Solution of Internal Forces in Statically Indeterminate Structures Under Localized Distributed Moments

Abstract

Classical methods for manually solving internal forces in statically indeterminate structures mainly include force and displacement methods. While the force method involves substantial work when solving the internal forces of structures with higher degrees of indeterminacy, the displacement method offers a fixed and easily understood approach. However, the displacement method requires prior knowledge of load constant formulas. Common methods for deriving load constant formulas include the force method, virtual beam method, and energy method. Nevertheless, deriving load constant formulas for localized distributed moments using these methods proves to be highly challenging. This study aims to derive load constant formulas for localized distributed moments. Firstly, the load constant formula for a single concentrated moment is derived using the formula for a single concentrated force. Then, the load constant formulas for localized uniform moments and localized linearly distributed moments are derived via the integral method, leveraging the load constant formula for a single concentrated moment. This approach addresses the problem of solving internal forces in statically indeterminate structures under distributed moments via the displacement method. Finally, the proposed approach is verified using three typical examples. The promotion of the research results in this article in teaching can deepen students’ understanding of load constants and the displacement method, enrich teaching content, and have certain engineering applications and teaching practical significance.

1. Introduction

Many engineering structures, such as H-type anti-sliding piles [1,2,3] and row piles [4,5], are statically indeterminate structures. The accurate calculation of their internal forces directly affects engineering safety. Solving internal forces in statically indeterminate structures is a key topic in structural mechanics [6]. Many scholars have researched methods by which to solve internal forces in statically indeterminate structures [7,8,9,10,11], but these studies typically consider common loads, such as concentrated forces, concentrated moments, and distributed loads. There is limited research on solving internal forces in structures under distributed moments. Shan J discovered “anomalies” in the bending moment diagram of a simply supported beam under localized uniform moments. He explained the reason for these “anomalies” by deriving the differential relationship of distributed moment loads and proposed a method for the equivalent transformation of distributed moments [12]. Bin He systematically studied the deflection and stress distribution laws of statically indeterminate slender beams under uniform moments [13,14]. However, these studies did not address the solution of internal forces in statically indeterminate structures under more general full-span or localized distributed moments.

The force and displacement methods are classical approaches by which to solve internal forces in statically indeterminate structures. When the degree of indeterminacy is high, using the force method requires substantial work. In contrast, the displacement method provides a fixed and easily understood approach. It is simpler than the force method for solving internal forces in structures with high degrees of indeterminacy. When using the displacement method to solve internal forces in statically indeterminate structures under localized distributed moments, it is essential to know the load constant formulas for the moments. Currently, none of the structural mechanics literature provides load constant formulas for arbitrary localized distributed moments, though some of the literature uses the energy method to provide load constant formulas for specific cases of localized uniform moments [6]. Generally, load constant formulas can be derived using the force method, the virtual beam method, and the energy method. While these methods are relatively easy for deriving load constant formulas for single concentrated moments and full-span distributed moments, they are very challenging for localized distributed moments. In fact, a distributed moment m(x) can be considered as consisting of numerous concentrated moments m(x)dx. Thus, the integral method can be employed by using the load constant formula for a single concentrated moment to transform the derivation of the load constant formula under localized distributed moments into solving a definite integral of a polynomial function, thereby significantly simplifying the derivation process.

For planar rod structures, this article focuses on the problem of solving the internal forces of statically indeterminate structures under localized distributed moments. Firstly, the load constant formula for a single concentrated moment was derived using the load constant formula for a single concentrated force. Then, the integral method was employed to derive the load constant formula for localized uniform moments and localized linearly distributed moments, leveraging the previously derived load constant formula for a single concentrated moment. This approach allows for solving internal forces in statically indeterminate structures under localized distributed loads using the displacement method. If the direction of the distributed moments acting on the statically indeterminate structure is opposite to that given in this paper, a negative sign can be added to the relevant load constant formulas. The formulas for the shear forces of fixed ends can be obtained from the fixed-end moment formulas combined with the equilibrium conditions of the members and are not further elaborated. This paper also provides examples of solving internal forces in a statically indeterminate multi-span beam, a super-static rigid frame with sidesway, and a super-static rigid frame without sidesway, offering practical teaching and engineering application significance.

