On the Question of Eccentric Compression Beyond the Elastic Range

V. P. Shubin

When calculating members operating under conditions of eccentric compression (Fig. 1), in the region of elastic deformations one uses the well-known formula

\[ \sigma = -\frac{P}{F} \pm \frac{M}{W_{\text{n.o.}}}. \tag{1} \]

Assuming that the compressive force is located in the plane of the principal axis \(Z\), with eccentricity “\(c\),” formula (1) takes the form

\[ \sigma = -\frac{P}{b \cdot h}\left(1 \mp \frac{6c}{h}\right). \tag{2} \]

Fig. 1

As is known, in this case the stress diagram depends on the value of the factor \(\left(1 \mp \dfrac{6c}{h}\right)\).

There are 3 possible cases of stress distribution over the cross-section of the bar.

First case:

The value \(\dfrac{6c}{h} < 1\), i.e. \(c < \dfrac{h}{6}\). In this case the entire cross-section of the bar works only in compression, and the neutral axis of the section lies outside the section (Fig. 2).

Fig. 2

Second case:

The value \(\dfrac{6c}{h} = 1\), i.e. \(c = \dfrac{h}{6}\).

In this case

\[ \text{for } \frac{6c}{h}=+1 \qquad \sigma=-\frac{2P}{b.h.} \]

\[ \text{for } \frac{6c}{h}=-1. \qquad \sigma=0. \]

One of the possible forms of the stress diagram may be the diagram shown in Fig. 3.

Fig. 3

Fig. 4

Third case:

The value \(\dfrac{6c}{h} > 1\), i.e. \(c > \dfrac{h}{6}\). In this case one part of the cross-section of the bar works in compression, and the other part in tension. The stress diagram takes the form shown in Fig. 4.

The position of the neutral axis in all these cases depends on the magnitude of the eccentricity “\(c\)” and is found from the general equation of the zero line, for \(y_0 = 0\).

\[ 1+\frac{Z_0 Z}{i_y^2}=0, \tag{3} \]

where \(Z_0=c\).

\[ i_y^2=\frac{h^2}{12} \]

is the radius of inertia of the cross section. The coordinate \(Z\) of the zero line is represented by the expression

\[ Z=-\frac{h^2}{12c}. \tag{4} \]

The value of “\(K\)”—the distance from the neutral axis to the extreme fibers (Fig. 4)—is readily found from the relation

\[ K=\frac{h}{2}-|Z|=\frac{h}{2}\left(1-\frac{h}{6c}\right). \tag{5} \]

From formula (2), assuming that the bending stress has a minus sign, we determine

\[ P=-\frac{\sigma \cdot b \cdot h}{\left(1-\frac{6c}{h}\right)}. \]

The plastic state is possible when \(\sigma=\sigma_S\)—the yield limit of the material.

Fig. 5

Denoting the quantity \(\sigma_S \cdot b \cdot h\), for values close to the yield limit, by \(P_S'\), we find the condition determining the upper boundary of the elastic stage

\[ P_S=-\frac{P_S'}{\left(1-\frac{6c}{h}\right)}. \tag{6} \]

Under a load \(P>P_S\), the second, elastic-plastic stage begins, characterized by the development of plastic deformation on one side of the section. This phenomenon is represented graphically by the diagrams shown in Figs. 5a and 5b.

The position of the neutral axis (Fig. 5a) can be determined by considering the equilibrium condition

\[ P=\int_F \sigma_x \cdot dF, \]

or

\[ P+\frac{1}{2}\sigma \cdot k \cdot b-\frac{1}{2}\sigma_S(h-a-k)\cdot b-\sigma_S \cdot b \cdot a=0, \]

whence, taking \(\sigma_S \cdot b \cdot h=P_S'\), we have:

\[ a=\frac{2P-P_s'}{\sigma_s b}+\frac{k(\sigma+\sigma_s)}{\sigma_s}. \tag{7} \]

