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index 92562de..6a283cb 100644 --- a/docs/help/theory.html +++ b/docs/help/theory.html @@ -20,115 +20,143 @@

Bernoulli-Euler beam theory

June 27, 2019
+

Contents

+
+ 1 Introduction +
2 Kinematics +
3 Constitutive relations +
4 Stress resultants +
5 Equilibrium +
6 Governing equation +
7 Finding moment, shear force, and slope from the displacement function +
8 Examples +
 8.1 Single span beam with constant distributed force +
 8.2 Single span beam with a single concentrated force +
 8.3 Single span beam with a concentrated force and distributed +load using the stiffness method +
+

1 Introduction

-

Introduction +

+ id="x1-20011">
-

-

+

PIC

Figure 1: Deformation of the Bernoulli-Euler beam. Definition of coordinate -axes and components of displacement.
+axes and components of displacement.
-

-

This tool employs the Bernoulli-Euler beam theory. This theory, also +

+

This tool employs the Bernoulli-Euler beam theory. This theory, also known as shear rigid beam theory, is based on the kinematic assumption that

-

Any plane cross section perpendicular to the undeformed beam’s +

Any plane cross section perpendicular to the undeformed beam’s axis remain plane and perpendicular to the axis throughout the deformation.

-

This allows us to reduce the three-dimensional problem to a single unknown +

This allows us to reduce the three-dimensional problem to a single unknown function, v(x), known as deflection of the beam.

2 Kinematics

-

Navier’s assumption leads to + id="x1-30002">Kinematics +

Navier’s assumption leads to
+ id="x1-3001r1">
u(x,y) = - yv′(x) v(x,y) = v(x )
(1)
-

+

This displacement field induces an axial strain of
+ id="x1-3002r2">
ε(x,y) = ∂u(x,y)-= - yv′′(x) . ∂x
(2)
-

+

Equation (2) states that a fiber parallel to the beam axis stretches in the bottom +href="#x1-3002r2">2) states that a fiber parallel to the beam axis stretches in the bottom portion of the beam (y < 0) and contracts if the fiber is located above the beam axis (y > 0). The beam axis itself does not stretch. -

+

3 Constitutive relations

-

For a slender beam, we can ignore stress components acting perpendicular to the + id="x1-40003">Constitutive relations +

For a slender beam, we can ignore stress components acting perpendicular to the beam’s axis. Thus, the constitutive relations can be simplified as the 1D-version of Hooke’s law:
+ id="x1-4001r3">
σ(x,y) = Eε(x,y)
(3)
-

+

where E is the modulus of elasticity. -

The imposed state of deformation induces normal stress proportional to the +

The imposed state of deformation induces normal stress proportional to the strain field (2) as +href="#x1-3002r2">2) as
+ id="x1-4002r4">
σ (x, y) = - Eyv ′′(x) .
(4)
-

+

This relation states that (i) the stress varies linearly with the distance from the beam’s axis, vanishing at the axis, and (ii) the stress is proportional to the curvature of the beam. -

+

4 Stress resultants

-

The beam sees two stress resultants: the internal moment, + id="x1-50004">Stress resultants +

The beam sees two stress resultants: the internal moment,
+ id="x1-5001r5">
∫ M (x) = - A yσ(x,y)dA
(5)
-

+

and the transverse shear force,
+ id="x1-5002r6">
∫ @@ -136,35 +164,35 @@ <h3 class=4
(6)
-

+

Substituting (4) into (5) yields +href="#x1-4002r4">4) into (5) yields
+ id="x1-5003r7">
∫ 2 ′′ ′′ ∫ 2 ′′ M (x ) = Ey v (x)dA = Ev (x) y dA = EIv (x) A A
(7)
-

+

where
+ id="x1-5004r8">
∫ I = y2dA A
(8)
-

+

is the area moment of inertia or, short, moment of inertia. -

Note that the modulus of elasticity,

Note that the modulus of elasticity, E, characterizes the material, the moment of inertia, I, characterized the shape of the cross section, and the second @@ -174,10 +202,10 @@

4 x), characterizes the deformation (curvature) of the beam. -

+

5 Equilibrium

-

Equilibrium is formulated in terms of shear forces, Equilibrium

+

Equilibrium is formulated in terms of shear forces, V (x), and internal moments, 5 y-direction, yields
+ id="x1-6001r9">
V ′(x) = - w(x)
(9)
-

