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1 +Calculations of the geometrical properties
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1 +Manuals.Theory Manual.WebHome
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1 -XWiki.XWikiGuest
1 +XWiki.StruSoft
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1 +(% style="text-align: justify;" %)
2 +Beam finite elements of arbitrary cross sections are important members of finite element models for a variety of practical applications. In the effective use of such elements, at the modelling stage, the user is faced with the time consuming problem of accurate determination of beam cross sectional properties.
3 +
4 +(% style="text-align: justify;" %)
5 +In the analysis of beams, the following local coordinate systems and notations are used (see Fig. below):
6 +
7 +* **N** node of finite element mesh,
8 +* **G** centroid,
9 +* **S** shear centre,
10 +* **x, y, z** local coordinates, where x is passing through element nodes,
11 +* **x', y', z'** central axes, parallel to x, y, z (i.e. x' is the beam axis),
12 +* **y,,G,,, z,,G,,** coordinates of **G** centroid, relative to N node,Fig. below
13 +* **y,,S,,, z,,S,,** coordinates of **S** shear (torsion) centre, relative to G centroid,
14 +* **e,,xmax,,, e,,xmin,,, e,,ymax,,, e,,ymin,,, e,,1max,,, e,,1min,,, e,,2max,,, e,,2min,,** distances of extreme fibres.
15 +
16 +(% style="text-align: justify;" %)
17 +[[image:1558007716236-122.png||height="514" width="553"]]
18 +The cross sectional properties are related to internal forces, namely tension, bending, torsion (free or constrained) and shear. The general solution of elastic beam problems can be found in many textbooks, for example in references (% style="color:#e74c3c" %)[3](%%) and (% style="color:#e74c3c" %)[4](%%), only the definitions and final results will be presented.
19 +
20 +(% style="text-align: justify;" %)
21 += Tension, bending =
22 +
23 +
24 +The well known formula calculating the normal stress distribution due to tension and bending:
25 +
26 +(% style="text-align: justify;" %)
27 +[[image:1558007957069-112.png||height="49" width="298"]]
28 +
29 +(% style="text-align: justify;" %)
30 +The cross sectional properties appearing here are:
31 +
32 +(% style="text-align: justify;" %)
33 +area:
34 +
35 +(% style="text-align: justify;" %)
36 +[[image:1558425292716-797.png||height="30" width="88"]]
37 +
38 +(% style="text-align: justify;" %)
39 +co-ordinates of centroid:
40 +
41 +(% style="text-align: justify;" %)
42 +[[image:1558425317826-189.png||height="32" width="237"]]
43 +
44 +(% style="text-align: justify;" %)
45 +moment of inertia with respect to centroidal axes:
46 +
47 +(% style="text-align: justify;" %)
48 +[[image:1558425668429-574.png||height="27" width="481"]]
49 +
50 +(% style="text-align: justify;" %)
51 +The principal moments of inertia and the angle α of the 1 principal axis are:
52 +
53 +(% style="text-align: justify;" %)
54 +[[image:1558425695510-439.png||height="33" width="114"]]
55 +
56 +(% style="text-align: justify;" %)
57 +[[image:1558425717716-657.png||height="29" width="475"]]
58 +
59 +(% style="text-align: justify;" %)
60 +Using the principal axes as local coordinate directions, the maximums of bending stress components are:
61 +
62 +(% style="text-align: justify;" %)
63 +[[image:1558425762893-809.png||height="36" width="193"]]
64 +
65 +In many tension/compression and bending related problems (eq. eccentric compression, the slenderness calculation) the radius of inertia is used. The definitions of radius of principal inertia are:
66 +
67 +(% style="text-align: justify;" %)
68 +[[image:1558425852341-521.png||height="41" width="142"]]
69 +
70 +(% style="text-align: justify;" %)
71 += Elastic-plastic bending =
72 +
73 +(% style="text-align: justify;" %)
74 += [[image:1558425911074-371.png||height="265" width="416"]] =
75 +
76 +(% style="text-align: justify;" %)
77 +The normal stress distribution due to simple bending of section, supposing the material is perfectly elastic-plastic (no strain hardening) is shown in Fig. 2. The axis ξ is passing trough the **G** centroid, while the line **η** divides the area into halves. The ξ axis of bending is either one of principal axes or parallel to one of the symmetry lines.
