How to Draw Stress Distribution Diagram

SHEAR STRESS DISTRIBUTION

E.J. HEARN Ph.D., B.Sc. (Eng.) Hons., C.Eng., F.I.Mech.E., F.I.Prod.Due east., F.I.Diag.E. , in Mechanics of Materials 1 (Third Edition), 1997

Problems

seven.1

(A/B). A uniform I-section beam has flanges 150 mm wide by 8 mm thick and a web 180 mm wide and 8 mm thick. At a certain section there is a shearing force of 120 kN. Draw a diagram to illustrate the distribution of shear stress across the section as a result of angle. What is the maximum shear stress?

[86.7 MN/m².]

7.2

(A/B). A girder has a uniform T cross-department with flange 250 mm × 50 mm and web 220 mm × 50 mm. At a sure department of the girder there is a shear force of 360 kN.

Plot neatly to calibration the shear–stress distribution across the section, stating the values:

(a)

where the web and the flange of the section meet;

(b)

at the neutral axis. [B.P.] [seven.47, 37.iv, 39.6 MN/grand².]

7.three

(A/B). A beam having an inverted T cross-section has an overall width of 150 mm and overall depth of 200 mm. The thickness of the crosspiece is 50 mm and of the vertical web 25 mm. At a certain section along the beam the vertical shear force is found to be 120 kN. Draw neatly to calibration, using xx mm spacing except where closer intervals are required, a shear–stress distribution diagram beyond this section. If the mean stress is calculated over the whole of the cross-sectional surface area, determine the ratio of the maximum shear stress to the mean shear stress.

[B.P.] [3.37.]

7.four

(A/B). The channel department shown in Fig. seven.17 is simply supported over a span of 5 m and carries a uniformly distributed load of xv kN/thou run over its whole length. Sketch the shearing-stress distribution diagram at the indicate of maximum shearing force and marking important values. Determine the ratio of the maximum shearing stress to the average shearing stress.

Fig. 7.17.

[B.P.] [iii, 9.2, 9.three MN/m²; two.42.]

vii.5

(A/B). Fig. seven.18 shows the cross-section of a axle which carries a shear force of 20 kN. Plot a graph to scale which shows the distribution of shear stress due to bending beyond the cross-department.

Fig. 7.eighteen.

[I.Mech.East.] [21.vii, 5.ii, 5.23 MN/thou².]

7.6

(B). Show that the difference between the maximum and hateful shear stress in the spider web of an I-section beam is $\frac{Q{h}^2}{24I}$ where Q is the shear forcefulness on the cross-department, h is the depth of the web and I is the second moment of area of the cantankerous-section about the neutral axis of bending. Assume the I-department to be congenital of rectangular sections, the flanges having width B and thickness t and the spider web a thickness b. Fillet radii are to be ignored.

[I.Mech.E.]

seven.vii

(B). Deduce an expression for the shearing stress at whatsoever point in a section of a axle owing to the shearing force at that section. Land the assumptions made.

A but supported beam carries a primal load W. The cantankerous-department of the axle is rectangular of depth d. At what distance from the neutral axis volition the shearing stress exist equal to the mean shearing stress on the section?

[U.L.C.I.] [d/√12.]

seven.8

(B). A steel bar rolled to the section shown in Fig. 7.xix is subjected to a shearing force of 200 kN applied in the direction YY. Making the usual assumptions, determine the boilerplate shearing stress at the sections A, B, C and D, and find the ratio of the maximum to the mean shearing stress in the section. Describe to calibration a diagram to show the variation of the boilerplate shearing stress across the section.

Fig. seven.19.

[U.L.] [Clue: care for equally equivalent section like to that of Instance 7.3.] [7.2, 12.3, 33.6, 43.8 MN/m², iii.93.]

7.9

(C). Using customary notation, show that the shear stress over the cantankerous-section of a loaded beam is given by τ = $\tau =\frac{QA\overline{y}}{Ib}$ .

The cantankerous-department of a axle is an isosceles triangle of base B and peak H, the base beingness arranged in a horizontal airplane. Find the shear stress at the neutral centrality owing to a shear force Q acting on the cross-section and express it in terms of the mean shear stress.

(The second moment of area of a triangle about its base is BH ³/12.) [U.L.C.I.] $\frac{\mathrm{viii}Q}{3BH};\frac{\mathrm{four}}{3}{\tau }_{\text{mean}}$

7.10

(C). A hollow steel cylinder, of 200 mm external and 100 mm internal diameter, acting as a axle, is subjected to a shearing force Q = 10 kN perpendicular to the axis. Determine the hateful shearing stress across the department and, making the usual assumptions, the average shearing stress at the neutral axis and at sections 25, 50 and 75 mm from the neutral axis equally fractions of the mean value.

Describe a diagram to evidence the variation of average shearing stress across the section of the cylinder.

[U.L.] [0.425 MN/m²; 1.87, one.65, 0.viii, 0.47 MN/m².]