+

where w(x) is the distributed lateral load per length. 5 x) is defined positive if pointing against the (upward) positive y-axis. -

Moment equilibrium around the out-of-plane axis on the same element +

Moment equilibrium around the out-of-plane axis on the same element yields
+ id="x1-6002r10">
M ′(x) = V (x) .
(10)
-

+

A system for which equations (9) and (10) are sufficient to determine the +href="#x1-6001r9">9) and (10) are sufficient to determine the internal moment and shear functions is called statically determinate. Otherwise, the system is called statically indeterminate. Solving these equations for the latter requires consideration of the kinematic relation (7) and respective +href="#x1-5003r7">7) and respective boundary conditions. -

Equations (9) and (10) may be combined into one equation as +

Equations (9) and (10) may be combined into one equation as
+ id="x1-6003r11">
′′ ′ M (x) = V (x) = - w (x )
(11)
-

+

Equation (11) replaces both equilibrium equations (9) and (10). -

+href="#x1-6003r11">11) replaces both equilibrium equations (9) and (10). +

6 Governing equation

-

The governing equation is obtained by assuming the displacement function, + id="x1-70006">Governing equation +

The governing equation is obtained by assuming the displacement function, v(x), as the primary unknown and expressing M(x) in (11) using (7) to +href="#x1-6003r11">11) using (7) to obtain
+ id="x1-7001r12">
( ) EI (x)v′′(x )′′ + w (x ) = 0
(12)
-

+

This equation is known as the governing equation of the Bernoulli-Euler beam. -

If the beam possesses a constant cross section and is made of one material, +

If the beam possesses a constant cross section and is made of one material, then EI(x) = EI = const. and (12) simplifies to +href="#x1-7001r12">12) simplifies to
+ id="x1-7002r13">
′′′′ EIv (x)+ w(x) = 0
(13)
-

+

Equation (13) is what is implemented in this program. -

+href="#x1-7002r13">13) is what is implemented in this program. +

7 Finding moment, shear force, and slope from the displacement + id="x1-80007">Finding moment, shear force, and slope from the displacement function

-

Solving (13) and applying suitable boundary conditions yields the displacement +

Solving (13) and applying suitable boundary conditions yields the displacement function, v(x), for the beam. The slope, 7 + id="x1-8001r14">

θ(x) = v′(x) .
(14) -

+

It is positive if the cross section rotates counter-clockwise during deformation. -

The moment follows from (7) as +

The moment follows from (7) as
+ id="x1-8002r15">
M (x) = EI (x )v′′(x) = EI (x)θ′(x) .
(15)
-

+

The transverse shear force follows from (11) as +href="#x1-6003r11">11) as
+ id="x1-8003r16">
V (x ) = M ′(x) = (EI (x)v′′(x))′
(16)
-

+

or, for constant EI, simplifies to
+ id="x1-8004r17">
V(x) = EIv ′′′(x) .
(17)
-

-

-

8 Examples

+

+

8 Examples

+

8.1 Single span beam with constant distributed force

-

Both bending stiffness, Single span beam with constant distributed force +

Both bending stiffness, EI, and distributed load, w(x) = w0, are constant over the length of the beam. Thus, (13) simplifies to +href="#x1-7002r13">13) simplifies to
+ id="x1-10001r18">
w0 v ′′′′(x) = - --- EI
(18)
-

+

+ id="x1-10002r19">
w0x @@ -366,59 +394,59 @@ <h4 class=8.1
(19)
-

+

+ id="x1-10003r20">
w x2 v′′(x) = - -0---+ c1x+ c2 2EI
(20)
-

+

+ id="x1-10004r21">
3 θ(x) = v′(x) = - w0x--+ 1c1x2 + c2x+ c3 6EI 2
(21)
-

+

+ id="x1-10005r22">
4 v (x ) = - w0x-+ 1c1x3 + 1c2x2 + c3x + c4 24EI 6 2
(22)
-

+

+ id="x1-10006r23">
′′ w0x2- M (x) = EIv (x ) = - 2 + EIc1x + EIc2
(23)
-

+

+ id="x1-10007r24">
V(x) = M ′(x) = EIv ′′′(x ) = - w x + EIc 0 1
(24)
-

+

Pinned on both ends yields the boundary conditions
+ id="x1-10008r25">
8.1
(25)
-