78 +
79 +(% style="text-align: justify;" %)
80 +There are two specific moments, **M,,e,,** elastic limit moment when the maximum stress equals to the **σ,,Y,,** yield stress,
81 +
82 +(% style="text-align: justify;" %)
83 +[[image:1558425955946-250.png||height="46" width="273"]]
84 +
85 +(% style="text-align: justify;" %)
86 +and the **M,,Y,,** plastic limit (ultimate) moment:
87 +
88 +(% style="text-align: justify;" %)
89 +[[image:1558426039571-380.png||height="27" width="94"]]
90 +
91 +(% style="text-align: justify;" %)
92 +where:
93 +
94 +(% style="text-align: justify;" %)
95 +**S,,0ξ,,** is the static (linear) moment of half area with respect to the axis ξ. The elastic-plastic moment capacity is defined as:
96 +
97 +(% style="text-align: justify;" %)
98 +[[image:1558426079223-242.png||height="49" width="194"]]
99 +
100 +(% style="text-align: justify;" %)
101 += Free torsion =
102 +
103 +(% style="text-align: justify;" %)
104 +Due to a torque **M,,x,,** each section of a straight beam undergoes a rotation about the point **S**, here called of torsion centre. In case of free torsion-when the torsional warping of the cross section is not constrained-the rate of twist is constant. Assuming homogeneous, elastic material, the shear stress distributions are given as:
105 +
106 +(% style="text-align: justify;" %)
107 +[[image:1558426145913-956.png||height="42" width="297"]]
108 +
109 +(% style="text-align: justify;" %)
110 +The cross sectional property is the torsion moment of inertia:
111 +
112 +(% style="text-align: justify;" %)
113 +[[image:1558426169454-888.png||height="31" width="259"]]
114 +
115 +(% style="text-align: justify;" %)
116 +and the co-ordinates of **S** torsion centre:
117 +
118 +(% style="text-align: justify;" %)
119 +[[image:1558426227749-795.png||height="34" width="535"]]
120 +
121 +(% style="text-align: justify;" %)
122 +The maximum of torsion stresses can be calculated with the **W,,t,,** torsion section modulus as:
123 +
124 +(% style="text-align: justify;" %)
125 +[[image:1558426265202-360.png||height="48" width="224"]]
126 +
127 +(% style="text-align: justify;" %)
128 +It follows from the internal equilibrium of linear elasticity and the boundary conditions that the ϕ(y,z) warping function is the solution of the following differential equation:
129 +
130 +(% style="text-align: justify;" %)
131 +[[image:1558426307163-384.png||height="45" width="402"]]
132 +
133 +(% style="text-align: justify;" %)
134 +where: **m** and **n** are the components of outward unit normal vector of section contour.
