Eng. D. E. Romanov.

Determination of the Sum of Shear Forces in a Segment of a Reinforced-Concrete Bar with Variable Depth from Bending Moments.

Reading the most recent literature on reinforced concrete, which considers the question of finding shear forces from moments¹, I arrived at several new, exact, and simple conclusions for bars of variable depth, which I bring to attention. For clarity and completeness of exposition, let us begin with results already known from Löser’s works for bars of constant depth \(h\), simplifying and in many respects clarifying the question.

1) Relation between \(M\) and \(Q\).

From the theory of structures it is known that in a section of a bent bar

\[ \frac{dM}{dx}=Q,\ldots\ldots\ldots\ldots\ldots\ldots\ldots (1) \]

i.e. \(dM=Q\,dx\), where \(Q\) is the shear force.

Taking any segment of the bar with abscissae \(a\) and \(b\), of length \(b-a=c\), we find that (Fig. 1)

\[ \int_a^b dM=\int_a^b Q\,dx, \]

i.e.

\[ M_b-M_a \quad \Delta M=\int_a^b Q\,dx=\Omega_a,\ldots\ldots\ldots (2) \]

since \(\int_a^b Q\,dx\) is the area of the diagram \(Q\) within the limits of the segment. Thus, the area of the diagram of shear forces over the segment is equal to the increment of the moment within the limits of this segment.

2) Shear forces and tangential stresses in a bar of constant section.

In any section the sum of the tensile stresses, or the so-called internal tensile force (Fig. 2),

\[ Z=\frac{M}{z},\ldots\ldots\ldots\ldots\ldots\ldots\ldots (3) \]

where \(z\) is the distance between the center of tension and the center of compression, or the so-called lever arm of the internal forces. If the tension in the concrete is neglected, this is the distance from the center of compression to the center of the reinforcement. The magnitude \(z\) depends on the shape of the section and on the degree of reinforcement. With a constant section it is constant.

Taking the derivative with respect to the abscissa \(x\), we obtain:

\[ \frac{dZ}{dx}=\frac{1}{z}\cdot\frac{dM}{dx}; \]

whence

\[ dZ=\frac{1}{z}\cdot dM. \]

But the increment of the tensile force is equal to the shearing force along the neutral layer over the length \(dx\) (Fig. 3), i.e.,

\[ T\cdot dx=dZ=\frac{dM}{z}. \]

The tangential stress will be equal to

\[ \tau=\frac{T}{b}=\frac{dM}{b\cdot z\cdot dx}=\frac{Q}{b\cdot z}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (4) \]

It is usually determined from this formula.

In reinforced-concrete bars it is important to find the sum of the shearing forces over a segment. It will be equal to

\[ \sum_a^b T=\int_a^b T\,dx=\int_a^b dZ=\frac{1}{z}\int_{M_a}^{M_b} dM, \]

whence

\[ \boxed{\sum_a^b T=Z_b-Z_a=\frac{1}{z}(M_b-M_a)=\frac{\Delta M}{z}} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (5) \]

Thus, the sum of the shearing forces along the neutral layer over a segment of a bent bar of constant section is equal to the increment of the internal forces over the segment, or, what is the same, to the increment of the moment divided by the lever arm of the internal forces.

This proposition, derived by Löser for bars of constant section, considerably facilitates the work of calculating reinforced-concrete structures, since the sum of the shearing forces needed for the calculation of stirrups and bent-up bars is computed directly from the bending moments, without calculating the transverse forces; that is, the work is considerably reduced. The quantities \(M\), however, must in any case be found for selecting the sections.

Knowing \(\Sigma T\), it is easy to find the required area of stirrups and bent-up bars. If the action of the concrete in shear is not taken into account, as is required by our standards for reinforced concrete, then

\[ \boxed{\frac{\Sigma T}{\sigma_e}=F_b+F'\cdot\frac{\cos \xi}{0.707}} \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (6) \]

\(^{1}\) B. Löser, Bemessungsverfahren… 1925, p. 72.