+

+ id="x1-10009r26">
( ) @@ -443,10 +471,10 @@ <h4 class=8.1
(26)
-

+

+ id="x1-10010r27">
( w ℓ ) @@ -458,29 +486,29 @@ <h4 class=8.1
(27)
-

+

+ id="x1-10011r28">
4 ( )( 2 ) v(x) = -w0ℓ- x- 1- x- x--- x-- 1 24EI ℓ ℓ ℓ2 ℓ
(28)
-

+

+ id="x1-10012r29">
′′ w0-ℓ2x-( x) M (x) = EIv (x) = 2 ℓ 1- ℓ
(29)
-

+

+ id="x1-10013r30">
8.1
(30)
-

+

Shear vanishes at x = ℓ and, thus,
+ id="x1-10014r31">
2 max M = M (ℓ∕2) = w0-ℓ 8
(31)
-

+

By symmetry, rotation vanishes at x = ℓ and, thus,
+ id="x1-10015r32">
5w0ℓ4- max |v| = - min v = - v(ℓ∕2) = 384EI
(32)
-

+

-

+

8.2 Single span beam with a single concentrated force

+ id="x1-110008.2">Single span beam with a single concentrated force +

+
+ v′1′′′(x) = 0 (33) + v′′′(x) = c (34) + 1′′ 1 + v1(x) = c1x + c2 (35) + ′ 1- 2 + θ1(x ) = v1(x) = 2c1x + c2x + c3 (36) + 1- 3 1- 2 + v1(x) = 6c1x + 2c2x + c3x + c4 (37) + M (x) = EIv ′′(x) = EIc x + EIc (38) + 1′ 1′′′ 1 2 +V1(x) = M 1(x) = EIv 1 (x) = EIc1 (39) +
+
+
+ ′′′′ + v2 (x) = 0 (40) + v′2′′(x) = d1 (41) + ′′ + v2(x) = d1x + d2 (42) + θ2(x) = v′(x) = 1d1x2 + d2x + d3 (43) + 2 2 + v (x) = 1d x3 + 1-d x2 + d x + d (44) + 2 6 1 2 2 3 4 + M2 (x) = EIv ′′2(x) = EId1x + EId2 (45) + ′ ′′′ +V2(x) = M 2(x) = EIv 2 (x) = EId1 (46) +
+
Boundary conditions +
+
+( v(0) = v (0) ) ( 0 ) +||{ 1 ||} ||{ ||} + M (0) = M1 (0) = 0 +||( v(ℓ) = v2(ℓ) ||) ||( 0 ||) + M (ℓ) = M2 (ℓ) 0 +
(47)
+

+Continuity conditions:
+ id="x1-11004r48">
-
(33)
-

+src="theory35x.png" alt="v1(a) = v2(a) and θ1(a) = θ2(a) +" class="math-display" >(48) +

+Equilibrium of forces for the interval [a - ϵ,a + ϵ]:
+ id="x1-11005r49">
-
(34)
-

+src="theory36x.png" alt="liϵ→m0 [V(a - ϵ)- P - V (a+ ϵ)] = 0 ⇒ V1(a)- V2(a) = P +" class="math-display" >(49) +

+Moment equilibrium for the interval [a - ϵ,a + ϵ]:
+ id="x1-11006r50">
- -
(35)
-

+src="theory37x.png" alt="lϵ→im0 [M (a- ϵ) + ϵV(a - ϵ)- M (a + ϵ)+ ϵV(a + ϵ)] = 0 ⇒ M1 (a) = M2 (a) +" class="math-display" >(50) +

