Problem 5.12 Draw the Free-body Diagram of the Beam

Roller Support

The roller support cannot exert a couple about the axis of the pin, and it cannot exert a force parallel to the surface on which information technology translates.

From: Mechanical Engineer'due south Handbook , 2001

Easily-On Applications of Structural Optimization

Osvaldo Thousand. Querin , ... Pascual Martí , in Topology Pattern Methods for Structural Optimization, 2017

five.6 Michell Beam With Roller Support

The Michell axle with a roller support [vi] consists of a rectangular design domain of width two   50 and summit L, the nominal length used is $L_{\mathrm{nom}}=L$ . Information technology has one fixed supports at the bottom left-manus corner and one roller support at the right corner, with a single vertical applied load in the centre of the bottom edge, Fig. 5.12.

Figure 5.12. Design domain for the Michell beam with one roller support.

The optimal nondimensional mass of this beam for the case of equal stress limits for the beams in tension and pinch is $\overline{\mathrm{Yard}}=\pi =3.14159$ [6]. The Growth method of Chapter 3, Detached Method of Structural Optimization, was used to solve this problem, the settings used by the TTO program are given in Table 5.11.

Table 5.11. TTO Settings for the Michell Axle With Ane Roller Support

L/H	σ c / σ t	Load	Angle	Iteration	Grid	Symmetry
ii	ane	1	−90	160	50	No

The optimal topology consists of a hemispherical shape made from a fan like region emerging from the point of application of the load consisting of direct radially positioned bars, Fig. v.13, where the tensile bars are in red (black in print versions) and the compressive bars in blue (grey in print versions). The summary of the number of joints, confined, and nondimensional masses are given in Table 5.12.

Figure five.13. Optimal design of the Michell beam with ane roller back up.

Table v.12. Optimal Parameters for the Michell Beam With One Roller Back up

50/H	Joints	Bars	$\overline{\mathrm{Thou}}$
2	160	317	iii.14169618

The ITD method of Chapter 4, Continuous Method of Structural Optimization, was also used to solve this problem. The dimensions of the model were: Fifty=100   mm, thickness=1   mm, as half the structure was modeled using symmetry weather condition, the applied load was F=500   N and the mesh used was of 100×100 FE, Fig. 5.14. The modulus of elasticity used for I elements was 210×10ⁱⁱⁱ  MPa, the Poisson'south ratio was 0.three, and the modulus of elasticity of O elements was 21   MPa.

Figure 5.fourteen. Design domain for Michell beam with one roller back up for employ in ITD.

The resulting topology for a volume fraction of 29.66% is shown in Fig. five.fifteen. The maximum vertical displacement $(D_{\mathrm{Max}})$ in mm; maximum $({\sigma }_{vM}^{\max })$ , mean $({\sigma }_{5G}^{\mathrm{thousand}ean})$ , and minimum $({\sigma }_{5M}^{\min })$ von Mises stresses in MPa are given in Table five.13. Its resulting nondimensional mass was $\overline{M}=\mathrm{ii.81462}$ .

Figure v.15. Michell beam with fixed supports design for a book fraction of 29.66%, showing the full design space.

Table 5.13. Michell Axle With Stock-still Supports Results

$(5/V_0)$	$(D_{\mathrm{Max}})$	$({\mathit{\sigma }}_{5\mathrm{1000}}^{\max })$	$({\mathit{\sigma }}_{v\mathrm{1000}}^{\mathrm{mean}})$	$({\mathit{\sigma }}_{vM}^{\min })$
0.thirty	−0.08	665.17	43.thirteen	22.47

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Relationships for Inhomogeneous Plates

C.M. Wang , ... Grand.H. Lee , in Shear Deformable Beams and Plates, 2000

Roller-supported round plate

Consider a solid circular plate with a roller support at r = a, a being the radius of the plate. The boundary weather are

(13.2.45a) $\text{At}r=0:u=0,\varphi =0,\frac{d{\mathrm{westward}}_0^{\mathrm{Thousand}}}{dr}=0,Q_r=0$

(xiii.2.45b) $\text{At}r=a:w=0,{\mathrm{Northward}}_{rr}=0,{\mathrm{Grand}}_{rr}=0$

The in a higher place boundary conditions requite

(xiii.2.46a) ${\mathrm{One\; thousand}}_1=-\frac{\mathrm{ii}{\Omega }_{\mathrm{v}}}{a{\Omega }_6}\left(-\widehat{D}\frac{d{\mathrm{west}}_0^{\mathrm{Thou}}(a)}{dr}+\frac{{\widehat{C}}_{2}a}{2}\right),K_{\mathrm{ii}}=0$

