ORIGINAL PAPER Yu-Hsi Huang ... · Kuang-Ming Hung · Yu-Hsi Huang ... around the circumference of...

Acta Mech 228, 1597–1619 (2017)DOI 10.1007/s00707-016-1766-3

ORIGINAL PAPER

Kuang-Ming Hung · Yu-Hsi Huang

Theoretical analysis and photoelastic experiment on full-fieldstress from multi-concentrated forces on circumference

Received: 29 August 2016 / Revised: 14 October 2016 / Published online: 11 January 2017© Springer-Verlag Wien 2016

Abstract This study derived a full-field stress series solution to multi-loads with small contact area uniformlydistributed around the circumference of a disk, and an analytical solution to concentrated forces spaced evenlyaround the circumference of a disk using the contribution of the full-field analytical solution of the Braziliantest model. Relative errors between the analytic solution and the series solution revealed that when 2α, theangle between the uniformly-distributed loads with small contact area, is less than 5◦, the full-field stressseries solution can be reliably replaced with the full-field stress analytical solution. Based on a photoelasticexperiment, we derived a formula bywhich to determine the full-field stress, resulting from concentrated forcesisotopically arranged around the circumference of the disk. The numerical findings are in good agreement withthose obtained in the photoelastic experiment, thereby verifying the efficacy of the simple theoretical formuladerived in this study. This study provides a full-field stress analytical solution for forces evenly spaced aroundthe circumference of a disk. This solution can be used to evaluate the internal stress distribution of diskstructures and optical components clamped at the edges.

1 Introduction

Changing the load on a disk from a concentrated force to uniformly-distributed radial pressure acting on twoopposite rims can be represented using the Brazilian test model. In 1959, Hondros [1] derived a full-fieldstress series solution for an analysis model under uniform compression, stress in the center of the disk, andsolutions to the stress functions of vertical and horizontal lines passing through the center of the disk. Thestress function solutions and strain gauge data were then used to inversely calculate the Young’s modulusand Poisson’s ratio of concrete. Recent progress in digital imaging technology and finite element analysis ishelping to advance experimental research [2] and numerical analyses [3,4] into the photoelasticity of disksunder applied loads at the top and bottom. Nonetheless, theoretical analysis and experimental measurementsare relatively easy. Except for a few specific instances, such as rock mechanics, most disk structures aresubject to only two symmetric loads, i.e., at the top and bottom. This study performs theoretical analysis ondisks subject to multi-forces evenly spaced around the circumference and compares the results with the seriessolution.

K.-M. HungDepartment of Mechanical Engineering, Hwa Hsia University of Technology, No. 111, Gongzhuan Rd., Zhonghe Dist.,235 New Taipei City, Taiwan, ROCE-mail: [email protected]

Y.-H. Huang (B)Department of Mechanical Engineering, National Taiwan University of Science and Technology, No. 43, Sec. 3, Keelung Rd.,Daan Dist., Taipei City, Taiwan, ROCE-mail: [email protected]

http://crossmark.crossref.org/dialog/?doi=10.1007/s00707-016-1766-3&domain=pdf

1598 K.-M. Hung, Y.-H. Huang

In engineering applications, the three-jaw ormulti-jaw universal chucks usedwithmachine tools [5] and theuniversal lens mounts used in optical components are equivalent to plane strain problems with disk structuresunder load. Stress affects the photoelastic properties of load-bearing optical components, such as the indexof refraction, as well as the imaging results [6,7]. However, research has been limited to the influence ofsurface stress and load on the optical properties of optical components. Little research has been conductedon the influence of the forces applied by supports at the rims of optical components. Sawyer [8] used thetheory of elasticity to discuss stress distribution caused by contact forces applied to lenses. Chaudhury etal. [9] investigated the deformation and optical imaging characteristics of cylindrical lenses subjected toconcentrated forces. Research on the application of force to multiple points on a disk began in 1969, whenFrocht [10] performed photoelastic experiments to measure stress, resulting from three-point and four-pointloads. Surendra and Simha [11,12] recently employed photoelasticity in the examination of friction effects indisks subjected to three-point loads with a focus on the location of isotropic points where the principal stressfulfills σ11 = σ22. This study discussed and compared discrepancies between experiment and theory withregard to three-point contact forces and contact angles.