2. Derivation of the Load Constant Formula for a Single Concentrated Moment

Most of the structural mechanics literature provides load constant formulas for a single concentrated force. The fixed-end moments for a beam fixed at both ends, subjected to a single concentrated force, are given by Equations (1) and (2) and shown in Figure 1. Similarly, for a beam with one end fixed and the other simply supported, the fixed-end moments are given by Equation (3) and depicted in Figure 2. For a beam fixed at one end and sliding at the other, the fixed-end moments are provided by Equations (4) and (5) and illustrated in Figure 3. To derive the fixed-end moment for a single concentrated moment, the force method can be employed. Alternatively, the problem of finding the fixed-end moment under a concentrated moment can be transformed into a polynomial simplification problem by leveraging the fixed-end moment for a single concentrated force. This is possible because a concentrated moment can be considered as a pair of equal, opposite, and parallel concentrated forces.

$$M_{AB}^{\mathrm{F}}=-{\displaystyle \frac{F_{\mathrm{P}}a{b}^2}{l^2}}$$

(1)

$$M_{BA}^{\mathrm{F}}={\displaystyle \frac{F_{\mathrm{P}}{a}^{2}b}{l^2}}$$

(2)

$$M_{AB}^{\mathrm{F}}=-{\displaystyle \frac{F_{\mathrm{P}}ab(l+b)}{2l^2}}$$

(3)

$$M_{AB}^{\mathrm{F}}=-{\displaystyle \frac{F_{\mathrm{P}}a(l+b)}{2l}}$$

(4)

$$M_{BA}^{\mathrm{F}}=-{\displaystyle \frac{F_{\mathrm{P}}{a}^2}{2l}}$$

(5)

Based on Equations (1)–(5), the load constant formulas for beams fixed at both ends, beams fixed at one end and simply supported at the other, and beams fixed at one end and sliding at the other are derived as follows. This establishes the foundation for deriving the load constant formulas under uniform moments and more complex distributed moments.

2.1. Beam Fixed at Both Ends

Figure 4 shows a beam fixed at both ends subjected to a single concentrated moment. The concentrated moment M can be transformed into two concentrated forces ${\displaystyle \frac{M}{d}}$ spaced at a distance d, as shown in Figure 5. For the upward concentrated force ${\displaystyle \frac{M}{d}}$, the bending moments of the fixed ends can be obtained by substituting $F_{\mathrm{P}}\to -{\displaystyle \frac{M}{d}}$, $a\to a-{\displaystyle \frac{d}{2}}$, and $b\to b+{\displaystyle \frac{d}{2}}$ into Equations (1) and (2). For the downward concentrated force, the bending moments of the fixed ends can be obtained by substituting $F_{\mathrm{P}}\to {\displaystyle \frac{M}{d}}$, $a\to a+{\displaystyle \frac{d}{2}}$, and $b\to b-{\displaystyle \frac{d}{2}}$ into Equations (1) and (2). By applying the superposition principle, the expressions for the load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be derived, as detailed below

$$M_{AB}^{\mathrm{F}}=-{\displaystyle \frac{-\frac{M}{d}(a-\frac{d}{2}){(b+\frac{d}{2})}^2}{l^2}}+\left[-{\displaystyle \frac{\frac{M}{d}(a+\frac{d}{2}){(b-\frac{d}{2})}^2}{l^2}}\right]={\displaystyle \frac{M}{l^2}}(2ab-b^2-{\displaystyle \frac{d^2}{4}})$$

(6)

$$M_{BA}^{\mathrm{F}}={\displaystyle \frac{-\frac{M}{d}{(a-\frac{d}{2})}^2(b+\frac{d}{2})}{l^2}}+{\displaystyle \frac{\frac{M}{d}{(a+\frac{d}{2})}^2(b-\frac{d}{2})}{l^2}}={\displaystyle \frac{M}{l^2}}(2ab-a^2-{\displaystyle \frac{d^2}{4}})$$

(7)

Furthermore, letting $d=0$ in Equations (6) and (7), and substituting $b=l-a$ yields the expressions for the load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ for a beam fixed at both ends under a single concentrated moment, as shown in Equations (8) and (9).

$$M_{AB}^{\mathrm{F}}={\displaystyle \frac{b(3a-l)}{l^2}}M$$

(8)

$$M_{AB}^{\mathrm{F}}={\displaystyle \frac{b(3a-l)}{l^2}}M$$

(9)

2.2. Beam Fixed at One End and Simply Supported at the Other

For a beam fixed at one end and simply supported at the other subjected to a single concentrated moment, as shown in Figure 6, the load constant can be derived using Equation (3) and the same substitutions outlined in Section 2.1, resulting in Equation (10). Hence, further details are omitted.

$$M_{AB}^{\mathrm{F}}={\displaystyle \frac{l^2-3b^2}{2l^2}}M$$

(10)

2.3. Beam Fixed at One End and Sliding at the Other

For a beam fixed at one end and sliding at the other subjected to a single concentrated moment, as shown in Figure 7, the load constants can be derived using Equations (4) and (5) and the same substitutions outlined in Section 2.1, resulting in Equations (11) and (12). Therefore, further details are omitted.

$$M_{AB}^{\mathrm{F}}=-{\displaystyle \frac{b}{l}}M$$

(11)

$$M_{BA}^{\mathrm{F}}=-{\displaystyle \frac{a}{l}}M$$

(12)

3. Derivation of the Load Constant Formula Under Localized Uniform Moments

To solve the internal forces in statically indeterminate structures under localized uniform moments, based on Equations (8)–(12), the load constant formulas for a beam fixed at both ends, a beam fixed at one end and simply supported at the other, and a beam fixed at one end and sliding at the other under localized uniform moments are derived as follows.