(13.2.46b) $C_{\mathrm{two}}=-\frac{2D}{a}\left(\frac{v-{\widehat{\Omega }}_{\mathrm{two}}+{\Omega }_8}{\mathrm{ane}+{\widehat{\Omega }}_2-{\Omega }_8}\right)\frac{d{w}_0^K(a)}{dr}$

(13.2.46c) $C_4=\frac{{\mathrm{Grand}}^{\mathrm{Chiliad}}(a)}{A_{55}}-\frac{{\widehat{C}}_2{a}^{\mathrm{two}}}{4},C_{\mathrm{i}}=C_3=0$

where Ω_eight = $\widehat{\Omega }$ ₃Ω_five/Ω₆. Hence, we have the following relations between the deflections, forces, and moments of the ii theories:

(thirteen.2.47) $w_0^F(r)=\widehat{D}{w}_0^K(r)+\frac{M^K(r)-M^K(a)}{A_{55}}+\frac{1}{4}{\widehat{C}}_2(a^2-r^{\mathrm{two}})$

(13.ii.48) $Q_r^F(r)=Q_r^K(r)$

(13.2.49) $N_{rr}^F(r)={\Omega }_5\widehat{D}\left(\frac{\mathrm{one}}{a}\frac{d{\mathrm{westward}}_0^K(a)}{dr}-\frac{1}{r}\frac{d{w}_0^K}{dr}\right)$

(thirteen.2.fifty) $N_{rr}^F(r)={\Omega }_{\mathrm{v}}\widehat{D}\left(\frac{\mathrm{ane}}{a}\frac{d{\mathrm{west}}_0^K(a)}{dr}-\frac{d^2{w}_0^K}{d{r}^2}\right)$

(13.2.51) $\begin{array}{l}{M}_{rr}^F(r)=M_{rr}^K(r)+D\frac{1}{r}\frac{d{\mathrm{westward}}_0^K}{dr}\left(\mathrm{five}-{\widehat{\Omega }}_{\mathrm{two}}\right)+\frac{1}{2}{C}_2\left(\mathrm{one}+{\widehat{\Omega }}_2\right)\hfill \\ +\frac{1}{2}{K}_1{\Omega }_3\hfill \end{array}$

(13.2.52) $\begin{array}{l}{M}_{\theta \theta }^F(r)=M_{\theta \theta }^G(r)+D\frac{d^2{w}_0^{\mathrm{Thousand}}}{d{r}^{\mathrm{ii}}}\left(v-{\widehat{\Omega }}_2\right)+\frac{1}{\mathrm{two}}{C}_2\left(1+{\widehat{\Omega }}_2\right)\hfill \\ +\frac{\mathrm{i}}{2}{\mathrm{Thousand}}_1{\Omega }_{\mathrm{three}}\hfill \end{array}$

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Statical indeterminacy

Alan Williams Ph.D., S.E., C.ENG. , in Structural Analysis, 2009

1.2 Indeterminacy in Pin-jointed Frames

In a pin-jointed frame, external reactions are provided past either roller supports or hinge supports, equally shown in Figure 1.1 (i) and (2). The roller support provides only ane degree of restraint in the vertical direction, and both horizontal and rotational displacements can occur. The hinge support provides two degrees of restraint in the vertical and horizontal directions, and just rotational displacement can occur. The magnitudes of the external restraints may be obtained from the three equations of equilibrium. Thus, a construction is externally indeterminate when it possesses more than than 3 external restraints and unstable when it possesses fewer than three.

Figure 1.1.

Figure 1.2 (i) and (ii) shows pin-jointed frames that have three degrees of restraint and are stable and determinate. Figure i.3 (i) shows a pin-jointed frame that has four degrees of restraint and is one caste indeterminate. The cutting-back structure is shown in Figure 1.3 (two). Figure 1.4 (i) shows a pin-jointed frame that is ii degrees indeterminate; the cut-back structure is shown in Figure ane.iv (ii).

Figure 1.2.

Figure i.3.

Figure 1.four.