In their research on the deformation of load-bearing disks, Sciammarella and Lamberti [13] used a semi-theoretical semi-analytical method with reference to experiment results for image analysis. Based on theBrazilian test model, Huang et al. [14] explored the strength distribution and damage to disks subjected toloads at the top and bottom with various contact angles. This study was based on the approach proposed byHung and Ma [15] in 2003, in which the full-field stress series solution of the Brazilian test model developedby Hondros[1] is replaced by a function solution. They performed photoelastic experiments with a proprietarydigital image device for the measurement of photoelasticity and compared the results with theoretical findings.In 2008, Ma and Hung [16] overcame the problem of integration in deriving a full-field displacement functionsolution from the stress function solution. Based on the above achievements, we firstly derived the stress seriessolution of a disk subject tomulti-uniformly-distributed loadswith small contact area around the circumference.After then, this was then simplified into an analytical solution of concentrated forces based on stress. Wesubsequently evaluated the accuracy of the results from numerical analysis and photoelastic experiments.

2 Theoretical analysis

Disk Model with Arbitrary Number of Loads with Uniform Radial DistributionThe model in this study presents a plane stress problem with isotropic material and no body forces. The

coordinate system and boundary conditions are presented in Fig. 1, involving k uniformly-distributed isotropicloads (pressure) p around the circumference of a disk with radius R, in which the contact angle between theloads is 2α. Figure1 illustrates the conditions with k = 5.

2.1 Series solution

In the following, we list the equilibrium equations with polar coordinates for the disk model used in this study.The model is a plane stress problem with linearly isotropic material disregarding the effects of body forces, asfollows:

∂σr

∂r+ σr − σθ

r+ 1

r

∂τrθ

∂θ= 0, (1)

1

r

∂σθ

∂θ+ ∂τrθ

∂r+ 2τrθ

r= 0, (2)

the boundary conditions of which are

σr |r=R = −2p

π

{α +

∞∑n=1

sin knα cos knθ

n

}, (3)

τrθ |r=R = 0. (4)

Suppose that the stress function is

φ (r, θ) = B0r2 +

∞∑n=1

fn(r) cos knθ, (5)

Theoretical analysis and photoelastic experiment on full-field stress 1599

Fig. 1 Boundary conditions of disk model

where

fn(r) = Anrkn + Bnr

kn+2 + C/rkn + D

/rkn−2. (6)

When r = 0, φ is bounded, such that C = D = 0, we obtain

φ (r, θ) = B0r2 +

∞∑n=1

[Anr

kn + Bnrkn+2]cos knθ. (7)

Based on these basic assumptions, the stress solution which satisfies the compatibility equation

(∂2

∂r2+ 1

r

∂

∂r+ 1

r2∂2

∂θ2

)(∂2

∂r2+ 1

r

∂

∂r+ 1

r2∂2

∂θ2

)φ = 0

is represented as

σr = ∂2φ

r2∂θ2+ ∂φ

r∂r, (8)

σθ = ∂2φ

∂r2, (9)

τrθ = − ∂

∂r

(1

r

∂φ

∂θ

)(10)


and automatically satisfies the stress balance equations. (1) and (2). Substituting Eq. (7) into Eqs. (8)–(10) givesthe following:

σr = ∂2φ

r2∂θ2+ ∂φ

r∂r= 2B0 −

∞∑n=1

[kn(kn − 1)Anr

kn−2 + (kn − 2)(kn + 1)Bnrkn]cos knθ, (11)

σθ = ∂2φ

∂r2= 2B0 +

∞∑n=1

[kn(kn − 1)Anr

kn−2 + (kn + 2)(kn + 1)Bnrkn]cos knθ, (12)

τrθ = − ∂

∂r

(∂φ

r∂θ

)=

∞∑n=1

kn[(kn − 1)Anr

kn−2 + (kn + 1)Bnrkn]sin knθ. (13)

To solve for the coefficients An and Bn in the stress function, we substitute Eqs. (11) and (13) into the boundaryconditions (3) and (4) to obtain the following:

σr |r=R = 2B0 −∞∑n=1

[kn(kn − 1)AnR

kn−2 + (kn − 2)(kn + 1)BnRkn]cos knθ

= −2p

π

{α +

∞∑n=1

sin knα cos knθ

n

}, (14)

τrθ |r=R = − ∂

∂r

(∂φ

r∂θ

)=

∞∑n=1

kn[(kn − 1)AnR

kn−2 + (kn + 1)BnRkn]sin knθ = 0. (15)