7.eleven

(C). A hexagonal-cross-section bar is used equally a beam with its greatest dimension vertical and only supported at its ends. The axle carries a fundamental load of lx kN. Draw a stress distribution diagram for a section of the beam at quarter span. All sides of the bar have a length of 25 mm.

(I _North.A. for triangle = bh ³/36 where b = base and h = superlative.)

[0, 9.ii, xiv.8, 25.9 MN/mⁱⁱ at 12.5 mm intervals above and below the Northward.A.]

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Response and Blueprint for Shear and Torsion

Naveed Anwar , Fawad Ahmed Najam , in Structural Cross Sections, 2017

Shear Stress Due to Torsion

The shear stress distribution for torsion is even more complex than that for shear forcefulness. In a homogeneous linear elastic fellow member, the mechanical behavior for torsion can exist estimated by using elasticity theory. The bending of twist per unit length is as follows:

(four.4) $\theta =\frac{\mathit{TL}}{\mathit{JG}}$

where T is the twisting moment, G is the shear modulus of material, L is the length of member, and J is the torsional constant. For a rectangular section, $J$ is given equally

(four.5) $J=c{b}^3\left[\frac{\mathrm{ane}}{3}-0.21\frac{b}{c}\left(\mathrm{one}-\frac{b^4}{12c^{\mathrm{four}}}\right)\right]$

where c and b are the two sides of the rectangle with b≤c (see Fig. 4.six). The maximum shear stress is at the middle of the longer side c and its value is given in the following equation:

Figure 4.half dozen. Shear stresses in a rectangular section, due to torsion T. Note that the maximum values are at the edges of section.

(four.6) ${\tau }_{\max }=\frac{T}{\mu c{b}^2}$

where μ is a dimensionless coefficient which varies with the attribute ratio c/b equally given in Table 4.1.

Table four.i. The Values of µ With Varying Attribute Ratios for a Rectangular Section

c/b	1.0	1.5	1.75	2.0	two.5	iii.0	iv.0	6.0	8.0	10.0	∞
µ	0.208	0.231	0.239	0.246	0.258	0.267	0.282	0.299	0.307	0.313	0.333

For simple, convex shapes, the shear stresses due to torsion course a shear flow or stress loop around the centroid. For flanges sections, several shear flows may be formed, providing the required stress resultant. Fig. 4.7 shows the shear stress distribution in various cantankerous-sections due to applied torsional moment.

Figure iv.seven. Shear stress distribution in various cross-sections due to applied torsional force (CSI Section Builder). (A) In solid, convex shapes, the shear flow due to torsion is formed near the outer edges. The highest stress occurs on the border nearest to the centroid. (B) In flanged shapes, the shear menstruation due to torsion tends to form in thicker parts first, and and then in thinner parts. For constant thickness, a uniform flow may be formed (greenlight gray indicates high stress, dark gray and medium greyness indicate negligible stress). (C) The shear flow in thin-walled open sections is localized in each rectangular element. So the torsional constant can be computed by summation of torsional constant of all rectangular strips (low-cal gray indicates high stress). (D) The shear menses in a steel tube embedded in a concrete department is concentrated in the steel tube. This ascertainment is relevant to concrete sections with hoop reinforcement, where almost the entire torsion is resisted by the reinforcement (light grayness indicates high stress). (E) The maximum shear flow in airtight shapes forms a loop near the outer surface. (F) The shear flow in square and box sections is developed virtually the outer edges (light gray indicates high stress).

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Impact of the Fluid-Structure Interaction Modeling on the Human Vessel Hemodynamics

Mauro Malvè , ... Miguel Angel Martínez , in Advances in Biomechanics and Tissue Regeneration, 2019

5.three.2 Instantaneous Wall Shear Stress Comparison

The WSS distributions computed past ways of the CFD and FSI techniques are similar in the 2 considered arteries. These distributions that, as discussed earlier, are due to the catamenia structures within the arteries, match well with the other results constitute in the literature studies [5, 13, 31]. Contrary menstruum and low or oscillating WSS correlate with atherosclerosis evolution in the carotid artery bifurcation, every bit documented in Ku et al. [twoscore]. In Fig. 5.5, a comparison between the temporal minimal shear stress for the aorta (upper panel) and for the carotid artery (lower panel), respectively, for rigid wall and FSI simulations is shown. In the effigy, the value of the WSS at the DA and at the ECA are plotted as a office of the cardiac cycle. These locations take been selected and postprocessed because they registered the minimum values for the variable. The computed curves for both arteries show the same trend equally the inlet condition (see Fig. 5.3), but no negligible differences were found in the WSS values. The temporal variations of the WSS show higher values if computed with CFD simulation rather than the FSI analysis. This tendency is reflected by the two cases.

Fig. 5.5. Temporal history of the minimum WSS in specific locations related to flow recirculation for the aorta (left panel) and carotid artery (right panel).