3.1. Beam Fixed at Both Ends

To derive the load constant for a beam fixed at both ends under localized uniform moments, as shown in Figure 8, the integral method can be used. By substituting M, a, and b in Equations (8) and (9) with M→mdx, a→x, and b→l − x, respectively, the expressions for $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be obtained as shown in Equations (13) and (14).

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}\frac{(l-x)\cdot (3x-l)}{l^2}}\cdot mdx=-{\displaystyle \frac{m}{l^2}}\left[(a+b){(a+b-l)}^2-a{(a-l)}^2\right]$$

(13)

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}\frac{x\cdot \left[3(l-x)-l\right]}{l^2}\cdot }mdx=-{\displaystyle \frac{M}{l^2}}\left[{(a+b)}^2(a+b-l)-a^2(a-l)\right]$$

(14)

when a = 0 and b = l, the load constant formula for a beam fixed at both ends under full-span uniform moments is obtained.

3.2. Beam Fixed at One End and Simply Supported at the Other

To derive the load constant for a beam fixed at one end and simply supported at the other under localized uniform moments, as shown in Figure 9, the integral method can be employed. By substituting M, a, and b in Equation (10) with the same substitutions outlined in Section 2.1, the expression for $M_{AB}^{\mathrm{F}}$ can be obtained as shown in Equation (15).

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}\frac{l^2-3{(l-x)}^2}{2l^2}\cdot mdx}=-{\displaystyle \frac{m}{2l^2}}\left\{\left[{(a+b)}^3-a^3\right]-3l\left[{(a+b)}^2-a^2\right]+2b{l}^2\right\}$$

(15)

when a = 0 and b = l, the load constant formula for a beam fixed at one end and simply supported at the other under full-span uniform moments is obtained.

3.3. Beam Fixed at One End and Sliding at the Other

To derive the load constant for a beam fixed at one end and sliding at the other under localized uniform moments, as shown in Figure 10, the integral method can be utilized. By substituting M, a, and b in Equations (11) and (12) with the same substitutions as those used in Section 2.1, the expressions for $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be obtained as shown in Equations (16) and (17).

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}-\frac{l-x}{l}\cdot mdx}=-{\displaystyle \frac{b(l-a-\frac{b}{2})}{l}}m$$

(16)

$$M_{BA}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}-\frac{x}{l}\cdot mdx}=-{\displaystyle \frac{b(2a+b)}{2l}}m$$

(17)

when a = 0 and b = l, the load constant formula for a beam fixed at one end and sliding at the other under full-span uniform moments is obtained.

4. Derivation of Load Constant Formulas for Localized Linearly Distributed Moments

Different orientations of linearly distributed moments will result in varying fixed-end moments. Therefore, the derivation should account for the cases in which the linearly distributed moments increase or decrease from left to right. For convenience, when the linearly distributed moments increase from left to right, the coordinate origin is set at the left support, with the vertical axis (load) positive upwards and the horizontal axis (section position) positive to the right. Conversely, when the linearly distributed moments decrease from left to right, the coordinate origin is set at the right support, with the vertical axis (load) positive upwards and the horizontal axis (section position) positive to the left. The meanings of parameters such as $M_{AB}^{\mathrm{F}}$, $M_{BA}^{\mathrm{F}}$, l, F_p, a, and b in the following equations are consistent with those previously mentioned and will not be reiterated here. m₀ represents the maximum intensity of the linearly distributed moments, and c denotes the length (as illustrated in the relevant figures). The following derivations are based on Equations (8)–(12) for a beam fixed at both ends, a beam fixed at one end and simply supported at the other, and a beam with one end fixed and the other end sliding under the influence of localized linearly distributed moments.

4.1. Beam Fixed at Both Ends

4.1.1. Case of Linearly Distributed Moments Increasing from Left to Right

To derive the load constant under the action of linearly distributed moments increasing from left to right, as shown in Figure 11, the integral method can be utilized. By substituting M, a, and b in Equations (8) and (9) with $M\to m_0{\displaystyle \frac{x-a}{b}}\mathrm{d}x$, $a\to x$, and $b\to l-x$, respectively, the expressions for load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be obtained as shown in Equations (18) and (19).

$$\begin{array}{ll}{M}_{AB}^{\mathrm{F}}\hfill & ={\displaystyle {\int }_a^{a+b}\frac{(l-x)\cdot (3x-l)}{l^2}}\cdot {\displaystyle \frac{m_0}{b}}(x-a)dx\hfill \\ & =-{\displaystyle \frac{m_0}{b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(a+b)}^4-a^4\right]-{\displaystyle \frac{4l+3a}{3}}\left[{(a+b)}^3-a^3\right]+{\displaystyle \frac{4la+l^2}{2}}\left[{(a+b)}^2-a^2\right]-ab{l}^2\right\}\hfill \end{array}$$