Here \(F_b\) is the cross-sectional area of the stirrups in the segment; \(F_s\) is the area of the bent-up bars within it; \(\sigma_e\) is the permissible tensile stress of the steel; and \(\beta\) is the angle between the axis of the bent-up bar and a line running at \(45^\circ\). Most often the bars are bent up at \(\angle 45^\circ\), i.e. \(\angle \beta = 0\), \(\cos \beta = 1\). By specifying the number of stirrups, i.e. their area \(F_b\), we find the required area of bent-up bars.

To facilitate the calculations, F. Weil compiled tables¹.

3) Shearing stresses in bars of variable depth.

In a reinforced-concrete bar, along its length there may vary the depth \(h\), the width \(b\), and the amount of steel \(F_e\). With respect to the calculation of shearing forces, what is important to us is only the change in the quantity \(z\)—the lever arm of the internal forces.

In a bar of rectangular section, \(z\) does not depend on \(b\). It depends chiefly only on the depth \(h\), and more weakly on \(F_e\). Moreover, the amount of steel always changes by jumps from segment to segment and, thus, generally speaking, it is always possible to choose a segment with \(F_e = \mathrm{const}\). We may take \(z = kh\), where \(k\) varies from 0.67 to 1.00 as extreme limits; usually \(k \simeq 7/8\).

For any section we shall have:

\[ Z=\frac{M}{z}=\frac{M}{kh}\ldots\ldots\ldots\ldots\ldots (7) \]

If the coefficient \(k\) is regarded as constant within the limits of the segment under investigation, then, differentiating expression (7), we obtain:

\[ T\,dx=dZ=\frac{k.h.dM-M.k\,dh}{k^2.h^2} =\frac{dM}{k.h}-\frac{M.dh}{k.h^2}\ldots\ldots (8) \]

Whence we directly obtain the known formula for shearing stresses:

\[ \tau=\frac{T.dx}{b.dx} =\frac{dM}{dx}\cdot\frac{1}{k.h.b} -\frac{M.dh}{k.h^2.b.dx},\ \text{i.e.} \]

\[ \boxed{\tau=\frac{Q}{b.z}-\frac{M.\tg\alpha}{h.z.b}}\ldots\ldots\ldots\ldots (9) \]

because

\[ \frac{dM}{dx}=Q,\quad \text{and}\quad \frac{dh}{dx}=\tg\alpha\ldots\ldots\ldots (10) \]

is the tangent of the angle between the tangents to the contour of the useful depth of the bar (Fig. 4), or, what is the same, to the contour of the full depth, since the reinforcement is almost always laid parallel to the edge of the bar. In the case of a straight haunch this (Fig. 5) will be its inclination, since

\[ h=h_0+x.\tg\alpha \]

and

\[ \frac{dh}{dx}=\operatorname{tg}\alpha=\frac{h_a-h_0}{a}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (11) \]

In formula (9) for \(T\) all the quantities \(Q\), \(M\), and \(\operatorname{tg}\alpha\) are substituted with their own signs.

If \(h\) increases, reckoning always from left to right, then the derivative

\[ \frac{dh}{dx}=\operatorname{tg}\alpha \]

will be positive, and conversely. Moreover, from formula (11) for a straight haunch it is immediately clear that if \(h_a > h_0\), i.e., the depth increases, then \(\operatorname{tg}\alpha\) is positive, etc.

For example, for the left fixed end of a beam (Fig. 6) the depth will be decreasing, i.e. \(\operatorname{tg}\alpha\) is negative, and if \(M\) is also negative, then, in numerical magnitudes, in formula (9) the minus sign in the second term will be retained. For the analogous right end (Fig. 7) the depth will be increasing, i.e. \(\operatorname{tg}\alpha\) is positive, and if the moment is negative, then in formula (9), in the second term, we shall obtain a plus sign. But since the shearing force will in this case be negative, then, taking the minus sign outside the parentheses, we shall obtain in the parentheses the difference between the magnitudes, etc. In the case of a haunch thickening not from below, but from above, or from above and below simultaneously (Fig. 4), the reasoning is unchanged. When the depth varies along a curve, \(\operatorname{tg}\alpha\) is obtained from the equation

\[ \frac{dh}{dx}=\operatorname{tg}\alpha . \]

For example (Fig. 8), a parabolic haunch. The depth \(h\) varies according to the equation:

\[ h=h_0-\frac{s}{a^2}\cdot x(2a-x). \]

Then

\[ \frac{dh}{dx}=-\frac{2s}{a^2}\cdot(a-x) \]

and, putting \(x=0\), we shall have

\[ \operatorname{tg}\alpha_0=-\frac{2s}{a} \]

(a negative quantity).