135 +
136 +(% style="text-align: justify;" %)
137 += Constrained torsion =
138 +
139 +(% style="text-align: justify;" %)
140 +If the torsional warping of the cross section of the straight beam is constrained, in addition to the shear stresses a secondary normal stress distribution appears:
141 +
142 +[[image:1558426380441-461.png||height="57" width="183"]]
143 +
144 +In this equation B( x) is the bimoment and the cross sectional property is the warping parameter defined as:
145 +
146 +[[image:1558426485761-763.png||height="30" width="210"]]
147 +
148 +For thin walled sections the warping function or the sector area function with pole S is:
149 +
150 +[[image:1558426493504-764.png||height="28" width="159"]]
151 +
152 += Shear =
153 +
154 +The beam is free of torsion if the **V,,y,,** and **V,,z,,** shear forces are passing through the S point. It follows from the Betti’s theorem, that the centre point of torsional rotation of the section and the shear centre are identical. The shear stress distributions from the **ψ,,1,,(y,z)** and **ψ,,2,,(y,z)** shear stress functions can be calculated as:
155 +
156 +[[image:1558426669259-295.png||height="45" width="365"]]
157 +
158 +In using the finite element method for elastic structures, the stiffness matrix is derived from the **U** internal energy. The **U,,s,,** shear contribution of the beam internal energy per unit length, with shear modulus **G**, is:
159 +
160 +[[image:1558426723712-824.png||height="46" width="318"]]
161 +
162 +The cross sectional properties are the shear factors:
163 +
164 +[[image:1558426756113-591.png||height="85" width="401"]]
165 +
166 +Transforming the **y, z** coordinates into the **1** and **2** principal directions, the shear factors are the principal shear factors:
167 +
168 +[[image:1558426799342-133.png||height="22" width="279"]]
169 +
170 +The quantities A1 = (A ρ,,1,,) and A2 = (A ρ,,2,,) are called by shear areas.
171 +From the condition of internal equilibrium two boundary value problems can be derived:
172 +
173 +if V,,y,, = 1 and V,,z,, = 0,
174 +
175 +[[image:1558426856532-292.png||height="33" width="337"]]
176 +
177 +if V,,y,, = 0 and V,,z,, = 1,
178 +
179 +[[image:1558426887009-629.png||height="34" width="333"]]
180 +
181 +where: **m** and **n** are the components of outward unit normal vector of section contour.
182 +
183 +The approximate distribution of shear stress in a thin walled section takes the form:
184 +
185 +[[image:1558426934875-527.png||height="30" width="451"]]
186 +
187 +where: **S,,y,,**, and **S,,z,,** are the static (linear) moment functions:
188 +
189 +[[image:1558426983048-596.png||height="235" width="196"]]
190 +
191 +
192 +[[image:1558427009663-143.png||height="28" width="286"]]
193 +
194 +Using the principal axes as local coordinate directions the shear stress distribution is:
195 +
196 +[[image:1558427077651-820.png||height="45" width="207"]]
197 +
198 += Lateral buckling =
199 +
200 +
201 +(% style="text-align: justify;" %)
202 +In the lateral-torsion buckling analysis of non symmetric section columns the so-called Wagner’s coefficients are used. They are defined as follows:
203 +
204 +(% style="text-align: justify;" %)
205 +[[image:1558428946937-752.png||height="80" width="436"]]
206 +
207 +(% style="text-align: justify;" %)
208 +In the local system of "1" and "2" principal axes:
209 +
210 +(% style="text-align: justify;" %)
211 +z,,1,, = β,,y,,, z,,2,, = β,,y,,, z,,ω,, = β,,ω,,
212 +
213 +
214 +(% style="text-align: justify;" %)
215 +These properties for a double symmetric section are zeros, so they can be termed as asymmetry properties. If the principal axis "**1**" is a symmetry line of the section than **z,,1,,** equals to zero and same holds for **z,,2,,**.
216 +
217 +(% style="text-align: justify;" %)
218 += Finite element method for sections =
219 +
220 +The (**I**.), (**II.**) and (**III.**) elliptic boundary value problems can be transformed into the following energy principles or weak forms:
221 +
222 +[[image:1558429255703-973.png||height="135" width="466"]]
223 +
224 +These problems can be solved by a 2D finite element method, where the stiffness matrix, derived from the quadratic part of principles, are the same. The only differences are in the linear parts, which are leading to three different right hand sides. Using a quadratic (8 or 6 node) isoparametric finite element formulation with one degree of freedom per node, the cross sectional properties can be calculated. The average element size is calculated automatically and the mesh is generated automatically too using this average element size. All integrals are calculated by Gauss quadratures.
225 +
226 +
227 +
228 +(% style="text-align: justify;" %)
229 +
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