(18)

$$\begin{array}{ll}{M}_{BA}^{\mathrm{F}}\hfill & ={\displaystyle {\int }_a^{a+b}\frac{x\cdot \left[3(l-x)-l\right]}{l^2}\cdot }{\displaystyle \frac{m_0}{b}}(x-a)dx\hfill \\ & =-{\displaystyle \frac{m_0}{b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(a+b)}^4-a^4\right]-{\displaystyle \frac{2l+3a}{3}}\left[{(a+b)}^3-a^3\right]+la\left[{(a+b)}^2-a^2\right]\right\}\hfill \end{array}$$

(19)

when a = 0 and b = l, the load constant formula for a beam fixed at both ends under full-span linearly distributed moments increasing from left to right is obtained.

4.1.2. Case of Linearly Distributed Moments Decreasing from Left to Right

To derive the load constant under the action of linearly distributed moments decreasing from left to right, as shown in Figure 12, the integral method can be utilized. By substituting M, a, and b in Equations (8) and (9) with the same substitutions outlined in Section 4.1.1, the expressions for the load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be derived. Given the symmetry between the left and right supports of the statically indeterminate beam in Figure 11, the load constant formulas, as shown in Equations (20) and (21), can be obtained by making simple substitutions in Equations (18) and (19): $a\to c$, $M_{AB}^F\to -M_{BA}^F$, and $M_{BA}^F\to -M_{AB}^F$.

$$M_{AB}^{\mathrm{F}}={\displaystyle \frac{m_0}{b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(c+b)}^4-c^4\right]-{\displaystyle \frac{2l+3c}{3}}\left[{(c+b)}^3-c^3\right]+lc\left[{(c+b)}^2-c^2\right]\right\}$$

(20)

$$M_{BA}^{\mathrm{F}}={\displaystyle \frac{m_0}{b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(c+b)}^4-c^4\right]-{\displaystyle \frac{4l+3c}{3}}\left[{(c+b)}^3-c^3\right]+{\displaystyle \frac{4lc+l^2}{2}}\left[{(c+b)}^2-c^2\right]-bc{l}^2\right\}$$

(21)

Setting c = 0 and b = l yields the load constant formula for a beam fixed at both ends subjected to full-span linearly distributed moments decreasing from left to right.

4.2. Beam Fixed at One End and Simply Supported at the Other

4.2.1. Case of Linearly Distributed Moments Increasing from Left to Right

To derive the load constant under the action of linearly distributed moments increasing from left to right, as shown in Figure 13, the integral method can be employed. By substituting M, a, and b in Equation (10) with the same substitutions outlined in Section 4.1.1, the expression for the load constant $M_{AB}^{\mathrm{F}}$ can be obtained as shown in Equation (22).

$$\begin{array}{ll}{M}_{AB}^{\mathrm{F}}\hfill & ={\displaystyle {\int }_a^{a+b}\frac{l^2-3{(l-x)}^2}{2l^2}\cdot \frac{m_0}{b}(x-a)}dx\hfill \\ & =-{\displaystyle \frac{m_0}{2b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(a+b)}^4-a^4\right]-(2l+a)\left[{(a+b)}^3-a^3\right]+(l^2+3al)\left[{(a+b)}^2-a^2\right]+2ab{l}^2\right\}\hfill \end{array}$$

(22)

when a = 0 and b = l, the load constant formula for a beam fixed at one end and simply supported at the other subjected to full-span linearly distributed moments increasing from left to right can be derived.

4.2.2. Case of Linearly Distributed Moments Decreasing from Left to Right

To derive the load constant under the action of linearly distributed moments decreasing from left to right, as shown in Figure 14, the integral method can be used. By substituting M, a, and b in Equation (10) with the following terms, $M\to -m_0{\displaystyle \frac{x-a}{b}}\mathrm{d}x$, $a\to l-x$, and $b\to x$, the expression for the load constant can be obtained as shown in Equation (23).

$$\begin{array}{ll}{M}_{AB}^{\mathrm{F}}\hfill & ={\displaystyle {\int }_c^{c+b}\frac{l^2-3x^2}{2l^2}\cdot \left[-\frac{m_0}{b}(x-c)\right]}dx\hfill \\ & ={\displaystyle \frac{m_0}{2b{l}^2}}\left\{{\displaystyle \frac{3}{4}}\left[{(c+b)}^4-c^4\right]-c\left[{(c+b)}^3-c^3\right]-{\displaystyle \frac{l^2}{2}}\left[{(c+b)}^2-c^2\right]+bc{l}^2\right\}\hfill \end{array}$$

(23)

when c = 0 and b = l, the load constant formula for a beam fixed at one end and simply supported at the other subjected to full-span linearly distributed moments decreasing from left to right can be derived.