I have deliberately dwelt on the question of signs in formula (9) in greater detail, since in the literature there are no exhaustive indications, and therefore in practical work ambiguity is often encountered.

4) Determination of \(\Sigma T\) over a segment with an arbitrary variation of depth
(according to the author’s investigations).

Let us determine the relationship between the sum of the shearing forces over a segment of variable depth and the bending moments. The existing formula of Löser for this purpose (see the cited work by him, p. 72) is unsatisfactory, since it is approximate and complicated. Let us try to find an exact relationship. For this it is necessary to integrate our equation (8) within the limits of the segment, from \(0\) to \(a\).

We obtain:

\[ \sum_0^a T=\int_0^a T\,dx =\int_{M_0}^{M_a}\frac{dM}{k\cdot h} -\int_{h_0}^{h_a}\frac{M\cdot dh}{k\cdot h^2} \]

or, taking \(k=\mathrm{const.}\) (see above),

\[ \sum_{0}^{a} T = \frac{1}{k}\int_{M_0}^{M_a}\frac{dM}{h} - \frac{1}{k}\int_{h_0}^{h_a}\frac{M\cdot dh}{h^2} \qquad \ldots\ldots (12) \]

Unfortunately, we are deprived of the pleasant possibility of solving these integrals in the most general form, since \(M\) and \(h\), each separately, have their own functional dependences on \(x\). Moreover, \(M\) also depends on the nature of the distribution of the load. Therefore we shall specify these dependences, i.e., we shall solve the problem by proceeding from the particular to the general.

a) In the case of a bar of equal resistance (the more general case), the height \(h\) may be expressed, for \(b=\mathrm{const.}\), as

\[ h=c\cdot \sqrt{M}, \]

where \(c=\mathrm{const.}\)

Further,

\[ dh=c\cdot \frac{dM}{2\sqrt{M}}; \]

substituting into equation (12), we have:

\[ \sum T = \frac{1}{k}\int_{h_0}^{h_a}\frac{2\sqrt{M}}{c^2\sqrt{M}}\cdot dh - \frac{1}{k}\int_{h_0}^{h_a}\frac{h^2}{c^2h^2}\,dh = \frac{h_a}{kc^2} - \frac{h_0}{kc^2}. \]

But since

\[ c=\frac{h}{\sqrt{M}}; \qquad c^2=\frac{h^2}{M} = \frac{h_a^2}{M_a} = \frac{h_0^2}{M_0}. \]

Having made the substitution, we obtain

\[ \sum_{0}^{a}T = \frac{h_a\cdot M_a}{k\cdot h_a^2} - \frac{h_0\cdot M_0}{k\cdot h_0^2} = \frac{M_a}{kh_a} - \frac{M_0}{kh_0}. \]

But for us \(kh_a=z_a\); \(kh_0=z_0\), i.e.,

\[ \boxed{ \sum_{0}^{a}T=\frac{M_a}{z_a}-\frac{M_0}{z_0} } \qquad \ldots\ldots (13) \]

We have obtained a very important relation: the sum of the shearing forces over a segment of variable height is equal to the difference between the moments at its boundaries, divided by the lever arms of the internal forces; i.e., it is equal to the increment of the internal forces in each zone. Moreover, it is entirely independent of the intermediate variation of \(h\) and \(M\).