In a pin-jointed frame with j joints, including the supports, 2j equations of equilibrium may exist obtained, since at each joint:

$\Sigma H=0\text{and}\Sigma V=0$

Each member of the frame is subjected to an centric force, and if the frame has n members and r external restraints, the number of unknowns is (northward+r). Thus, the degree of indeterminacy is:

$D=n+r-\mathrm{ii}j$

Figure 1.5 (i) and (ii) shows pin-jointed frames that are determinate. For frame (i):

Effigy ane.5.

$D=5+\mathrm{iii}-(2\times 4)=0$

and for (2):

$D=2+4-(2\times 3)=0$

Figure ane.vi (i) and (ii) shows pin-jointed frames that are indeterminate. For frame (i)

Figure 1.6.

$D=10+3-(2\times \mathrm{vi})=1$

and for (ii):

$D=11+4-(\mathrm{two}\times 6)=3$

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Welding and Defect Acceptance

Qiang Bai , Yong Bai , in Subsea Pipeline Design, Analysis, and Installation, 2014

Abstract

During offshore pipeline installation, the occasional weld defect repair may be carried out just after the sternmost tensioner and before the side by side roller support in the repair station. The local stresses at the weld repair are intensified during the weld excavation process. Defects exceeding the project criteria are thoroughly removed; nevertheless, the removal of excess material should exist minimized to minimize local stresses. The determination of condom weld digging sizes for repair is 1 of the more difficult evaluations in pipeline installation engineering. This subject has attracted much attention because of personnel safety and the risk of the pipeline parting or buckling on the ramp during excavation. This chapter presents an analytical method of determining condom weld earthworks lengths, preventing both plastic collapse and fast fracture of the pipe during weld defect repair.

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Welding and Defect Credence

Yong Bai , Qiang Bai , in Subsea Pipelines and Risers, 2005

33.i Introduction

During offshore pipeline installation, the occasional weld defect repair may exist carried out only after the stern-well-nigh tensioner and before the adjacent roller support in the repair station. The weld defect is removed past grinding and subsequent re-welding. Aft of the tensioner, the pipe undergoing earthworks is highly loaded past angle and tension. The local stresses at the weld repair are intensified during the weld excavation process. Defects in the girth weld are located and measured by radiography or ultrasonics. Defects exceeding the projection criteria are thoroughly removed, however, the removal of excess material should be minimized so as to minimize local stresses. The determination of safe weld excavation sizes for repair is one of the more than difficult evaluations in pipeline installation engineering. This subject has attracted much attending from pipeline owners, installation contractors, and operating companies because of personnel safety and as well the risk of the pipeline parting or buckling on the ramp during earthworks. This paper presents an analytical method of determining safety weld excavation lengths preventing both plastic collapse (buckling) and fast fracture of the pipage during weld defect repair.

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Statically determinate pin-jointed frames

Alan Williams Ph.D., S.E., C.ENG. , in Structural Analysis, 2009

2.2 Statical Determinacy

A statically determinate truss is one in which all member forces and external reactions may exist determined by applying the equations of equilibrium. In a simple truss, external reactions are provided past either swivel supports or roller supports, as shown in Figure 2.iii (i) and (two). The roller support provides only one degree of restraint, in the vertical direction, and both horizontal and rotational displacements tin can occur. The swivel support provides two degrees of restraint, in the vertical and horizontal directions, and but rotational displacement tin occur. The magnitudes of the external restraints may exist obtained from the 3 equations of equilibrium. Thus, a truss is externally indeterminate when it possesses more than iii external restraints and is unstable when information technology possesses less than iii.

Effigy two.3.

In a simple truss with j nodes, including the supports, 2j equations of equilibrium may be obtained, since at each node:

$\Sigma H=0$

and

$\Sigma V=0$

Each member of the truss is subjected to an unknown axial forcefulness; if the truss has due north members and r external restraints, the number of unknowns is (due north +r). Thus, a uncomplicated truss is determinate when the number of unknowns equals the number of equilibrium equations or:

$\mathrm{north}+r=2j$

A truss is statically indeterminate, as shown in Figure 2.4, when:

Figure 2.4.

$n+r>2j$

The truss at (i) is internally redundant, and the truss at (two) is externally redundant.

A truss is unstable, as shown in Figure 2.5, when:

Figure 2.5.

$n+r<2j$

The truss at (i) is internally deficient, and the truss at (ii) is externally deficient.

However, a situation tin can occur in which a truss is deficient fifty-fifty when the expression n +r = 2j is satisfied. As shown in Figure 2.6, the left-hand side of the truss has a redundant member, while the right-hand side is unstable.

Effigy two.6.