For the aorta, at the DA (see Fig. 5.5, left panel) a gap between the temporal variations of the WSS computed by means of the FSI and the CFD techniques is clearly visible. The maximal difference is reached at fourth dimension t = 0.14   south. The WSS values obtained with the CFD technique reflect higher values forth the cardiac cycle. In particular, the comparing betwixt the CFD and FSI techniques shows that the maximal value of the WSS at peak flow computed with the CFD technique is almost two times that computed with the FSI approach. At peak systole (t = 0.14   south), the CFD-computed WSS is 1.631   Pa while the FSI-computed value is express to 0.8541. The latter highlights that the CFD not only tends to overestimate the WSS ciphering, only provides a value that is twice that computed with a compliant avenue. In the diastolic region of the wheel (see Fig. 5.5), a reduction of the gap betwixt the computed curves is visible. Nevertheless, but betwixt t = 0.55   s and t = 0.65   s do the two temporal histories overlap. For the carotid avenue, at the ECA (see Fig. 5.five, bottom panel), we can observe a gap between the minimal WSS values computed with FSI and with CFD. The maximal gap between both curves is reached at the systolic peak (t = 0.16   s) where the value computed through rigid wall analysis is approximately twice that calculated with FSI. The gap attenuates at time t = 0.38   s but recovers a considerable difference (0.1   Pa) until the end of the cardiac bike. In Fig. 5.sixA and B, the spatial distributions of the WSS at elevation menstruation during systole are depicted for the aorta and the carotid artery, respectively. The ii figures highlight nonuniform distributions and a part of the same differences registered in the numerical values. In both cases, it is clearly visible that the amplitude of the low WSS regions is reduced when computed using the CFD technique. The reduction is visible in the descending trunk of the aorta (see Fig. 5.6A), both in the frontal and dorsal view too equally in the respective views of the carotid artery (see Fig. five.6B). In the latter, the low WSS regions registered at the ECA, ICA, and CCA are strongly spatially less extended than those computed, including the compliance.

Fig. five.6. Comparing between FSI and CFD-computed WSS distribution at peak systolic period for the aorta (A) and the carotid artery (B).

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Flow Instabilities in Thin-wall Injection Molding of Thermoplastic Polyurethane

Christian D. Smialek , Christopher L. Simpson , in Specialized Molding Techniques, 2001

SHEAR STRESS

During flow a shear stress distribution develops in the cavity, with the shear stress at the freezing skin layer virtually the wall at a maximum value, retreating to a minimum value in the hot flowing heart. ⁵ The critical shear stress represents the value above which primary bonds in the polymer backbone can be broken during menstruum. If the shear force of menstruum is too high it will overcome the frictional forcefulness between the mold wall and the pare layer. This will human activity to tear some of the skin layer away from already frozen polymer (slip-stick phenomena). ⁵ This leads to serious cosmetic defects in the molding. This effect is magnified in thinwall molding.

Upon subsequent analysis of the molded parts it was institute that some of the samples molded at pressures greater than 109 MPa demonstrated this slip-stick phenomena. These parts were characterized by depressions in the superlative and bottom surfaces, along the whole length of the part. This pressure of 109 MPa was greater than the 84 MPa that was predicted to induce the onset of slippage.

To investigate this result, the cadre side of the mold was surfaced with a fine aluminum-oxide pulverization (vapor-honed) which acted to increment the roughness of the surface, and coefficient of sliding friction between the polymer and the cavity surface. Samples were then molded and analyzed. Information technology was found that the rougher surface led to a more stable flow, and a more than cosmetically appealing surface on the molding. Thus, from a practical standpoint, information technology is better to create a mold with a rougher surface to prevent slippage during filling.

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Methods for Evaluating Wheeled Vehicle Functioning

J.Y. Wong Ph.D., D.Sc. , in Terramechanics and Off-Road Vehicle Engineering (2d Edition), 2010

C. Stress Distributions on the Contact Patch of Towed Rigid Wheels

The radial and shear stress distributions under a towed rigid bike with diameter of 1.25  1000 and width of 0.15   m on compact sand are shown in Effigy 11.10 (Onafeko and Reece, 1967; Wong and Reece, 1967b). It shows that the maximum radial stress occurs at the point where the shear stress changes direction, which may be called the transition betoken. Information technology is found that the transition point is located at the junction of the two flow zones under a towed rigid wheel shown in Figure 2.18 (Wong, 1967; Wong and Reece, 1967b). Under the action of section AD of the rim of a towed rigid bicycle, the soil in the region ABD moves upward while the rim rotates around the instantaneous centre I, equally shown in Effigy 2.xviii. The soil therefore slides along section AD of the rim in such a fashion as to produce shear stresses in the direction opposite to that of cycle rotation, which is defined as positive, as shown in Figure 11.10. In the section betwixt A and Eastward in Figure two.18, the soil moves forrad slowly while the bicycle rim moves forrard relatively fast. In this region, the shear stress therefore acts in the management of bike rotation, which is defined every bit negative, every bit shown in Figure xi.ten. As the shear stress on part of the contact patch is acting in one direction (positive), while that on the other part acting in the other direction (negative), the resultant torque acting on a towed wheel is zero. This indicates that the towed bike is in free rolling condition.