+ id="x1-11007r51"> +
-
(36)
-

+src="theory38x.png" alt="⌊ 0 0 0 1 0 0 0 0 ⌋ ( ) ( ) +| | || c1 || || 0 || +| 0 EI 0 0 0 0 0 0 | ||| ||| ||| ||| +|| || |||| c2 |||| |||| 0 |||| +|| 0 0 0 0 ℓ3∕6 ℓ2∕2 ℓ 1 || |||| |||| |||| |||| +|| || ||| c3 ||| ||| 0 ||| +|| 0 0 0 0 EI ℓ EI 0 0 || ||{ c4 ||} ||{ 0 ||} +|| || = +| a3∕6 a2∕2 a 1 - a3∕6 - a2∕2 - a - 1 | ||| d1 ||| ||| 0 ||| +|| || |||| |||| |||| |||| +|| a2∕2 a 1 0 - a2∕2 - a - 1 0 || |||| d2 |||| |||| 0 |||| +|| || ||| d ||| ||| 0 ||| +|⌈ EIa EI 0 0 - EIa - EI 0 0 |⌉ |||| 3 |||| |||| |||| + ( d4 ) ( P ) + EI 0 0 0 - EI 0 0 0 +" class="math-display" >(51) +

+Using α = a∕ℓ, the integration constants are obtained as
+ id="x1-11008r52">
-
(37)
-

+src="theory39x.png" alt=" { ( ) } + (1---α)P- α--α2 --3α-+-2-Pℓ2- +{c1, c2, c3, c4} = EI , 0, - 6EI , 0 +" class="math-display" >(52) +

+and
+ id="x1-11009r53">
-
(38)
-

+src="theory40x.png" alt=" { ( 2) 2 3 3} +{d1, d2, d3, d4} = --αP-, αP-ℓ, - α--2+-α---Pℓ-, α--Pℓ- + 2EI EI 6EI 6EI +" class="math-display" >(53) +

+ id="x1-11010r54">
-
(39)
-

+src="theory41x.png" alt=" +" class="math-display" >(54) +

+ id="x1-11011r55">
-
(40)
-

+src="theory42x.png" alt=" +" class="math-display" >(55) +

+ id="x1-11012r56">
-
(41)
-

+src="theory43x.png" alt=" +" class="math-display" >(56) +

+ id="x1-11013r57">
-
(42)
-

-

+src="theory44x.png" alt=" +" class="math-display" >(57) +

+
+
+ +
(58)
+

+
+
+ +
(59)
+

+
+
+ +
(60)
+

+
+ +
+ +
(61)
+

+
+
+ +
(62)
+

+

8.3 Single span beam with a concentrated force and distributed load using + id="x1-120008.3">Single span beam with a concentrated force and distributed load using the stiffness method

