the vertical axis. Let this be the segment \(OD\). Projecting it onto the vertical axis, we obtain the segment

\[ OE = l \cos\varphi . \]

Multiplying \(OE\) by the scale of lengths \(m_l\) and by the magnitude of the force \(P\), we obtain the bending moment \(M_2\):

\[ M_2 = OE \cdot m_l P = l \cos\varphi \, m_l P = Pl \cos\varphi . \]

To obtain the position of the moment \(M_2\), rotate the segment \(OE\) through the angle \(\beta_2\). After this we obtain the segment \(OF\), characterizing the magnitude and position of the moment \(M_2\).

3) The segment \(CF\), as is clear from the drawing, characterizes the magnitude and direction of the resultant moment

\[ M = CF \cdot m_l P . \]

The plane of action of \(M\) passes through the centroid of the beam section and makes an angle \(\alpha\) with the vertical axis.

To determine \(P_1\) and \(P_2\), we proceed as follows: lay off the magnitude \(P\), on the scale of forces \(m_p\), on a line at an angle \(\varphi\) to the horizontal. Let this be the segment \(OT\). Projecting it onto the horizontal and vertical axes, we obtain two segments \(OR\) and \(OQ\).

Multiplying them by the scale of forces \(m_p\), we obtain:

\[ OR \cdot m_p = OT \cdot m_p \cos\varphi = P \cdot m_p \cos\varphi = P_2 m_p = P_2, \]

\[ OQ \, m_p = OT \cdot m_p \sin\varphi = P \cdot m_p \sin\varphi = P_1 m_p = P_1. \]

i.e.

\[ P_2 = OR \cdot m_p; \qquad P_1 = OQ \cdot m_p . \]

To determine the twisting moment

\[ M_t = P_2 \cdot d_2 \]

we lack the magnitude \(d_2\), which, as is not difficult to see, is equal to:

\[ d_2 = KN \cdot m_l . \]

Consequently:

\[ M_t = OR \cdot KN \cdot m_p \cdot m_l . \]

Thus, we have reduced the applied force \(P\) to a central force \(P_1\), a bending moment \(M\), and a twisting moment \(M_t\), which we can determine with the aid of the diagram shown in Fig. No. 10.

The normal stress at the dangerous point of the beam, when selecting cross-sections, must be allowable.

\[ R=M\left(\frac{y}{J_{x_0}}\cos\alpha+\frac{x}{J_{y_0}}\sin\alpha\right)+\frac{P_1}{F}. \]

Dividing this equation by \(M\) and transferring the term \(\dfrac{R}{M}\) to the right-hand side, and

\[ \frac{P_1}{FM}=\frac{1}{Fe} \]

(here \(M=P_1 e\)) to the left, we obtain the following equation:

\[ -\frac{1}{Fe}=\frac{y}{J_{x_0}}\cos\alpha+\frac{x}{J_{y_0}}\sin\alpha-\frac{R}{M}. \]

In our coordinate system \(\left(\dfrac{y}{J_{x_0}},\dfrac{x}{J_{y_0}}\right)\), the quantity \(-\dfrac{1}{Fe}\) is the distance from the point \(A\left(\dfrac{y_A}{J_{x_0}},\dfrac{x_A}{J_{y_0}}\right)\), lying on the side of the origin, as indicated by the minus sign, to the straight line with perpendicular \(\dfrac{R}{M}\), drawn from the origin at an angle \(\alpha\) from the axis \(\dfrac{y}{J_{x_0}}\).

For known values of \(M\), \(e\), \(\alpha\), and \(R\), the selection of sections is carried out as follows:

1. We determine the end of the quantity \(\dfrac{R}{M}\) on the line at an angle \(\alpha\) from the axis \(\dfrac{y}{J_{x_0}}\).