Figure 11.x. Measured radial (normal) and shear stress distributions on the contact patch of a towed rigid bike on meaty sand

(Reprinted by permission of ISTVS from Onafeko and Reece, 1967)

The location of the transition point may exist estimated based on the theory of plastic equilibrium discussed in Section 2.two of Affiliate two. Since at the transition point, shear stress is zero, the radial (normal) stress at that indicate must be the major principal stress at the boundary of the soil mass. Co-ordinate to the theory of passive world force per unit area, there must be two slip lines on either side of the centrality of the major principal stress with an angle of 45° − φ/2, as AF and AG shown in Figure eleven.xi. φ is the bending of internal shearing resistance of the soil. It therefore follows that if a signal on the bike–soil interface can be institute where the slip lines have an angle of 45° − φ/2 with the radial, and so this betoken will be the transition point. This indicates that to locate the transit point, the slip line field on the boundary of the soil mass adjacent to the rim should be examined. It is found that the tangent of the trajectory of soil particles at the junction of the two flow zones below a towed rigid wheel coincides with the direction of the absolute velocity of the corresponding point on the rim. In solving problems in soil mechanics, it is usually assumed that the trajectories of soil particles in the flow zones represent one ready of the slip lines. If this assumption is adopted, then information technology follows that at the transition bespeak the direction of the accented velocity of the rim coincides with one of the slip lines at the boundary of the soil mass. Based on the higher up consideration, it becomes axiomatic that the problem of locating the transition point is now reduced to determining where the direction of the absolute velocity on the rim has an angle of 45° − φ/ii with the radial.

Figure xi.11. Diagram illustrating the angle between the management of the absolute velocity vector and the radial at a betoken on the rim of a towed rigid wheel

(Reprinted past permission of ISTVS from Wong and Reece, 1967b)

Referring to Effigy 11.11, information technology can be seen that the angle θ_r betwixt the direction of the accented velocity and the radial at different points on the wheel rim can be determined from the kinematics of the wheel. For a towed rigid bicycle of radius r with skid i_s , the instantaneous eye I is at a distance of Δr beneath the bottom-dead-centre, equally shown in Figure 11.11. In the following assay, the skid i_s of a bicycle is defined equally follows:

(11.26) $i_{\mathrm{southward}}=1-\frac{V_t}{V}=1-\frac{r\omega }{r_{\mathrm{east}}\omega }=1-\frac{r}{r+\mathrm{\Delta }r}$

where r is the radius of the rigid wheel, r_east is usually chosen the effective rolling radius of the bicycle, which is equal to r + Δr shown in Figure 11.11, and Δr = ri_southward /(ane − i_s ), ω is the athwart speed of the wheel, V_t is the theoretical speed of the bike center and is equal to rω, and Five is the bodily forward speed of the cycle centre and is equal to r_e ω. For a bicycle completely locked up during braking (i.eastward. its athwart speed is zippo, while the forrard speed of the bike centre is not zero), the skid i_south is 100%, following the definition of i_s given by Eqn (11.26).

From the geometry shown in Figure 11.eleven, the post-obit relationships tin can be established (Wong and Reece, 1967b):

$\frac{r}{r+\mathrm{\Delta }r}=\frac{sin\left[90°-(\theta +{\theta }_r)\right]}{sin(90°+{\theta }_r)}=\frac{cos(\theta +{\theta }_r)}{cos{\theta }_r}$

$1-i_s=cos\theta -sin\theta tan{\theta }_r$

(eleven.27) $tan{\theta }_r=\frac{cos\theta -(\mathrm{one}-i_{\mathrm{due\; south}})}{sin\theta }$

Since at the transition bespeak θ_r = 45° − θ/2, the angular position of the transition betoken θ_tr can be obtained by solving the following equation:

(xi.28) $tan({45}^{\circ }-\varphi /2)=\frac{cos{\theta }_{tr}-(1-i_s)}{sin{\theta }_{tr}}$

Effigy 11.12 shows the variations of θ_r with θ at various values of skid i_south . The athwart position θ_tr of the transition signal on the wheel rim at a given skid i_s is determined past the intersection of the bend for a given skid i_s and the horizontal line for θ_r = +(45° − θ/2), as shown in Figure eleven.12. Incidentally, another horizontal line representing θ_r = −(45° − θ/2) may also be drawn in Effigy xi.12 (not shown in the effigy) and it may have another intersection indicate on each bend. This bespeak occurs at an angle much greater than that shown in the effigy. Experimental testify indicates that the actual transition point locates at θ_r = +(45°− θ/2).