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Principles of statics

Alan Williams Ph.D., S.E., C.ENG. , in Structural Analysis, 2009

1.five Triangle of forces

When a structure is in equilibrium under the action of three concurrent forces, the forces class a triangle of forces. As indicated in Effigy i.10 (i), the iii forces F ₁, F ₂, and F ₃ are concurrent. As shown in Figure ane.10 (ii), if the initial point of forcefulness vector F ₂ is placed at the concluding signal of forcefulness vector F ₁, and then the force vector F ₃ drawn from the terminal point of force vector F ₂ to the initial signal of force vector F ₁ is the equilibrant of F ₁ and F ₂. Similarly, as shown in Figure 1.x (3), if the forcefulness vector F _three is drawn from the initial betoken of force vector F ₁ to the terminal indicate of force vector F ₂, this is the resultant of F _ane and F ₂. The magnitude of the resultant is given algebraically past:

Figure 1.10.

${(F_{\mathrm{three}})}^{\mathrm{ii}}={(F_1)}^2+{(F_2)}^2-\mathrm{ii}{F}_1{F}_{\mathrm{ii}}cos{f}_3$

and:

$F_3=F_{\mathrm{i}}sin{f}_{\mathrm{iii}}csc{f}_{\mathrm{i}}$

or:

$\begin{array}{c}{F}_1/sin{f}_1=F_2/sin{f}_2\\ =F_3/sin{f}_3\end{array}$

Example i.2

Make up one's mind the angle of inclination and magnitude of the support reaction at stop i of the pin-jointed truss shown in Figure 1.11 . End one of the truss has a hinged support, and cease two has a roller support.

Figure one.11.

Solution

Taking moments about support ane gives:

$\begin{array}{c}{V}_2=\mathrm{viii}\times 20/\mathrm{xvi}\\ =\mathrm{ten}\text{kips}\dots \text{ acting vertically upward}\end{array}$

The triangle of forces is shown at (2), and the magnitude of the reaction at support i is given past:

$\begin{array}{c}{R}^2={(V_2)}^2+{(F_3)}^{\mathrm{two}}-25_2{F}_{\mathrm{three}}cos{f}_{\text{r}}\\ ={10}^2+{20}^{\mathrm{two}}-\mathrm{two}\times 10\times 20cos90°\\ R={(100+400)}^{\mathrm{0.v}}\\ =22.36\text{kips}\end{array}$

The bending of inclination of R is:

$\begin{array}{c}\theta =\text{atan}(10/\mathrm{xx})\\ =26.57°\end{array}$

Alternatively, since the 3 forces are concurrent, their betoken of concurrency is at point six in Figure ane.eleven (i), and:

$\begin{array}{c}\theta =\text{atan}(8/\mathrm{sixteen})\\ =26.57°\end{array}$

and

$\begin{array}{c}R=20sin90°/sin63.43°\\ =22.36\text{kips}\end{array}$

The reaction R may besides be resolved into its horizontal and vertical components:

$\begin{array}{c}{H}_{\mathrm{i}}=Rcos\theta \\ =20\text{kips}\\ 5_1=Rsin\theta \\ =\mathrm{x}\text{kips}\end{array}$

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Elements in flexure

Alan Williams Ph.D., S.E., C.ENG. , in Structural Analysis, 2009

Publisher Summary

A statically determinate beam or rigid frame is 1 in which all the member forces and external reactions can be determined by applying the equations of equilibrium. In a axle or rigid frame, external reactions are provided by either hinge or roller supports, or past a fixed end. The roller back up provides merely one degree of restraint, in the vertical direction, and both horizontal and rotational displacements can occur. The hinge support provides two degrees of restraint, in the vertical and horizontal directions, and but rotational displacement tin occur. The stock-still stop provides iii degrees of restraint—vertical, horizontal, and rotational. Identifying whether a construction is determinate depends on the configuration of the structure. If the rigid frame has n members and r external restraints, the number of unknowns is (threen + r). Thus, a beam or frame is determinate when the number of unknowns equals the number of equilibrium equations.

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Normal Force, Shear Force, Angle Moment and Torsion

Dr. T.H.K. Megson , in Structural and Stress Assay (Fourth Edition), 2019

3.iii Normal force

Case iii.one

Construct a normal force diagram for the beam AB shown in Fig. 3.viii(a).

Figure iii.8. Normal force diagram for the beam of Ex. 3.1.