2. To find the point \(A\left(\dfrac{y_A}{J_{x_0}},\dfrac{x_A}{J_{y_0}}\right)\), we need to know the quantity \(\dfrac{1}{Fe}\), which is undetermined, since it depends on the as yet unknown point \(A\left(\dfrac{y_A}{J_{x_0}},\dfrac{x_A}{J_{y_0}}\right)\). To resolve this indeterminacy, we choose the quantity \(\dfrac{1}{Fe}\) so that it is equal to the distance between the end of the segment \(\dfrac{R}{M}\) and the point of intersection of the circle characterizing the chosen profile number with the segment \(\dfrac{R}{M}\). For convenience in selecting the quantity \(\dfrac{1}{Fe}\), we use a diagram. As an example, we have drawn a diagram for an I-beam profile (Nos. 24, 26 … 45), see Fig. No. 9. Using this diagram, we mark on the edge of a paper strip a series of segments with the profile numbers indicated.

3. We place this strip on the segment \(\dfrac{R}{M}\) so that the beginning of the segments \(\dfrac{1}{Fe}\) marked on it coincides with the end of \(\dfrac{R}{M}\), and the quantities \(\dfrac{1}{Fe}\) are directed toward the origin.

1. The coincidence of one of the circles characterizing the profile numbers with the corresponding point plotted on the paper strip will indicate to us the suitable profile number.

We carry out the check by the formula

\[ \sigma = \frac{a}{b} R, \]

where:

\(a\) is the magnitude of the distance from the origin of coordinates to the point of intersection of the circle with the segment \(\frac{-R}{M}\),

\(b\) is the magnitude of the distance from the origin of coordinates to the end of \(\frac{1}{Fe}\).

Having selected the profile number according to the normal stress, we determine, if necessary, the dangerous point with respect to shear stresses and check it by the condition \(\tau_{\max} \leq R_{\tau}\).

Fig. 31.

Fig. 2.

WOODEN BEAMS

Fig. 3.

WOODEN BEAMS

Fig. 4.

[Two nomographic charts are shown. The charts contain families of curved and radial grid lines with angle markings in degrees and labels including \(S_i\), \(\alpha\), \(R\), and \(m_{\max}\). Visible curve labels include, among others, \(12\!-\!16\), \(14\!-\!14\), \(14\!-\!16\), \(16\!-\!16\), \(16\!-\!18\), \(16\!-\!20\), \(18\!-\!18\), \(8\!-\!14\), \(8\!-\!12\), \(10\!-\!12\), \(10\!-\!10\), and \(8\!-\!10\). Angle markings visible include \(5^\circ\), \(10^\circ\), \(15^\circ\), \(20^\circ\), \(25^\circ\), \(30^\circ\), \(35^\circ\), \(40^\circ\), \(45^\circ\), \(50^\circ\), \(55^\circ\), \(60^\circ\), \(65^\circ\), \(70^\circ\), \(75^\circ\), and \(80^\circ\). Vertical numerical scales visible include values such as \(0.0009\), \(0.0011\), \(0.0013\), \(0.0015\), \(0.0017\), \(0.0019\), \(0.0021\), \(0.0023\), \(0.0026\), \(0.0029\), \(0.0032\), \(0.0040\), \(0.0048\), \(0.0056\), \(0.0064\), \(0.0072\), \(0.0080\), \(0.0088\), \(0.0096\), \(0.0108\), and \(0.0120\).]

I-beam steel, St. 16

\(y\)

\(x\)

\(M_{\max}\)

\(\alpha\)

\(1/W_y\)

\(1/W_x\)

Fig. 5.

Fig. 6.

Drawing 5a.

Drawing 6a.

Channel iron
channel sections
OST 17

Chart 7.

Chart 8.

Drawing 9.

Fig. 10.

Visible labels in the diagrams and chart: \(l\), \(P\), \(\varphi\), \(x_0\), \(y_0\), \(\alpha\), \(\beta\), \(\beta_1\), \(\beta_2\), \(\beta_3\), \(Q\), \(R\), \(K\), \(C\), \(O\), \(N\), \(E\), \(F\), \(D\); angular markings \(0^\circ\), \(15^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\), \(75^\circ\), \(90^\circ\); scale markings \(-35\), \(-30\), \(-25\), \(-20\), \(-15\), \(-10\), \(-5\), \(0\), \(5\), \(10\), \(15\), \(20\), \(25\), \(30\), \(35\), \(40\).