Figure xi.12. Relationships between the angles θ_r and θ at unlike skids

(Reprinted by permission of ISTVS from Wong and Reece, 1967b)

Table xi.2 shows a comparison between the measured angular positions of the transition point of two towed rigid wheels with different widths and the corresponding predicted ones obtained using the to a higher place-noted method on compact and loose sand (Wong and Reece, 1967b). Information technology shows that while there are differences in the absolute value between the predicted and measured athwart positions of the transition point, the variation of the predicted value shows a similar tendency to that of the measured. The differences are probably due to the fact that the prediction method is based on the consideration of a two-dimensional soil flow beneath a wheel, while the relatively narrow wheels used in the experiments actually cause a three-dimensional soil period.

Table 11.ii. Comparison of the measured and predicted locations of the transition points of towed rigid wheels of different dimensions on compact and loose sand

Soil type	Compact sand	Loose sand
Wheel dimensions	Diameter – one.25   thousand Width – 0.xv   m	Diameter – ane.25   m Width – 0.xv   m		Bore – 1.25   thou Width – 0.thirty   thousand
Sideslip, %	17.8	xxx.6	31.1	31.3	33.9	34.ix	35.3
Measured location of the transition point, θtr	11.6°	18.1°	xviii.i°	xx.0°	17.9°	17.nine°	17.nine°
Predicted location of the transition point, θtr	xv.3°	23.3°	23.six°	23.8°	25.5°	26.0°	26.two°

Source: Wong and Reece (1967b).

In summary, in the original method developed by Bekker, only the radial (normal) stress on the contact patch is taken into account and the shear stress is neglected. Experimental bear witness indicates that the shear stress is an essential gene that should exist taken into consideration in the assay of bicycle–terrain interaction. It also shows that the mechanics of interaction between a wheel and the terrain is much more complex than that causeless in the Bekker method. This suggests that in a more comprehensive analysis of the mechanics of bicycle–terrain interaction, both the radial (normal) and shear stresses on the contact patch should be taken into account. This is discussed in detail in the adjacent affiliate.

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Wings

T.H.G. Megson , in Aircraft Structures for Engineering science Students (7th Edition), 2022

23.iii Torsion

The chordwise pressure level distribution on an aerodynamic surface may be represented by shear loads (lift and drag loads) acting through the aerodynamic middle together with a pitching moment Chiliad ₀ (see Department 12.i). This system of shear loads may be transferred to the shear middle of the section in the form of shear loads S ₁₀ and Due south _y together with a torque T. Nosotros shall consider the pure torsion example offset, since this volition be found to be helpful in the assay of wings carrying shear loads. In the analysis, we assume that no axial constraint effects are present and that the shape of the wing section remains unchanged by the load application. In the absenteeism of axial constraint, there is no development of direct stress in the wing section, so that only shear stresses are nowadays. It follows that the presence of booms does not affect the analysis in the pure torsion case.

The wing section shown in Fig. 23.iv comprises N cells and carries a torque T, which generates private merely unknown torques in each of the Northward cells. Each cell therefore develops a constant shear catamenia q _I,q ₂, …, q _R , …, q _N given by Eq. (18.1).

Effigy 23.4. Multicell Fly Section Subjected to Torsion

The total torque is therefore

(23.four) $T=\sum _{R=\mathrm{i}}^N\mathrm{two}{A}_R{q}_R$

Although Eq. (23.4) is sufficient for the solution of the special case of a single-cell section, which is therefore statically determinate, additional equations are required for an N-jail cell section. These are obtained by considering the charge per unit of twist in each cell and the compatibility of displacement condition that all N cells possess the same rate of twist dθ/dz; this arises straight from the assumption of an undistorted cross-section.

Consider the Rth cell of the wing section, shown in Fig. 23.5. The charge per unit of twist in the jail cell is, from Eq. (17.22),

Figure 23.v. Shear Flow Distribution in the Rth Prison cell of an N-Cell Wing Section

(23.5) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}{A}_{R}G}{\oint }_{R}q\frac{\text{d}s}{t}$

The shear menstruum in Eq. (23.5) is constant forth each wall of the cell and has the values shown in Fig. 23.5. Writing ∫ dsouth/t for each wall as δ, Eq. (23.5) becomes

$\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{\mathrm{one}}{\mathrm{two}{A}_{R}G}[q_R{\mathrm{\delta }}_{12}+(q_R-q_{R-\mathrm{i}}){\mathrm{\delta }}_{23}+q_R{\mathrm{\delta }}_{34}+(q_R-q_{R+1}){\mathrm{\delta }}_{41}]$

or, rearranging the terms in foursquare brackets,

$\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}{A}_R\mathrm{One\; thousand}}[-q_{R-1}{\mathrm{\delta }}_{23}+q_R({\mathrm{\delta }}_{12}+{\mathrm{\delta }}_{23}+{\mathrm{\delta }}_{34}+{\mathrm{\delta }}_{41})-q_{R+1}{\mathrm{\delta }}_{41}]$

In general terms, this equation may be rewritten in the form

(23.6) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}{A}_{R}G}(-q_{R-\mathrm{i}}{\mathrm{\delta }}_{R-1}{}_{,R}+q_R{\mathrm{\delta }}_R-q_{R+1}{\mathrm{\delta }}_{R+1,R})$

in which δ _R-1,R is ∫ ds/t for the wall common to the Rth and (R – 1)th cells, δ _R is ∫ ds/t for all the walls enclosing the Rthursday cell, and δ _{R+   1,R} is ∫ ds/t for the wall common to the Rth and (R + 1)th cells.