4.3. Beam Fixed at One End and Sliding at the Other

4.3.1. Case of Linearly Distributed Moments Increasing from Left to Right

To derive the load constant under the action of linearly distributed moments increasing from left to right, as shown in Figure 15, the integral method can be utilized. By substituting M, a, and b in Equations (11) and (12) with the same substitutions outlined in Section 4.1.1, the expressions for the load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be obtained as shown in Equations (24) and (25).

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}-\frac{l-x}{l}\cdot \frac{m_0}{b}(x-a)dx}={\displaystyle \frac{m_0}{bl}}\left\{{\displaystyle \frac{1}{3}}\left[{(a+b)}^3-a^3\right]-{\displaystyle \frac{l+a}{2}}\left[{(a+b)}^2-a^2\right]+abl\right\}$$

(24)

$$M_{BA}^{\mathrm{F}}={\displaystyle {\int }_a^{a+b}-\frac{x}{l}\cdot \frac{m_0}{b}(x-a)dx}=-{\displaystyle \frac{m_0}{bl}}\left\{{\displaystyle \frac{1}{3}}\left[{(a+b)}^3-a^3\right]-{\displaystyle \frac{a}{2}}\left[{(a+b)}^2-a^2\right]\right\}$$

(25)

when a = 0 and b = l, the load constant formula for a beam fixed at one end and sliding at the other subjected to full-span linearly distributed moments increasing from left to right can be derived.

4.3.2. Case of Linearly Distributed Moments Decreasing from Left to Right

To derive the load constant under the action of linearly distributed moments decreasing from left to right, as shown in Figure 16, the integral method can be employed. By substituting M, a, and b in Equations (11) and (12) in the same substitutions outlined in Section 4.2.2, the expressions for the load constants $M_{AB}^{\mathrm{F}}$ and $M_{BA}^{\mathrm{F}}$ can be obtained as shown in Equations (26) and (27).

$$M_{AB}^{\mathrm{F}}={\displaystyle {\int }_c^{c+b}-\frac{x}{l}\cdot \left[-\frac{m_0}{b}(x-a)\right]dx}={\displaystyle \frac{m_0}{bl}}\left\{{\displaystyle \frac{1}{3}}\left[{(c+b)}^3-c^3\right]-{\displaystyle \frac{c}{2}}\left[{(c+b)}^2-c^2\right]\right\}$$

(26)

$$M_{BA}^{\mathrm{F}}={\displaystyle {\int }_c^{c+b}-\frac{l-x}{l}\cdot \left[-\frac{m_0}{b}(x-a)\right]dx}=-{\displaystyle \frac{m_0}{bl}}\left\{{\displaystyle \frac{1}{3}}\left[{(c+b)}^3-c^3\right]-{\displaystyle \frac{l+c}{2}}\left[{(c+b)}^2-c^2\right]+bcl\right\}$$

(27)

when c = 0 and b = l, the load constant formula for a beam fixed at one end and sliding at the other subjected to full-span linearly distributed moments decreasing from left to right can be derived.

5. Derivation of Load Constant Formula Using the Force Method

In order to illustrate the advantages of this paper’s method in deriving the download constants for the action of locally distributed moments and the reasonableness of this paper’s method, the simplest single statically indeterminate beam (beam fixed at one end and simply supported at the other) subjected to the simplest uniform moment m is taken as an example (Figure 9) by utilizing the most commonly used method of deriving the load constants, the force method.

(1)

Determine the number of times the super-static beam shown in Figure 9 has been super-static for 1 time.

(2)

Selection of the basic system (Figure 17).

The selection of the basic system is not unique; in order to facilitate the calculation, the basic system is obtained as follows after withdrawing the roller support at B and replacing it with an unknown reaction force X₁:

(3)

Write the force method equation

$${\delta }_{11}{X}_1+{\Delta }_{1\mathrm{P}}=0$$

(28)

(4)

Solve the coefficients δ₁₁ and Δ_1P of the force law equation.

The graph of $\overline{M}$ is shown below (Figure 18)

The graph of $M_{\mathrm{P}}$ is shown below.

The graph multiplication by the graph of 18 itself yields

$${\delta }_{11}={\displaystyle \frac{1}{EI}}\left[{\displaystyle \frac{1}{2}}\times l\times l\times ({\displaystyle \frac{2}{3}}\times l)\right]={\displaystyle \frac{l^3}{3EI}}$$

(29)

To solve Δ_1P more easily, the area of Figure 19 can be divided into rectangles and triangles by the dotted line in the figure, which in turn can be obtained by multiplying the graphs of 18 and 19:

$${\Delta }_{1\mathrm{P}}=-{\displaystyle \frac{1}{EI}}\left[mb\times a\times (l-{\displaystyle \frac{a}{2}})+{\displaystyle \frac{1}{2}}\times b\times mb\times (l-a-{\displaystyle \frac{b}{3}})\right]={\displaystyle \frac{m}{EI}}\left[ab({\displaystyle \frac{a}{2}}-l)+{\displaystyle \frac{b^2}{2}}(a+{\displaystyle \frac{b}{3}}-l)\right]$$

(30)

(5)

Solving the fundamental unknown X₁.