Let us verify this proposition for other, more particular cases.

b) Let \(M\) vary according to a parabola of the 2nd degree, and \(h\) along a straight line (Fig. 9). In this case the moment in any section may be represented in the form

\[ M=\frac{x}{a}(M_a-M_0)+M_0+\frac{4m}{a^2}\cdot x(a-x). \]

Denoting

\[ M_a-M_0=\Delta M_1; \qquad \frac{4m}{a^2}=c, \]

we obtain

\[ M=\frac{\Delta M_1}{a}\cdot x+M_0+c(a-x)x, \]

and the differential \(M\) will be

\[ dM=\frac{\Delta M_1}{a}\,dx+c(a-2x)\,dx. \]

Further, in the case of a variation of \(h\) along a straight line,

\[ h=h_0+x\tg\alpha; \]

whence

\[ dh=\tg\alpha\cdot dx,\quad \text{and}\quad x=\frac{h-h_0}{\tg\alpha}. \]

Substituting the quantities obtained into the principal expression (12) for \(\Sigma T\) and solving each integral separately, we obtain:

\[ \frac{1}{k}\int_{M_0}^{M_a}\frac{dM}{h} = \frac{1}{k}\int_0^a \frac{\dfrac{\Delta M_1}{a}+c(a-2x)}{h}\,dx = \]

\[ = \frac{\Delta M_1}{k\,a}\int_{h_0}^{h_a}\frac{dh}{\tg\alpha\cdot h} + \frac{c}{k}\int_{h_0}^{h_a} \frac{a-\dfrac{2(h-h_0)}{\tg\alpha}}{h}\cdot \frac{dh}{\tg\alpha}. \]

Replacing

\[ a\cdot \tg\alpha=s=h_a-h_0, \]

we shall have

\[ \frac{1}{k}\int_{M_0}^{M_a}\frac{dM}{h} = \frac{\Delta M_1}{k\cdot s}\ln\frac{h_a}{h_0} + \frac{c\cdot a^2}{k\cdot s^2} \int_{h_0}^{h_a}\frac{h_a h_0+h_0-2h}{h}\,dh = \]

\[ = \frac{\Delta M_1}{k\cdot s}\ln\frac{h_a}{h_0} + \frac{4m}{k\cdot s^2} \left[ (h_a+h_0)\ln\frac{h_a}{h_0}-2s \right]. \]

We pass to the 2nd term of equation (12).

\[ \frac{1}{k}\int_{h_0}^{h_a}\frac{M\,dh}{h^2} = \frac{1}{k}\int_{h_0}^{h_a} \left[ \frac{\Delta M_1}{a}\cdot x+M_0+c\cdot x(a-x) \right]\frac{dh}{h^2} = \]

\[ = \frac{\Delta M_1}{a\cdot k} \int_{h_0}^{h_a}\frac{(h-h_0)}{\tg\alpha\cdot h^2}\,dh + \frac{M_0}{k}\int_{h_0}^{h_a}\frac{dh}{h^2} + \frac{c}{k}\int_{h_0}^{h_a} \frac{(h-h_0)}{\tg\alpha} \left(a-\frac{h-h_0}{\tg\alpha}\right) \frac{dh}{h^2} = \]

\[ = \frac{\Delta M_1}{k\cdot s} \left( \int_{h_0}^{h_a}\frac{dh}{h} - h_0\int_{h_0}^{h_a}\frac{dh}{h^2} \right) + \frac{M_0}{k}\int_{h_0}^{h_a}\frac{dh}{h^2} + \frac{c\cdot a^2}{k\cdot s^2} \int_{h_0}^{h_a} \frac{(h-h_0)(h_a-h)}{h^2}\,dh. \]

All the integrals are easily evaluated.

We obtain

\[ \frac{1}{k}\int_{h_0}^{h_a}\frac{M\,dh}{h^2} = \frac{\Delta M_1}{k\cdot s} \left( \ln\frac{h_a}{h_0}-\frac{s}{h_a} \right) + \frac{M_0\cdot s}{k\cdot h_a\cdot h_0} + \]

\[ + \frac{c\cdot a^2}{k\cdot s^2} \left[ h_a\int_{h_0}^{h_a}\frac{dh}{h} - \int_{h_0}^{h_a}dh - h_a\cdot h_0\int_{h_0}^{h_a}\frac{dh}{h^2} + h_0\int_{h_0}^{h_a}\frac{dh}{h} \right] = \]