The general form of Eq. (23.6) is applicable to multicell sections in which the cells are connected consecutively: that is, cell I is connected to jail cell Two, cell Two to cells I and III, and and then on. In some cases, cell I may be connected to cells II and III, and then forth (see Problem P.23.4), then that Eq. (23.6) cannot be used in its full general form. For this type of section, the term ∫ q(ds/t) should exist computed by considering ∫ q(ds/t) for each wall of a detail cell in turn.

In that location are N equations of the type (23.vi), which, with Eq. (23.4), comprise the N + 1 equations required to solve for the N unknown values of shear flow and the one unknown value of dθ/dz.

Frequently, in do, the skin panels and spar webs are fabricated from materials possessing different properties such that the shear modulus G is not constant. The analysis of such sections is simplified if the bodily thickness t of a wall is converted to a modulus-weighted thickness t* as follows. For the Rth cell of an N-cell fly department in which K varies from wall to wall, Eq. (23.5) takes the form

$\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}{A}_R}{\oint }_{R}q\frac{\text{d}\mathrm{due\; south}}{\mathrm{Thousand}t}$

This equation may exist rewritten as

(23.7) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}{A}_R{\mathrm{Thou}}_{\text{REF}}}{\oint }_{R}q\frac{\text{d}\mathrm{due\; south}}{(\mathrm{Yard}/G_{\text{REF}})t}$

in which K _REF is a convenient reference value of the shear modulus. Equation (23.7) is now rewritten as

(23.8) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{ii}{A}_R{G}_{\text{REF}}}{\oint }_{R}q\frac{\text{d}\mathrm{south}}{t^{*}}$

in which the modulus-weighted thickness t* is given by

(23.nine) $t^{*}=\frac{\mathrm{1000}}{G_{\text{REF}}}t$

And then, in Eq. (23.6), δ becomes ∫ ds/t*.

Example 23.two

Summate the shear stress distribution in the walls of the iii-cell wing department shown in Fig. 23.6 when it is subjected to a counterclockwise torque of eleven.iii kNm. The data are in Table 23.two.

Figure 23.6. Fly Department of Case 23.2

Tabular array 23.ii. Example 23.two

Wall	Length (mm)	Thickness (mm)	G (Northward/mm2)	Cell expanse (mm2)
12°	1650	1.22	24,200	A I = 258,000
12i	508	two.03	27,600	A II = 355,000
13, 24	775	1.22	24,200	A Iii = 161,000
34	380	1.63	27,600
35, 46	508	0.92	twenty,700
56	254	0.92	20,700

Note: The superscript symbols o and i are used to distinguish between outer and inner walls connecting the aforementioned two booms.

Since the wing section is loaded by a pure torque, the presence of the booms has no effect on the analysis.

Choosing G _REF = 27,600 N/mmⁱⁱ, then, from Eq. (23.ix),

$t_{12°}^{*}=\frac{\mathrm{24,200}}{\mathrm{27,600}}\times \mathrm{i.22}=1.07\text{mm}$

Similarly,

$t_{13}^{*}=t_{24}^{*}=\mathrm{one.07}\text{mm},t_{35}^{*}=t_{46}^{*}=t_{56}^{*}=0.69\text{mm}$

Hence,

${\delta }_{12°}={\int }_{12°}\frac{\text{d}s}{t^{*}}=\frac{\mathrm{1,650}}{\mathrm{one.07}}=1542$

Similarly,

${\mathrm{\delta }}_{{12}^{\text{i}}}=250,{\mathrm{\delta }}_{13}={\mathrm{\delta }}_{24}=725,{\mathrm{\delta }}_{34}=233,{\mathrm{\delta }}_{35}={\mathrm{\delta }}_{46}=736,{\mathrm{\delta }}_{56}=368$

Substituting the appropriate values of δ in Eq. (23.6) for each cell in plow gives the following:

•

For cell I,

(i) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{2\times \mathrm{258,000}{\mathrm{Grand}}_{\text{REF}}}[q_{\text{I}}(\mathrm{1,542}+250)-250q_{\text{Ii}}]$

•

For prison cell Ii,

(ii) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{two}\times \mathrm{355,000}{\mathrm{Grand}}_{\text{REF}}}[-250q_{\text{I}}+q_{\text{II}}(250+725+233+725)-233q_{\text{Three}}]$