- + id="x1-12001r63">
-
(43)
-

+src="theory50x.png" alt=" +" class="math-display" >(63) +

+ id="x1-12002r64">
-
(44)
-

+src="theory51x.png" alt=" +" class="math-display" >(64) +

+ id="x1-12003r65">
-
(45)
-

- +src="theory52x.png" alt=" +" class="math-display" >(65) +

+ id="x1-12004r66"> +
-
(46)
-

+src="theory53x.png" alt=" +" class="math-display" >(66) +

+ id="x1-12005r67">
-
(47)
-

+src="theory54x.png" alt=" +" class="math-display" >(67) +

+ id="x1-12006r68">
+src=
(68)
+

-" class="math-display" >(48) -

+ id="x1-12007r69">
-
(49)
-

+src="theory56x.png" alt=" +" class="math-display" >(69) +

+ id="x1-12008r70">
-
(50)
-

+src="theory57x.png" alt=" +" class="math-display" >(70) +

+ id="x1-12009r71">
-
(51)
-

+src="theory58x.png" alt=" + +" class="math-display" >(71) +

+ id="x1-12010r72">
-
(52)
-

+src="theory59x.png" alt=" +" class="math-display" >(72) +

- diff --git a/docs/help/theory.idv b/docs/help/theory.idv index 46a3713..75ec446 100644 Binary files a/docs/help/theory.idv and b/docs/help/theory.idv differ diff --git a/docs/help/theory.lg b/docs/help/theory.lg index f18e577..2936d62 100644 --- a/docs/help/theory.lg +++ b/docs/help/theory.lg @@ -166,23 +166,14 @@ File: beam.png --- needs --- theory.idv[32] ==> theory30x.png --- --- needs --- theory.idv[33] ==> theory31x.png --- --- needs --- theory.idv[34] ==> theory32x.png --- ---- empty picture --- theory.idv[34] --- --- needs --- theory.idv[35] ==> theory33x.png --- ---- empty picture --- theory.idv[35] --- --- needs --- theory.idv[36] ==> theory34x.png --- ---- empty picture --- theory.idv[36] --- --- needs --- theory.idv[37] ==> theory35x.png --- ---- empty picture --- theory.idv[37] --- --- needs --- theory.idv[38] ==> theory36x.png --- ---- empty picture --- theory.idv[38] --- --- needs --- theory.idv[39] ==> theory37x.png --- ---- empty picture --- theory.idv[39] --- --- needs --- theory.idv[40] ==> theory38x.png --- ---- empty picture --- theory.idv[40] --- --- needs --- theory.idv[41] ==> theory39x.png --- ---- empty picture --- theory.idv[41] --- --- needs --- theory.idv[42] ==> theory40x.png --- ---- empty picture --- theory.idv[42] --- --- needs --- theory.idv[43] ==> theory41x.png --- --- empty picture --- theory.idv[43] --- --- needs --- theory.idv[44] ==> theory42x.png --- @@ -205,6 +196,22 @@ File: beam.png --- empty picture --- theory.idv[52] --- --- needs --- theory.idv[53] ==> theory51x.png --- --- empty picture --- theory.idv[53] --- +--- needs --- theory.idv[54] ==> theory52x.png --- +--- empty picture --- theory.idv[54] --- +--- needs --- theory.idv[55] ==> theory53x.png --- +--- empty picture --- theory.idv[55] --- +--- needs --- theory.idv[56] ==> theory54x.png --- +--- empty picture --- theory.idv[56] --- +--- needs --- theory.idv[57] ==> theory55x.png --- +--- empty picture --- theory.idv[57] --- +--- needs --- theory.idv[58] ==> theory56x.png --- +--- empty picture --- theory.idv[58] --- +--- needs --- theory.idv[59] ==> theory57x.png --- +--- empty picture --- theory.idv[59] --- +--- needs --- theory.idv[60] ==> theory58x.png --- +--- empty picture --- theory.idv[60] --- +--- needs --- theory.idv[61] ==> theory59x.png --- +--- empty picture --- theory.idv[61] --- --- characters --- Font("cmr","10","10","109") Font("cmr","17","17","100") diff --git a/docs/help/theory.log b/docs/help/theory.log index 3e22831..97c703e 100644 --- a/docs/help/theory.log +++ b/docs/help/theory.log @@ -1,4 +1,4 @@ -This is pdfTeX, Version 3.14159265-2.6-1.40.18 (TeX Live 2017) (preloaded format=latex 2017.5.23) 27 JUN 2019 19:11 +This is pdfTeX, Version 3.14159265-2.6-1.40.18 (TeX Live 2017) (preloaded format=latex 