\[ = \frac{\Delta M_1}{k\cdot s} \left( \ln\frac{h_a}{h_0}-\frac{s}{h_a} \right) + \frac{M_0\cdot s}{k\cdot h_a\cdot h_0} + \frac{c\cdot a^2}{k\cdot s^2} \left[ (h_a+h_0)\ln\frac{h_a}{h_0}-2s \right] = \]

\[ = \frac{\Delta M_1}{k\cdot s}\ln\frac{h_a}{h_0} - \frac{1}{k\cdot h_a} \left( \Delta M_1-M_0\cdot\frac{s}{h_0} \right) + \frac{4m}{k\cdot s^2} \left[ (h_a+h_0)\ln\frac{h_a}{h_0}-2s \right]. \]

The complete sum of the shearing forces on the segment from \(o\) to \(a\)

\[ \sum_{0}^{a} T = \frac{1}{\kappa}\int_{M_0}^{M_a}\frac{dM}{h} - \frac{1}{\kappa}\int_{h_0}^{h_a}\frac{Mdh}{h^2} = \frac{\Delta M_1}{\kappa\cdot s}\ln\frac{h_a}{h_0} + \]

\[ + \frac{4m}{\kappa\cdot s^2} \left[ (h_a+h_0)\ln\frac{h_a}{h_0}-2s \right] - \frac{\Delta M_1}{\kappa\cdot s}\ln\frac{h_a}{h_0} + \]

\[ + \frac{1}{\kappa\cdot h_a} \left( \Delta M_1-M_0\frac{s}{h_0} \right) - \frac{4m}{\kappa s^2} \left[ (h_a+h_0)\ln\frac{h_a}{h_0}-2s \right] = \]

\[ = \frac{1}{\kappa h_a} \left( M_a-M_0-M_0\frac{s}{h_0} \right) = \frac{M_a}{\kappa h_a} - \frac{M_0}{\kappa h_0}. \]

But \(\kappa\cdot h_a=z_a,\quad \kappa\cdot h_0=z_0,\) i.e.

\[ \boxed{ \sum_{0}^{a} T=\frac{M_a}{z_a}-\frac{M_0}{z_0} } = Z_a-Z_0 \qquad\ldots\ldots (13). \]

We have obtained the same result as in case (a).

c) Having carried out the analogous integration for the case, for example, of variation of \(M\) along a straight line, we obtain the very same result¹. It may perhaps be said that, for any variation of \(M\) and \(h\), for \(\Sigma T\) the expression obtained (13) will hold.

For a bar, let us assume, consisting of two segments differing in the variation of \(h\) (Fig. 10), we obtain²

\[ \sum_{0}^{b}T = \sum_{0}^{a}T_{\mathrm{I}} + \sum_{a}^{b}T_{\mathrm{II}} = \frac{M_a}{z_a} - \frac{M_0}{z_0} + \frac{M_b}{z_b} - \frac{M_a}{z_a}. \]

\[ \boxed{ \sum_{0}^{b}T=\frac{M_b}{z_b}-\frac{M_0}{z_0} } \qquad\ldots\ldots (14). \]

5) General conclusion.

Summarizing the conclusions obtained, we shall state the rule:

In a bent reinforced-concrete bar, the sum of the shearing forces along the neutral layer over any segment, with arbitrary variation of the section, is equal to the difference between the moments at the boundaries of the segment divided by their arms of the internal forces. Or, in other words—the sum of the shearing forces is equal to the increment of the internal forces over a segment of any length and shape.

In general form this will be:

\[ \boxed{ \sum_{1}^{2}T=\frac{M_2}{z_2}-\frac{M_1}{z_1} } \qquad\ldots\ldots (15) \]

In order to obtain \(\Sigma T_{\max}\), it is necessary to take such a distribution of the load, i.e. such values of \(M\) at the boundaries of the segment, that their difference will be greatest. If the moments at the boundaries have different signs, then, of course, the difference is numerically transformed into the sum of the quantities \(\frac{M_2}{z_2}\) and \(\frac{M_1}{z_1}\).