•

For cell 3,

(iii) $\frac{\text{d}\mathrm{\theta }}{\text{d}z}=\frac{1}{\mathrm{ii}\times \mathrm{161,000}{\mathrm{Grand}}_{\text{REF}}}[-233q_{\text{II}}+q_{\text{III}}(736+233+736+368)]$

In addition, from Eq. (23.4),

(4) $\mathrm{xi.iii}\times {10}^{\mathrm{vi}}=\mathrm{ii}(\mathrm{258,000}{q}_{\text{I}}+\mathrm{355,000}{q}_{\text{II}}+\mathrm{161,000}{q}_{\text{3}})$

Solving Eqs. (i)–(iv) simultaneously gives

$q_{\text{I}}=7.1\text{N}/\text{mm},q_{\text{II}}=\mathrm{8.nine}\text{N}/\text{mm},q_{\text{III}}=\mathrm{four.2}\text{N}/\text{mm}$

The shear stress in any wall is obtained by dividing the shear flow by the actual wall thickness. Hence, the shear stress distribution is as shown in Fig. 23.7.

Figure 23.7. Shear Stress (N/mm²) Distribution in the Fly Department of Instance 23.ii

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Introduction to the rheometry of circuitous suspensions

G. Ovarlez , in Understanding the Rheology of Concrete, 2012

Shear localisation

In all geometries, the shear stress distribution is heterogeneous (even if the heterogeneity is slight in cone and plate and thin gap Couette geometries). In yield stress fluids, when the practical macroscopic stress is close to the yield stress τ_Y , this implies that the stress is lower than τ_Y and that the material does not flow in some regions of the gap of the geometry. This phenomenon is called 'shear localisation' as period occurs only in some regions.

This is illustrated in Fig. two.18a where local velocities measured in a concentrated emulsion flowing in a wide gap Couette geometry are depicted. It is observed that the fabric does not catamenia in the whole gap, the velocity existence nothing in a region near the outer cylinder. Moreover, the fraction of the material that is sheared decreases when the rotational velocity Ω of the inner cylinder decreases. Shear localisation is hither attributed to the 1/R ² subtract of the shear stress τ(r,Ω), which passes below τ_Y at the interface R_c (Ω) between the flowing and the non flowing regions. When Ω tends to 0, R_c (Ω) tends to R_i , i.eastward. the thickness of the flowing region tends to zero and the stress τ(R_i ,Ω) at the inner cylinder tends to τ _Y .

ii.18. (a) Dimensionless velocity (ratio between the local azimuthal velocity and the azimuthal velocity at the inner cylinder surface) every bit a function of the radial position inside the gap of a wide gap Couette geometry (R_i = 4.15   cm, R_o = vi   cm) for a full-bodied emulsion (NMR measurements). (b) Local constitutive police inferred from the velocity profiles of Fig. ii.18a (empty symbols), and data obtained using Eq. ii.eleven (crosses) (Ovarlez et al., 2008).

In standard macroscopic experiments, this phenomenon of shear localisation prevents us from calculating the true (local) shear charge per unit inside the gap with the standard equations of rheometry (see Department ii.1) and may thus lead to wrong evaluation of the rheological properties of the material. Information technology is nevertheless possible to take this artefact into business relationship thanks to Eq.2.xi, which is supposed to exist valid without any hypothesis about the flow. When shear is localised, the shear rate at the outer cylinder is nix and this equation reduces to $\mathrm{two}T=\frac{\partial \Omega }{\partial T}=\dot{\gamma }\left(\tau \left(R_i\right)\right),$ , which is easy to handle. From a whole set of macroscopic measurements T(Ω), the cloth constitutive police can and so exist computed. Figure 2.18b shows a comparison of data obtained using this approach and data measured locally from the velocity profiles thanks to the approach presented in Section ii.4.2, in the concentrated emulsion of Fig. two.18a. Information technology is observed that the data obtained with Eq.two.eleven match perfectly the local behaviour, which is a strong validation of this equation (Ovarlez et al., 2008; Coussot et al., 2009).

In conclusion, one has to be careful when studying yield stress fluids. Shear localisation forbids the use of unproblematic standard equations of rheometry, which lead to underestimation of the true shear rate. All the same, shear localisation can be properly taken into business relationship using more elaborated equations such as Eq. 2.xi in a Couette geometry.

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Computer-Aided Method NTVPM for Evaluating the Functioning of Vehicles with Flexible Tracks

J.Y. Wong Ph.D., D.Sc. , in Terramechanics and Off-Road Vehicle Technology (2nd Edition), 2010

8.4 Prediction of Motility Resistance and Drawbar Pull

When the normal pressure level and shear stress distributions under a tracked vehicle at a given slip have been determined, the tractive performance of the vehicle tin then exist predicted. The tractive performance of an off-road vehicle is ordinarily characterized by its move resistance, tractive effort, and drawbar pull (the difference between tractive endeavor and motion resistance) equally functions of slip.