From the geometric similarity of the triangles of elastic deformations and of tensile and compressive stresses, we obtain

\[ \frac{\sigma}{k}=\frac{\sigma_s}{h-k-a}, \quad \text{whence} \]

\[ a=\frac{\sigma h-k(\sigma+\sigma_s)}{\sigma}. \tag{8} \]

Equating the values (7) and (8) and solving them with respect to \(k\), we have:

\[ k=\frac{P_s'-P}{b}\cdot \frac{2\sigma}{(\sigma+\sigma_s)^2}. \tag{9} \]

From equation (7), for a known value of \(k\), one can find the quantity \(a\), which determines the depth of propagation of the zone of plastic deformations,

\[ a=\frac{(2P-P_s')\sigma_s+P_s'\sigma}{b\sigma_s(\sigma+\sigma_s)}. \tag{10} \]

The value of the elastic stress \(\sigma\) in the extreme fiber can be determined from consideration of the equilibrium equation

\[ M=\int_F Z\cdot \sigma_x\,dF, \]

or

\[ P\left(\frac{h}{2}+c\right)+\frac{\sigma k^2 b}{6} -\sigma_s\frac{(h-a-k)b}{2} \]

\[ \cdot\left[k+\frac{2}{3}(h-k-a)\right] -\sigma_sab\left(h-\frac{a}{2}\right)=0. \]

Substituting here the values of \(k\) and \(a\) from formulas (9) and (10) and solving the equation with respect to \(\sigma\), we obtain

\[ A\sigma^4+B\sigma^3+C\sigma^2+D\sigma+e=0, \tag{11} \]

where the coefficients:

\[ A=3Phb+\sigma Pc \]

\[ B=8P_s'^2+20PP_s'+24Pc\,\sigma_s-4P^2 \]

\[ C=6P^2s'\sigma_s-12P^2\sigma_s+36Pc\,\sigma_s^2+42PP_s'\sigma_s \]

\[ D=36PP_s'\sigma_s^2-12P^2\sigma_s^2+24Pc\,\sigma_s^3 \]

\[ E=2P_s'^2\sigma_s^3+8PP_s'\sigma_s^3-4P\sigma_s^3 \]

are known quantities. In these coefficients \(P_s'=bh\cdot\sigma_s\).

Solving equation (11) in general form is very difficult; therefore one may proceed as follows. Introduce the notation

\[ \sigma=y;\quad \frac{B}{A}=a;\quad \frac{C}{A}=b;\quad \frac{D}{A}=c;\quad \frac{E}{A}=d, \]

then

\[ y^4+ay^3+by^2+cy+d=0. \tag{12} \]

We solve equation (12) according to the following plan. The substitution \(y=x-\dfrac{a}{4}\) reduces equation (12) to the form

\[ x^4+px^2+qx-r=0. \tag{13} \]

The solution of this equation will be obtained if one finds the roots of the equation

\[ Z^3+2pZ^2+(p^2-4r)Z-q^2=0. \tag{14} \]

If the roots of the equation are \(Z_1;\ Z_2;\ Z_3\), then the roots of the reduced equation (13) will be:

\[ x_1=\frac{1}{2}\left(\sqrt{Z_1}+\sqrt{Z_2}+\sqrt{Z_3}\right), \]

\[ x_2=\frac{1}{2}\left(\sqrt{Z_1}-\sqrt{Z_2}-\sqrt{Z_3}\right), \]

\[ x_3=\frac{1}{2}\left(-\sqrt{Z_1}+\sqrt{Z_2}-\sqrt{Z_3}\right), \]

\[ x_4=\frac{1}{2}\left(-\sqrt{Z_1}-\sqrt{Z_2}+\sqrt{Z_3}\right). \]

Here

\[ \sqrt{Z_1}\cdot\sqrt{Z_2}\cdot\sqrt{Z_3}=-q. \]

Knowing the quantities \(x_1,\ x_2,\ x_3,\ x_4\), it is easy to find the values of \(y\), i.e. \(\sigma\).