The relation obtained by us is simple to calculate and has complete accuracy, which cannot be said, for example, of Löser’s approximate formula or analogous formulas obtained from the area of the diagram of shearing stresses¹.

Let us explain the use of formula (15) by a numerical example.

b) Example.

a) Conditions.

A beam with straight haunches, elastically fixed at the ends, has a span \(l = 8.0\ \text{m}\), a useful depth at the supports \(h_A = h_B = 1.20 - 0.05 = 1.15\ \text{m}\), and in the span \(h_o = 0.75 - 0.05 = 0.70\ \text{m}\) (Fig. 11).

At the ends act the moments:

\[ M_A = - 6.0\ \text{t}\cdot\text{m}. \]

\[ M_B = -12.0\ \text{t}\cdot\text{m}. \]

The load is concentrated: \(P = 12\ \text{t}\) at a distance \(c = 3\ \text{m}\); in this case

\(M_{\max} = +12.64\ \text{t}\cdot\text{m}\). Determine the greatest sums of shearing forces?

b) Solution by formula (15).

We have two segments with a boundary in the section with \(M_{\max}\).

Taking approximately \(z = \frac{7}{8}h\) (it may also be taken exactly), we obtain, by formula (15):

\[ \Sigma T_I = \frac{M_{\max}}{z_o} - \frac{M_A}{z_A} = \frac{12.64}{\frac{7}{8}\cdot 0.70} + \frac{6.0}{\frac{7}{8}\cdot 1.15} = +26.60\ \text{t}. \]

\[ \Sigma T_{II} = \frac{M_B}{z_B} - \frac{M_{\max}}{z_o} = -\frac{12.0}{\frac{7}{8}\cdot 1.15} - \frac{12.64}{\frac{7}{8}\cdot 0.70} = -32.57\ \text{t}. \]

The different signs obtained indicate the difference in the direction of bending of the bars in the segments.

c) Solution according to Löser.

For comparison, let us solve this example by Löser’s formula. In doing so, it is first necessary to determine the following additional data.

The difference between the depths in the span and at the support is

\[ s = h_A - h_o = 1.15 - 0.70 = 0.45\ \text{m}. \]

The length of the haunch for an inclination \(\operatorname{tg}\alpha = \frac{1}{3}\) is

\[ a = \frac{s}{\operatorname{tg}\alpha} = 0.45 \cdot 3 = 1.35\ \text{m}. \]

The depth at the middle of the haunch is

\[ h_m^A = h_m^B = \frac{h_A + h_o}{2} = \frac{1.15 + 0.70}{2} = 0.925\ \text{m}. \]

Moment at the beginning of the left haunch

\[ M_a^A=\frac{P(l-c)}{l}\,a+M_A+M_A\,\frac{a}{l} =\frac{12(8-3)}{8}\cdot 1.35-6-12\cdot\frac{1.35}{8} =+2.18\ \text{t m}. \]

Moment at the beginning of the right haunch

\[ M_a^B=\frac{P\cdot c}{l}\cdot a+M_B+M_A\cdot\frac{a}{l} =\frac{12\cdot3\cdot1.35}{8}-12-6\cdot\frac{1.35}{8} =-6.95\ \text{t m}. \]

Moment at the middle of the left haunch

\[ M_m^A=\frac{M_A+M_a^A}{2} =\frac{-6+2.18}{2} =-1.91\ \text{t m}. \]

And, finally, the moment at the middle of the right haunch

\[ M_m^B=\frac{M_B+M_a^B}{2} =\frac{-12-6.95}{2} =-9.475\ \text{t m}. \]

Further, by Löser’s formula, we find:

in the left portion

\[ \sum T_1= \frac{M_{\max}-M_a^A}{z_0} + \frac{M_a^A-M^A+M_m^A\,\dfrac{s}{h_m}}{z_m} = \]

\[ = \frac{12.64-2.18}{\tfrac{7}{8}\cdot0.70} + \frac{2.18+6-1.91\,\dfrac{0.45}{0.925}}{\tfrac{7}{8}\cdot0.925} =+26.05\ \text{t}; \]

in the right portion

\[ \sum T_1= -\left[ \frac{M_{\max}-M_a^B}{z_0} + \frac{M_a^B-M^B+M_m^B\,\dfrac{s}{h_m}}{z_m} \right] = \]

\[ = -\left[ \frac{12.64+6.95}{\tfrac{7}{8}\cdot0.70} + \frac{-6.95+12-9.475\,\dfrac{0.45}{0.925}}{\tfrac{7}{8}\cdot0.925} \right] =-32.54\ \text{t}; \]

d) Comparison of the solutions.