The external motility resistance R_t of the runway tin can be determined from the horizontal component of the normal pressure acting on the track in contact with the ground. For a vehicle with two tracks, R_t can be described past

(eight.31) $R_t=2b{\int }_0^{L_t}psin\alpha \text{d}l$

where b is the contact width of the rails, L_t is the length of track in contact with the terrain, p is normal pressure, and α is the angle of the runway element with respect to the horizontal.

If the rail sinkage is greater than the ground clearance of the vehicle, the belly (hull) will be in contact with the terrain, giving ascension to an additional drag, known equally abdomen drag R_be . It can be determined from the horizontal components of the normal and shear stresses interim on the belly–terrain interface and is described by

(8.32) $R_{be}=b_b\left[{\int }_0^{L_b}{p}_{b}sin{\alpha }_b\text{d}\mathrm{fifty}+{\int }_0^{L_b}{s}_{b}cos{\alpha }_b\text{d}l\right]$

where b_b is the contact width of the abdomen, 50_b is the contact length of the belly, α_b is the angle of the abdomen with respect to the horizontal, and p_b and southward_b are the normal and shear stresses on the abdomen–terrain interface, respectively.

The tractive effort F of the vehicle can be calculated from the horizontal component of the shear stress acting on the rails in contact with the terrain. For a vehicle with two tracks, F is given by

(eight.33) $F=2b{\int }_0^{L_t}scos\alpha \text{d}l$

where due south is the shear stress on the track–terrain interface.

Since both the normal pressure level p and shear stress s are functions of rails sideslip, the rails motion resistance R_t , belly drag R_exist (if any) and tractive effort F vary with slip.

For a rail with safety pads, part of the total tractive effort is generated past the safe–terrain shearing. To predict the tractive effort developed past the rubber pads, the portion of the vehicle weight supported by the safety pads should exist estimated and the characteristics of rubber–terrain shearing should be taken into account.

It should be mentioned that the tractive endeavor F calculated by Eqn (8.33) is due to the shearing action of the track beyond the grouser tips (or due to safety–terrain shearing for a track with condom pads). For a track with long grousers, additional thrust volition be adult due to the shearing action on the vertical surfaces on either side of the rails. Co-ordinate to Reece (1964), the maximum shear force per unit rails length Due south_{five max} from the two sides of a track is given by

(8.34) ${\mathrm{Due\; south}}_{v\text{undefined}max}=\mathrm{iv}ch{sin}^2\left(\frac{\pi }{\mathrm{four}}+\frac{\varphi }{2}\right)+2h{z}_m\gamma tan\left(\frac{\pi }{4}+\frac{\varphi }{2}\right)cos\left(\frac{\pi }{2}-\varphi \right)$

where c, γ and ϕ are the cohesion, weight density, and angle of shearing resistance of the terrain, respectively, h is the grouser height, and z_m is the mean sinkage of the grouser.

The relationship between the shear strength adult on the ii sides of a track and shear displacement may be causeless to be like to that betwixt the shear stress and shear displacement described in Chapter 5. Therefore, the distribution of the shear force on the two vertical sides forth the length of the rail tin exist predicted in the same style as that described in Section eight.2, and the thrust F₅ from the two sides of a rails can exist expressed by

(eight.35) $F_v={\int }_0^{{\mathrm{Fifty}}_t}{\mathrm{Due\; south}}_{v}cos\alpha \text{d}l$

where S_v is the shear force per unit length of the track.

The drawbar pull F_d of the vehicle tin can be considered equally the difference betwixt the tractive endeavour and the total motion resistance (including the belly elevate, if whatsoever), and tin be expressed by

(viii.36) $\begin{array}{l}{F}_d=F+2F_{\mathrm{five}}-R_t-R_{be}\\ =\mathrm{two}b{\int }_0^{L_t}\mathrm{south}cos\alpha \text{d}\mathrm{fifty}+2{\int }_0^{L_t}{S}_{5}cos\alpha \text{d}l-2b{\int }_0^{L_t}psin\alpha \text{d}\mathrm{50}\\ -b_b\left[{\int }_0^{{\mathrm{Fifty}}_b}{p}_{b}sin{\alpha }_b\text{d}l+{\int }_0^{L_b}{\mathrm{due\; south}}_{b}cos{\alpha }_b\text{d}l\right]\end{array}$

From this equation, the human relationship between drawbar pull F_d and rail slip i can be adamant.

The analyses and procedures described in a higher place are implemented in the estimator-aided method NTVPM. In its development, particular attention has been paid to its user-friendliness. For instance, the dialogue box format for inputting all vehicle and terrain information is adopted for the convenience of the user. It runs on Microsoft Windows operating systems. Effigy 8.six shows the control centre, as displayed on the figurer monitor screen, for the operation of the latest version of NTVPM.

Figure 8.6. Command center for the operation of the computer-aided method NTVPM, every bit displayed on the monitor screen

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How to Draw Stress Distribution Diagram

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