Let us next consider the case when plastic deformations develop simultaneously from the two edges of the cross section, as is shown in Fig. 6.

Fig. 6

Setting up the equilibrium conditions, we have:

\[ \sum_y \left| \begin{aligned} &\sigma_s\cdot b\cdot(a-a_1)=P; \qquad (a-a_1)=\frac{P}{\sigma_s\cdot b}, \end{aligned} \right. \]

\[ \sum M_{n.o} \left| \begin{aligned} &\sigma_s\cdot a_1\cdot b\left(k+\frac{a_1}{2}\right) +\sigma_s k b\,\frac{2}{3}k \\ &\quad +\sigma_s ab\left(k+\frac{a}{2}\right)-P(c+x)=0. \end{aligned} \right. \]

From the geometric relations it is easy to show that

\[ k+\frac{a_1}{2}=\frac{h-a}{2}; \qquad k=\frac{h-a_1-a}{2}; \]

\[ k\cdot \frac{2}{3}k=\frac{1}{3}(h^2-2ha_1-2ha+2aa_1+a_1^2+a^2) \]

\[ k+\frac{a}{2}=\frac{h-a_1}{2}, \]

\[ x=\frac{a-a_1}{2}=\frac{P}{2\sigma_s b}. \]

For these values, the second equilibrium equation assumes the form:

\[ -\frac{\sigma_s bha_1}{6}-\frac{\sigma_s bha}{6} +\frac{\sigma_s bh^2}{3} +\frac{\sigma_s ba_1^2}{3} + \]

\[ +\frac{\sigma_s ba^2}{3}-Pc-\frac{P^2}{2\sigma_s b}=0. \]

Taking into account that \(\sigma_s bh=P_s'\), this equation is transformed into

\[ -\frac{P_s'}{6}(a_1+a)+\frac{P_s'}{3}h+\frac{\sigma_s b}{3}(a_1^2+a^2)-Pc-\frac{P^2}{2\sigma_s b}=0. \tag{15} \]

From the first equilibrium condition, the value of \(a\) can be expressed in terms of \(a_1\)

\[ a=\frac{P+\sigma_s ba_1}{\sigma_s b}. \tag{16} \]

After substituting \(a\) in terms of \(a_1\), substituting into equation (15), and carrying out the corresponding transformations, we obtain:

\[ a_1^2+a_1\left(\frac{2P\sigma_s b-P_s'}{2\sigma_s b}\right) +\frac{2P_s'^2-PP_s'-P^2-6P\sigma_s bc}{4\sigma_s^2b^2}=0, \]

whence the value of \(a_1\) assumes the form:

\[ a_1=\frac{1}{4\sigma_s b}\left[ P_s'(1-2\sigma_s b)\pm \right. \]

\[ \left. \pm\sqrt{ 4Pb\sigma_s(P\sigma_s b+6c)+4PP_s'(1-\sigma_s b)+7P_s'^2+4p^2 } \right]. \tag{17} \]

From equation (16), \(a=\dfrac{P}{\sigma_s b}+a_1\), or

\[ a=\frac{P}{\sigma_s b}+\frac{1}{4\sigma_s b}\left[ P_s'(1-2\sigma_s b)\pm \right. \]

\[ \left. \pm\sqrt{ 4Pb\,\sigma_s(P\sigma_s b+6c)+4PP_s'(1-\sigma_s b)+7P_s'^2+4P^2 } \right]. \tag{18} \]

Thus, we have determined the depths of propagation of the zone of plastic deformations.

From Fig. 6 it follows that the position of the neutral axis, determined by the distance \(Z\), will be:

\[ Z=a_1+k. \tag{19} \]

Making the substitution \(k=\dfrac{h-a-a_1}{2}\), it is easy to obtain

\[ Z=\frac{h}{2}-\frac{P}{2\sigma_s b} =\frac{\sigma_s bh-P}{2\sigma_s b}, \]

whence

\[ Z=\frac{P_s'-P}{2\sigma_s b}. \tag{20} \]