The results obtained are very close to the results computed by our formula (15), but the calculation by Löser’s method itself, as is evident, is much more complicated.

In this case, the calculation of \(\sum T\) from the area of the diagram of shear stresses will likewise be comparable in complexity with respect to the number of operations, as is often done, since there it would also be necessary to compute the shearing forces \(Q\) and the stresses \(\tau\) in several sections, then determine the area of the \(\tau\)-diagram, and so on.

Thus, the practical usefulness of the formula proposed above (16) becomes obvious.

For the article by Eng. D. E. Rotinov

Fig. 1.

Visible labels: \(a\), \(b\), \(c\), \(g\), \(h\), \(y\), \(h-y\), \(x\), \(x_0\), \(\delta\), \(q_a\), \(q_b\), \(M_a\), \(\Delta M\), \(M\), \(M_b\), \(M_{\max}\).

Fig. 2.

Visible labels: \(g\), \(h\), \(y\), \(x\), \(z\), \(M\).

Fig. 3.

Visible labels: \(y\), \(z\), \(h-y\), \(T_{0d}\), \(z=z_0\).

Fig. 4.

Visible labels: \(a\), \(b\), \(x\), \(h_a\), \(h_b\), \(h_c=h_1\), \(M_a\), \(M_b\), \(M_0\), \(y\), \(T\), \(I\), \(II\).

Fig. 5.

Visible labels: \(a\), \(x\), \(y\), \(h_c\), \(h_y\), \(h_0\), \(\alpha\), \(\alpha\).

Fig. 6.

Visible labels: \(a\), \(b\), \(x\), \(y\), \(h_a\), \(h_b\), \(M_a\), \(M_b\), \(m\), \(n\), \(\rho\), \(\rho\), \(S\).

Fig. 7.

Visible labels: \(a\), \(b\), \(\rho\), \(\rho\).

Fig. 8.

Visible labels: \(a\), \(x\), \(y\), \(h_a\), \(h_b\), \(h_0\), \(\alpha_0\), \(\alpha_x\), \(\alpha\), \(\delta\).

Fig. 9.

Visible labels: \(a\), \(x\), \(y\), \(h_y\), \(\alpha\), \(\alpha\).

Fig. 10.

Visible labels: \(a\), \(b\), \(h_a\), \(h_b\), \(M_a\), \(M_b\), \(M_0\), \(y\), \(T\), \(I\), \(II\).

Fig. 11.

Visible labels and numerical annotations:

\[ c=3.0 \]

\[ P=10\ \text{t} \]

\[ \delta=0.45 \]

\[ \ell=8.0\ \text{m} \]

\[ a=1.35 \]

\[ 1:3 \]

\[ 1.20 \]

\[ 0.75 \]

\[ M_A=-6\ \text{tm} \]

\[ M_B=-12.0\ \text{tm} \]

\[ M_{\max}=+18.64\ \text{tm} \]

Additional visible labels: \(A\), \(B\), \(I\) sec., \(II\) sec., \(M_A\), \(M_B\), \(M_a^A\), \(M_a^B\), \(M_{\max}\).

1. When calculating \(\Sigma T\) from the area of the diagram of shearing stresses \(\tau\), the variation of \(\tau\) within the limits of the variable depth is usually taken as linear, which, of course, is not always correct (see formula 9). ↩↩↩↩

2. Löser’s formula \(\Sigma T=\dfrac{\Delta M}{z}\) for constant depth is, as is evident, only a special case of formula (13). ↩