Substituting ${\delta }_{11}$ and ${\Delta }_{1\mathrm{P}}$ obtained in step (4) into the force method equation in step (3) yields X₁ as follows:

$$X_1=-{\displaystyle \frac{3m}{l^3}}\left[ab({\displaystyle \frac{a}{2}}-l)+{\displaystyle \frac{b^2}{2}}(a+{\displaystyle \frac{b}{3}}-l)\right]$$

(31)

(6)

Solve the bending moment at end A using the section method (Figure 19)

Take AB rods as the object of study and take the isolation body as shown in Figure 20. In order to facilitate the comparison with the solid end bending moment expression derived using the method of this paper at the end of the A end, remember the bending moment at the end of the A end and the moment equilibrium equations can be obtained by listing moments on the A point as follows:

$$M_{AB}^F=-m\left\{b+{\displaystyle \frac{3}{l^2}}\left[{\displaystyle \frac{b^2}{2}}(a+{\displaystyle \frac{b}{3}}-l)\right]+ab({\displaystyle \frac{a}{2}}-l)\right\}$$

(32)

Comparison of Equations (15) and (32) shows that they are completely equivalent. This shows the correctness of the method of this paper to derive the download constants for the action of locally distributed force moments. In addition, from the derivation in this section, it can be seen that even the simplest single-span super-static beam subjected to the simplest locally uniformly distributed force moments, there is still a great deal of workload in applying the force method for the derivation of the loading constants, which shows the superiority of the method in this paper.

6. Example Verification

This article presents three typical examples, and the use of force methods to calculate the internal forces of a statically indeterminate structure under the action of localized distributed moments is a huge workload. But using the formula derived in the previous text, combined with the displacement method, it can be solved simply and conveniently.

6.1. Example of a Statically Indeterminate Beam

A continuous beam subjected to a distributed load is illustrated in Figure 21. Each member has a bending stiffness of EI. The bending moment diagram of the beam is determined as follows.

(1)

Determine the primary unknown quantity: the angular displacement θ_B at node B.

(2)

Calculate the fixed-end moment for each member (load constant) induced by the load.

For member AB, the fixed-end moment can be determined by substituting $F_{\mathrm{P}}=4 \mathrm{kN}$, $l=3 \mathrm{m}$, $a=1.5 \mathrm{m}$, and $b=1.5 \mathrm{m}$ into Equations (1) and (2) to obtain $M_{AB}^F=-{\displaystyle \frac{3}{2}} \mathrm{kN}\cdot \mathrm{m}$ and $M_{BC}^F={\displaystyle \frac{3}{2}} \mathrm{kN}\cdot \mathrm{m}$.

Member BC is a beam fixed at one end and simply supported at the other. When it is subjected to localized uniform moments, its fixed-end moment can be calculated by substituting $m=20 \mathrm{kN}/\mathrm{m}$, $4=6 \mathrm{m}$, $a=0 \mathrm{m}$, and $b=3 \mathrm{m}$ into Equation (15) to obtain the following: $M_{BC}^F=-{\displaystyle \frac{45}{32}} \mathrm{kN}\cdot \mathrm{m}$.

(3)

Calculate the bending moments at the ends of the members owing to the load and end displacements (load constants + shape constants × end displacements).

For member AB: $M_{AB}=-{\displaystyle \frac{3}{2}}+{\displaystyle \frac{2}{3}}EI{\theta }_B$ and $M_{BA}={\displaystyle \frac{3}{2}}+{\displaystyle \frac{4}{3}}EI{\theta }_B$;

For member BC: $M_{BC}=-{\displaystyle \frac{45}{32}}+{\displaystyle \frac{3}{4}}EI{\theta }_B$.

(4)

Establish the displacement method equations and solve the primary unknown quantity.

Considering the equilibrium at node B (Figure 22), i.e., $\sum M_B=0$, we obtain $M_{BA}+M_{BC}=0$, leading to ${\theta }_B=-{\displaystyle \frac{9}{200EI}}$.

(5)

Determine the end moments of each member and construct the bending moment diagram:

By substituting ${\theta }_B=-{\displaystyle \frac{9}{200EI}}$ into the expressions for the end moments of each member obtained in Step (3), we have the following: $M_{AB}=-1.53 \mathrm{kN}\cdot \mathrm{m}$; $M_{BA}=1.44 \mathrm{kN}\cdot \mathrm{m}$; and $M_{BC}=-1.44 \mathrm{kN}\cdot \mathrm{m}$. Then, use the section method and superposition principle to construct the bending moment diagram as shown in Figure 23 (units of moment in kN·m).