The first step is to calculate the back up reactions using the methods described in Section ii.v . In this example, since the beam is on a roller support at B, the horizontal load at B is reacted at A; conspicuously R _A,H=10   kN acting to the left.

Generally the distribution of an internal strength will alter at a loading discontinuity. In this case there is no loading discontinuity at any section of the beam then that we can determine the complete distribution of the normal force by calculating the normal force at whatever section X, a altitude x from A.

Consider the length AX of the beam as shown in Fig. three.8(b) (every bit we could consider the length XB). The internal normal force acting at X is N _AB which is shown acting in a positive (tensile) direction. The length AX of the beam is in equilibrium under the action of R _A,H (=10   kN) and N _AB. Thus, from Department 2.4, for equilibrium in the x direction

${\mathrm{Due\; north}}_{\mathrm{AB}}–R_{\mathrm{A},\mathrm{H}}={\mathrm{North}}_{\mathrm{AB}}–10=0$

which gives

$N_{\mathrm{AB}}=+10\mathrm{kN}$

N _AB is positive and therefore acts in the assumed positive management; the normal strength diagram for the complete beam is then as shown in Fig. 3.viii(c).

When the equilibrium of a portion of a construction is considered as in Fig. 3.8(b) nosotros are using what is termed a free body diagram.

Example 3.2

Draw a normal force diagram for the beam ABC shown in Fig. 3.nine(a).

Figure 3.9. Normal force diagram for the axle of Ex. iii.2.

Once again past because the overall equilibrium of the axle we encounter that R _A,H  =   x   kN acting to the left (C is the roller back up).

In this case there is a loading discontinuity at B so that the distribution of the normal strength in AB will be different to that in BC. We must therefore determine the normal forcefulness at an arbitrary section X_one between A and B, so at an arbitrary section X₂ between B and C.

The free body diagram for the portion of the axle AX₁ is shown in Fig. 3.9(b). (Alternatively we could consider the portion X₁C). As before, we describe in a positive normal force, Northward _AB. Then, for equilibrium of AX_one in the x direction

$N_{\mathrm{AB}}–\mathrm{x}=0$

so that

$N_{\mathrm{AB}}=+10\text{kN}(\text{tension})$

Now consider the length ABX_ii of the beam; again we draw in a positive normal strength, N _BC. So for equilibrium of ABX₂ in the x direction

$N_{\mathrm{BC}}+\mathrm{x}–\mathrm{x}=0$

which gives

${\mathrm{Due\; north}}_{\mathrm{BC}}=0$

Notation that we would have obtained the aforementioned result by because the portion X₂C of the beam.

Finally the complete normal force diagram for the beam is drawn as shown in Fig. 3.nine(d).

Example iii.3

Figure three.10(a) shows a axle ABCD supporting three concentrated loads, two of which are inclined to the longitudinal axis of the axle. Construct the normal strength diagram for the beam and determine the maximum value.

Figure 3.x. Normal force diagram for the beam of Ex. 3.iii.

In this example we are just concerned with determining the normal forcefulness distribution in the beam, then that it is unnecessary to calculate the vertical reactions at the supports. Further, the horizontal components of the inclined loads can only be resisted at A since D is a roller support. Thus, considering the horizontal equilibrium of the axle

$R_{\mathrm{A},\mathrm{H}}+\mathrm{half-dozen}\cos 60°–4\cos 60°=0$

which gives

$R_{\mathrm{A},\mathrm{H}}=–\mathrm{ane}\mathrm{kN}$

The negative sign of R _A,H indicates that the reaction acts to the right and not to the left as originally assumed. However, rather than alter the direction of R _A,H in the diagram, information technology is simpler to retain the causeless direction and so insert the negative value as required.

Although in that location is an apparent loading discontinuity at B, the 2   kN load acts perpendicularly to the longitudinal centrality of the axle and volition therefore non bear upon the normal forcefulness. We may therefore consider the normal forcefulness at any department Ten₁ betwixt A and C. The free body diagram for the portion AX₁ of the beam is shown in Fig. 3.10(b); again we draw in a positive normal forcefulness N _Ac. For equilibrium of AX₁

${\mathrm{Northward}}_{\mathrm{AC}}-R_{\mathrm{A},\mathrm{H}}=0$

so that

$N_{\mathrm{AC}}=R_{\mathrm{A},\mathrm{H}}=-1\mathrm{kN}(\text{pinch})$

The horizontal component of the inclined load at C produces a loading discontinuity then that nosotros now consider the normal force at whatsoever section 10_two betwixt C and D. Here it is slightly simpler to consider the equilibrium of the length Ten₂D of the beam rather than the length AX₂. Thus, from Fig. 3.10(c)

$N_{\mathrm{CD}}-\mathrm{iv}\cos 60°=0$

which gives

$N_{\mathrm{CD}}=+2\mathrm{kN}(\text{tension})$

From the completed normal force diagram in Fig. 3.10(d) we run across that the maximum normal force in the beam is 2   kN (tension) acting at all sections between C and D.