2017.5.23) 27 JUN 2019 23:03 entering extended mode restricted \write18 enabled. %&-line parsing enabled. @@ -482,100 +482,113 @@ LaTeX Font Info: Try loading font information for U+msb on input line 21. (/usr/local/texlive/2017/texmf-dist/tex/latex/amsfonts/umsb.fd File: umsb.fd 2013/01/14 v3.01 AMS symbols B -) [3] [4 +) (./theory.4ct) +[3] [4 ] -l.26 --- TeX4ht warning --- File `"beam.xbb"' not found --- -l.26 --- TeX4ht warning --- Cannot determine size of graphic in "beam.xbb" (no +l.28 --- TeX4ht warning --- File `"beam.xbb"' not found --- +l.28 --- TeX4ht warning --- Cannot determine size of graphic in "beam.xbb" (no BoundingBox) --- -l. 26 --- needs --- beam.png --- +l. 28 --- needs --- beam.png --- [5 ] -l. 39 Writing theory.idv[1] (theory0x.png) -l. 46 Writing theory.idv[2] (theory1x.png) +l. 41 Writing theory.idv[1] (theory0x.png) +l. 48 Writing theory.idv[2] (theory1x.png) [6 ] -l. 54 Writing theory.idv[3] (theory2x.png) -l. 61 Writing theory.idv[4] (theory3x.png) +l. 56 Writing theory.idv[3] (theory2x.png) +l. 63 Writing theory.idv[4] (theory3x.png) [7] -l. 69 Writing theory.idv[5] (theory4x.png) -l. 74 Writing theory.idv[6] (theory5x.png) +l. 71 Writing theory.idv[5] 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492996 - 62368 string characters out of 6132730 - 253644 words of memory out of 5000000 - 8886 multiletter control sequences out of 15000+600000 + 5481 strings out of 492996 + 63062 string characters out of 6132730 + 256704 words of memory out of 5000000 + 8938 multiletter control sequences out of 15000+600000 11558 words of font info for 44 fonts, out of 8000000 for 9000 1141 hyphenation exceptions out of 8191 30i,10n,32p,874b,506s stack positions out of 5000i,500n,10000p,200000b,80000s -Output written on theory.dvi (27 pages, 55112 bytes). +Output written on theory.dvi (31 pages, 68352 bytes). diff --git a/docs/help/theory.pdf b/docs/help/theory.pdf index e2eaf0c..d34f1ad 100644 Binary files a/docs/help/theory.pdf and b/docs/help/theory.pdf differ diff --git a/docs/help/theory.tex b/docs/help/theory.tex index 0b01d22..fe1636c 100644 --- a/docs/help/theory.tex +++ b/docs/help/theory.tex @@ -20,6 +20,8 @@ \begin{document} \maketitle +\tableofcontents + \section{Introduction} \begin{figure}[h] \begin{center} @@ -287,36 +289,169 @@ \subsection{Single span beam with constant distributed force} \subsection{Single span beam with a single concentrated force} -\begin{equation} +\begin{eqnarray} + v_1''''(x) &=& 0 \\ + v_1'''(x) &=& c_1 \\ + v_1''(x) &=& c_1\,x + c_2 \\ + \theta_1(x) = v_1'(x) &=& \frac{1}{2} \,c_1\,x^2 + c_2\,x + c_3 \\ + v_1(x) &=& \frac{1}{6} \,c_1\,x^3 + \frac{1}{2} \,c_2\,x^2+ c_3\,x + c_4 \\ + M_1(x) = EI \, v_1''(x) &=& EI\,c_1\,x + EI\,c_2 \\ + V_1(x) = M_1'(x) = EI \, v_1'''(x) &=& EI\,c_1 \label{B1} -\end{equation} -\begin{equation} +\end{eqnarray} +\begin{eqnarray} + v_2''''(x) &=& 0 \\ + v_2'''(x) &=& d_1 \\ + v_2''(x) &=& d_1\,x + d_2 \\ + \theta_2(x) = v_2'(x) &=& \frac{1}{2} \,d_1\,x^2 + d_2\,x + d_3 \\ + v_2(x) &=& \frac{1}{6} \,d_1\,x^3 + \frac{1}{2} \,d_2\,x^2+ d_3\,x + d_4 \\ + M_2(x) = EI \, v_2''(x) &=& EI\,d_1\,x + EI\,d_2 \\ + V_2(x) = M_2'(x) = EI \, v_2'''(x) &=& EI\,d_1 \label{B2} -\end{equation} +\end{eqnarray} +Boundary conditions \begin{equation} + \left\{ + \begin{array}{c} + v(0) = v_1(0) \\ + M(0) = M_1(0) \\ + v(\ell) = v_2(\ell) \\ + M(\ell) = M_2(\ell) + \end{array} + \right\} + = + \left\{ + \begin{array}{c} + 0 \\ + 0 \\ + 0 \\ + 0 + \end{array} + \right\} \label{B3} \end{equation} +Continuity conditions: \begin{equation} + v_1(a) = v_2(a) + \qquad\hbox{and}\qquad + \theta_1(a) = \theta_2(a) \label{B4} \end{equation} +Equilibrium