The bending moment diagram in Figure 23 agrees with that obtained using the structural mechanics solver, a computer-aided analysis software package developed by Tsinghua University for educators, students, and engineering professionals [15].

6.2. Example of a Super-Static Rigid Frame Without Sidesway

A super-static rigid frame without sidesway subjected to loading is shown in Figure 24. Each member has a bending stiffness of EI, and axial deformations are ignored. The bending moment diagram of the frame is determined as follows.

(1)

Determine the primary unknown quantity: the angular displacement θ_B at node B.

(2)

Calculate the fixed-end moment for each member (load constant) induced by the load.

Member AB is a beam fixed at both ends subjected to localized uniform moments. Its fixed-end moment can be calculated by substituting $m=2 \mathrm{kN}\cdot \mathrm{m}/\mathrm{m}$, $l=6 \mathrm{m}$, $a=1 \mathrm{m}$, and $b=3 \mathrm{m}$ into Equations (13) and (14) to obtain $M_{AB}^F={\displaystyle \frac{1}{2}} \mathrm{kN}\cdot \mathrm{m}$ and $M_{BA}^F={\displaystyle \frac{3}{2}} \mathrm{kN}\cdot \mathrm{m}$. For member BC, which is subjected to no load, the fixed-end moments are zero.

(3)

Calculate the bending moments at the ends of the members owing to the load and end displacements (load constants + shape constants × end displacements).

For member AB: $M_{AB}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}EI{\theta }_B$ and $M_{BA}={\displaystyle \frac{3}{2}}+{\displaystyle \frac{2}{3}}EI{\theta }_B$;

For member BC: $M_{BC}={\displaystyle \frac{3}{4}}EI{\theta }_B$.

(4)

Establish the displacement method equations and solve the primary unknown quantity.

Considering the equilibrium at node B (Figure 25), i.e., $\sum M_B=0$, we obtain the following: $M_{BA}+M_{BC}=0$, leading to ${\theta }_B=-{\displaystyle \frac{18}{17EI}}$.

(5)

Determine the end moments of each member and construct the bending moment diagram.

Substitute ${\theta }_B=-{\displaystyle \frac{18}{17EI}}$ into the expressions for the end moments of each member obtained in Step (3) to obtain $M_{AB}=0.15 \mathrm{kN}\cdot \mathrm{m}$ and $M_{BA}=0.79 \mathrm{kN}\cdot \mathrm{m}$. Then, use the section method and superposition principle to construct the bending moment diagram as shown in Figure 26 (units of moment in kN·m).

The bending moment diagram in Figure 26 agrees with that obtained using the structural mechanics solver.

6.3. Example of a Super-Static Rigid Frame with Sidesway

A super-static rigid frame with sidesway subjected to loading is shown in Figure 27. Each member has a bending stiffness of EI, and axial deformations are ignored. The bending moment diagram of the frame is determined as follows.

(1)

Determine the primary unknown quantities: the angular displacement θ_B at node B and the horizontal displacement Δ of end B of member AB and end C of member CD.

(2)

Calculate the fixed-end moment for each member induced by the load.

Members AB and CD have no external loads. Their fixed-end moments are therefore zero. Member BC is a beam fixed at both ends subjected to linearly distributed moments increasing from left to right. Its fixed-end moment can be calculated by substituting $m_0=4 \mathrm{kN}\cdot \mathrm{m}/\mathrm{m}$, $l=6 \mathrm{m}$, $a=1 \mathrm{m}$, and $b=3 \mathrm{m}$ into Equations (18) and (19) to obtain the following: $M_{BC}^F={\displaystyle \frac{5}{4}} \mathrm{kN}\cdot \mathrm{m}$ and $M_{CB}^F={\displaystyle \frac{5}{4}} \mathrm{kN}\cdot \mathrm{m}$.

(3)

Calculate the bending moments at the ends of the members owing to the load and end displacements (load constants + shape constants × end displacements).

For simplification, set EI = 6, then each member’s linear stiffness meets i_AB = i_BC = i_CD = 1. The end moments for each member are calculated as follows:

For member AB: $M_{AB}=2i_{AB}{\theta }_B-{\displaystyle \frac{6i_{AB}}{l_{AB}}}\Delta =2{\theta }_B-\Delta$ and $M_{BA}=4i_{AB}{\theta }_B-{\displaystyle \frac{6i_{AB}}{l_{AB}}}\Delta =4{\theta }_B-\Delta$;

For member BC: $M_{BC}=4i_{BC}{\theta }_B+2i_{BC}{\theta }_C+M_{BC}^F=4{\theta }_B+2{\theta }_C+{\displaystyle \frac{5}{4}}$ and $M_{CB}=2i_{BC}{\theta }_B+4i_{BC}{\theta }_C+M_{CB}^F=2{\theta }_B+4{\theta }_C+{\displaystyle \frac{5}{4}}$;

For member CD: $M_{CD}=4i_{CD}{\theta }_C-{\displaystyle \frac{6i_{CD}}{l_{CD}}}\Delta =4{\theta }_C-\Delta$ and $M_{DC}=2i_{CD}{\theta }_C-{\displaystyle \frac{6i_{CD}}{l_{CD}}}\Delta =2{\theta }_C-\Delta$.