Example 3.4

Construct the normal force diagram for the cranked cantilever beam shown in Fig. 3.11(a).

Effigy 3.11. Normal forcefulness diagram for the beam of Ex. 3.4.

Annotation that in this case there will exist two components of support reaction at the congenital-in stop of the beam, R _A,H, and R _A,V (there will besides be a moment reaction but since we are concerned simply with normal force this is irrelevant). However, if nosotros consider the equilibrium of portions of the axle away from the built-in stop it will not be necessary to calculate them. Annotation also that there is a loading discontinuity at B and structural discontinuities at C and B. Initially, therefore, nosotros consider the normal force, North _DC, at the section X_i as shown in Fig. three.eleven(b).

For horizontal equilibrium of the length DX_i of the axle

${\mathrm{Due\; north}}_{\mathrm{DC}}+5=0$

and then that

$N_{\mathrm{DC}}=-\mathrm{v}\mathrm{kN}(\text{compression})$

The vertical 10   kN load acting at D will produce a normal force in CB. Then, because the vertical equilibrium of the portion DCX₂ of the beam in Fig. iii.11(c)

${\mathrm{Northward}}_{\mathrm{CB}}-10=0$

which gives

${\mathrm{Northward}}_{\mathrm{CB}}=+10\mathrm{kN}(\text{tension})$

Finally we consider the horizontal equilibrium of the portion DCBX₃ of the beam in Fig. 3.eleven(d).

$N_{\mathrm{BA}}-\mathrm{eight}+5=0$

from which

${\mathrm{Northward}}_{\mathrm{BA}}=+3\mathrm{kN}(\text{tension})$

The normal strength diagram for the complete beam is then as shown in Fig. 3.xi(due east). The normal force for the vertical portion CB may exist drawn on either side of CB equally is convenient.

Example 3.5

The cranked cantilever ABC shown in Fig. three.12(a) carries a uniformly distributed load of 10 kN/m of horizontal length. Construct the normal strength diagram for the cantilever.

Effigy 3.12. Normal force diagrams for the beam of Ex. 3.5.

Consider whatsoever section 10_i of the cantilever betwixt C and B equally shown in Fig. 3.12(b).

Resolving forces horizontally for this length of axle gives

$N_{\mathrm{BC}}=0$

At present consider any section 10₂ between B and A where X₂ is perpendicular to the axis of the beam as shown in Fig. three.12(c).

Resolving forces parallel to the centrality of the beam gives

$N_{\mathrm{B}\mathrm{A}}-10x\cos 60°-10\times \mathrm{i}\times \cos 60°=0$

and so that Northward _BA = 5(x + 1)

When 10 = 0, Due north _BA = 5 kN (tension) and when 10 = i.5m, Northward _BA = 12.five kN (tension).

The complete normal strength distribution is shown in Fig. 3.12(d).

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Volume one

Chang-Koon Choi , Tae-Yeol Lee , in Computational Mechanics–New Frontiers for the New Millennium, 2001

Movable shoe

To testify the applicability of the variable-node hexahedral elements in local mesh refinement, the upper part of a movable shoe is considered. For simplicity simply a quadrant of the shoe is actually modeled using 'MR-H10' series (Choi et al. 2001 ), and the boundary condition for lower surface is given as a roller support, i.e., displacements are restricted just in the radial direction. A uniform pressure level of full 6000 tf is interim on the elevation surface of the model. Material properties are given every bit Young'southward modulus Due east  =   2.1   ×   l0^iv  kgf/mm², Poisson'south ratio v   =  0.iii. Analysis outcome shows the vertical displacement as 0.11230   mm at a middle point of superlative surface, and the constructive stress distribution on the surface of mesh are shown in Figure ix(d). This practical instance shows that as the mesh gradation by using variable-node elements can maintain the mesh regularity in refinement procedure, the variable node element is very effective for adaptive mesh refinement where the finer mesh in connected to the coarser mesh.

Figure 9. Movable shoe

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