of forces for the interval $[a-\epsilon,a+\epsilon]$: \begin{equation} + \lim_{\epsilon\to 0} \left[ V(a-\epsilon) - P - V(a+\epsilon) \right] = 0 + \qquad\Rightarrow\quad + V_1(a) - V_2(a) = P \label{B5} \end{equation} +Moment equilibrium for the interval $[a-\epsilon,a+\epsilon]$: \begin{equation} + \lim_{\epsilon\to 0} \left[ M(a-\epsilon) + \epsilon V(a-\epsilon) - M(a+\epsilon) + \epsilon V(a+\epsilon) \right] = 0 + \quad\Rightarrow\quad + M_1(a) = M_2(a) \label{B6} \end{equation} \begin{equation} + \left[ + \begin{array}{cccccccc} + 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\[2ex] + 0 & EI & 0 & 0 & 0 & 0 & 0 & 0 \\[2ex] + 0 & 0 & 0 & 0 & \ell^3/6 & \ell^2/2 & \ell & 1 \\[2ex] + 0 & 0 & 0 & 0 & EI\,\ell & EI & 0 & 0 \\[2ex] + a^3/6 & a^2/2 & a & 1 & -a^3/6 & -a^2/2 & -a & -1 \\[2ex] + a^2/2 & a & 1 & 0 & -a^2/2 & -a & -1 & 0 \\[2ex] + EI \,a & EI & 0 & 0 & -EI\,a & -EI & 0 & 0 \\[2ex] + EI & 0 & 0 & 0 & -EI & 0 & 0 & 0 + \end{array} + \right] + \left\{ + \begin{array}{c} + c_1 \\[1.5ex] + c_2 \\[1.5ex] + c_3 \\[1.5ex] + c_4 \\[1.5ex] + d_1 \\[1.5ex] + d_2 \\[1.5ex] + d_3 \\[1.5ex] + d_4 + \end{array} + \right\} + = + \left\{ + \begin{array}{c} + 0 \\[1.5ex] + 0 \\[1.5ex] + 0 \\[1.5ex] + 0 \\[1.5ex] + 0 \\[1.5ex] + 0 \\[1.5ex] + 0 \\[1.5ex] + P + \end{array} + \right\} \label{B7} \end{equation} +Using $\alpha=a/\ell$, the integration constants are obtained as \begin{equation} + \left\{ + c_1 \,,~ + c_2 \,,~ + c_3 \,,~ + c_4 + \right\} + = + \left\{ + \frac{ (1-\alpha) P}{{EI}},~ + 0,~ + -\frac{ \alpha \left(\alpha ^2-3 \alpha +2\right) P \ell^2 }{6 \,{EI}},~ + 0 + \right\} \label{B8} \end{equation} +and \begin{equation} + \left\{ + d_1 \,,~ + d_2 \,,~ + d_3 \,,~ + d_4 + \right\} + = + \left\{ + \frac{ -\alpha P}{2 {EI}},~ + \frac{\alpha P \ell }{{EI}},~ + -\frac{ \alpha \left(2+\alpha^2\right) P \ell^2 }{6 \,{EI}},~ + \frac{\alpha^3 P \ell^3}{6 \,{EI}} + \right\} \label{B9} \end{equation} \begin{equation} \label{B10} \end{equation} +\begin{equation} + \label{B11} +\end{equation} +\begin{equation} + \label{B12} +\end{equation} +\begin{equation} + \label{B13} +\end{equation} +\begin{equation} + \label{B14} +\end{equation} +\begin{equation} + \label{B15} +\end{equation} +\begin{equation} + \label{B16} +\end{equation} +\begin{equation} + \label{B17} +\end{equation} +\begin{equation} + \label{B18} +\end{equation} \subsection{Single span beam with a concentrated force and distributed load using the stiffness method} diff --git a/docs/help/theory.toc b/docs/help/theory.toc new file mode 100644 index 0000000..9cc5c79 --- /dev/null +++ b/docs/help/theory.toc @@ -0,0 +1,11 @@ +\contentsline {section}{\numberline {1}Introduction}{1} +\contentsline {section}{\numberline {2}Kinematics}{2} +\contentsline {section}{\numberline {3}Constitutive relations}{2} +\contentsline {section}{\numberline {4}Stress resultants}{3} +\contentsline {section}{\numberline {5}Equilibrium}{3} +\contentsline {section}{\numberline {6}Governing equation}{4} +\contentsline {section}{\numberline {7}Finding moment, shear force, and slope from the displacement function}{4} +\contentsline {section}{\numberline {8}Examples}{5} +\contentsline {subsection}{\numberline {8.1}Single span beam with constant distributed force}{5} +\contentsline {subsection}{\numberline {8.2}Single span beam with a single concentrated force}{6} +\contentsline {subsection}{\numberline {8.3}Single span beam with a concentrated force and distributed load using the stiffness method}{8} diff --git a/docs/help/theory.xref b/docs/help/theory.xref index ff29d43..7fc8691 100644 --- 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