(4)

Establish the displacement method equations and solve the primary unknown quantities.

Consider the equilibrium at node B (Figure 28), i.e., $\sum M_B=0$, we obtain the following: $M_{BA}+M_{BC}=0$. Thus, the equilibrium equation can be established as follows:

$$8{\theta }_B+2{\theta }_C-\Delta +{\displaystyle \frac{5}{4}}=0$$

(33)

Consider the equilibrium at node C (Figure 29), i.e., $\sum M_C=0$, we obtain the following: $M_{CB}+M_{CD}=0$. Thus, the equilibrium equation can be established as follows:

$$2{\theta }_B+8{\theta }_C-\Delta +{\displaystyle \frac{5}{4}}=0$$

(34)

Cutting through the top of the column, consider the equilibrium of the crossbeam BC above the column (Figure 30), i.e., $\sum F_x=0$, we obtain $F_{\mathrm{Q}BA}+F_{\mathrm{Q}CD}=0$.

Next, consider the equilibrium of columns AB and CD (Figure 31) to calculate F_QBA and F_QCD. From $\sum M_A=0$, it follows that $F_{\mathrm{Q}BA}=-{\displaystyle \frac{M_{AB}+M_{BA}}{l_{AB}}}$; from $\sum M_D=0$, it follows that $F_{\mathrm{Q}CD}=-{\displaystyle \frac{M_{CD}+M_{DC}}{l_{CD}}}$. Thus, the equilibrium equation for crossbeam BC becomes the following:

$$3{\theta }_B+3{\theta }_C-2\Delta =0$$

(35)

Solving Equations (33)–(35) simultaneously yields the following: ${\theta }_B=-{\displaystyle \frac{5}{28}}$, ${\theta }_C=-{\displaystyle \frac{5}{28}}$, and $\Delta =-{\displaystyle \frac{15}{28}}$.

(5)

Determine the end moments of each member and construct the bending moment diagram.

Substitute ${\theta }_B=-{\displaystyle \frac{5}{28}}$, ${\theta }_C=-{\displaystyle \frac{5}{28}}$, and $\Delta =-{\displaystyle \frac{15}{28}}$ into the expressions for the end moments of each member obtained in Step (3). We have $M_{AB}=0.18 \mathrm{kN}\cdot \mathrm{m}$, $M_{BA}=-0.18 \mathrm{kN}\cdot \mathrm{m}$, $M_{BC}=0.18 \mathrm{kN}\cdot \mathrm{m}$, $M_{CB}=0.18 \mathrm{kN}\cdot \mathrm{m}$, $M_{CD}=-0.18\mathrm{kN}\cdot \mathrm{m}$, and $M_{DC}=0.18\mathrm{kN}\cdot \mathrm{m}$. Then, use the section method and superposition principle to construct the bending moment diagram as shown in Figure 32 (units of moment in kN·m).

The bending moment diagram in Figure 32 agrees with that obtained using the structural mechanics solver.

7. Conclusions

The accurate calculation of internal forces in statically indeterminate structures is the key to structural design and an important research topic in structural mechanics. In order to more conveniently solve the internal forces of statically indeterminate structures under the action of localized distributed moments, this paper uses certain technical means to derive relevant load constants, thereby solving the problem of using the displacement method to solve the internal forces of statically indeterminate structures under the action of localized distributed moments. The conclusions of this paper are as follows:

(1)

The derivation of load constants under the action of a single concentrated moment has been simplified. Using the load constant formula for a single concentrated force, the derivation of the load constant formula for a single concentrated moment is reduced to combining like terms.

(2)

The integral method allows the derivation of load constants for distributed moments to be transformed into solving a definite integral of a polynomial function, which is more efficient and requires less effort, compared with the force, virtual beam, and energy methods. The advantages of the integral method become more pronounced with the increasing complexity of the distributed moment, such as quadratic and higher-order parabolic distributions.

(3)

The results of this study provide significant convenience for solving internal forces in statically indeterminate structures under distributed moments using the displacement method. The approach is of great significance for both educational practice and engineering applications.

(4)

In the future, the concept of substructures can be applied to derive the commonly used formulas for the fixed-end moment of substructures under the action of localized distributed moments, thereby facilitating the solution of internal forces in certain specific hyperstatic structures.