Comparison of the coefficients in Eqs. (14) and (15) results in

B0 = − p

πα, (16)

kn(kn − 1)Rkn−2An + (kn − 2)(kn + 1)Rkn Bn = 2p

π

sin knα

n, (17)

(kn − 1)Rkn−2An + (kn + 1)Rkn Bn = 0. (18)

Solving Eqs. (17) and (18) to obtain An and Bn results in

An = 2p

π

R2−kn sin knα

2n(kn − 1), (19)

Bn = −2p

π

R−kn sin knα

2n(kn + 1). (20)

By substituting Eqs. (16), (19), and (20) into Eqs. (11), (12), and (13), we obtain

σr = −2p

π

{α +

∞∑n=1

[k

2ρ−2 −

(k

2− 1

n

)]ρkn sin knα cos knθ

}, (21)

σθ = −2p

π

{α −

∞∑n=1

[k

2ρ−2 −

(k

2+ 1

n

)]ρkn sin knα cos knθ

}, (22)

τrθ = kp

π

∞∑n=1

[ρ−2 − 1

]ρkn sin knα sin knθ, (23)

where ρ = rR , |ρ ≤ 1|.


Equations (21), (22), and (23) are the full-field stress series solutions for the application ofmulti-uniformly-distributed loads applied to the circumference of the disk. When k = 2,

σr = −2p

π

{α +

n=∞∑n=1

[1 −(1 − 1

n

)ρ2]ρ2n−2 sin 2nα cos 2nθ

}, (24)

σθ = −2p

π

{α −

n=∞∑n=1

[1 −(1 + 1

n

)ρ2]ρ2n−2 sin 2nα cos 2nθ

}, (25)

τrθ = 2p

π

{n=∞∑n=1

[1 − ρ2]ρ2n−2 sin 2nα sin 2nθ

}, (26)

Eqs. (24), (25), and (26) perfectly match the full-field stress series solutions derived by Hondros [1] for theBrazilian test model in 1959.

When α = π/k, the rim of the disk is subject to uniform compression q = −2p/k. Substituting α = π/kinto Eqs. (21), (22), and (23) indicates that the stress state at this time is σr = σθ = q and τrθ = 0, whichis referred to as hydrostatic compression. The derivations in this study are consistent with those obtained byTimoshenko and Goodier [17].

2.2 Analytical solution

Using the characteristics of trigonometric functions,

sin knα cos knθ = 1

2[sin kn(α + θ) + sin kn(α − θ)] = 1

2Im[eikn(α+θ) + eikn(α−θ)

], (27)

sin knα sin knθ = 1

2[cos kn(α − θ) − cos kn(α + θ)] = 1

2Re[eikn(α−θ) − eikn(α+θ)

]. (28)

Substituting Eqs. (27) and (28) into Eqs. (24)–(26) produces

σr = − p

π

⎧⎪⎪⎨⎪⎪⎩2α + k

2

(ρ−2 − 1

)Im

∞∑n=1

[ρei(α+θ)

]kn + k2

(ρ−2 − 1

)Im

∞∑n=1

[ρei(α−θ)

]kn+Im

∞∑n=1

1n

[ρei(α+θ)

]kn + Im∞∑n=1

1n

[ρei(α−θ)

]kn⎫⎪⎪⎬⎪⎪⎭ , (29)

σθ = − p

π

⎧⎪⎪⎨⎪⎪⎩2α − k

2 (ρ−2 − 1)Im

[ρei(α+θ)

]kn − k2 (ρ

−2 − 1)Im∞∑n=1

[ρei(α−θ)

]kn+Im

∞∑n=1

1n

[ρei(α+θ)

]kn + Im∞∑n=1

1n

[ρei(α−θ)

]kn⎫⎪⎪⎬⎪⎪⎭ , (30)

τrθ = kp

2π(ρ−2 − 1)

{Re

∞∑n=1

[ρei(α−θ)

]kn − Re∞∑n=1

[ρei(α+θ)

]kn}. (31)

Next, we recall that∑∞

n=0 Xn = 1

1−X , where |X | < 1. Due to the fact that∣∣∣[ρei(α±θ)

]k∣∣∣ < 1, we can derive

the following:

∞∑n=1

[ρei(α±θ)

]kn =∞∑n=0

[ρei(α±θ)

]kn − 1 = 1

1 − [ρei(α±θ)]k − 1 =

[ρei(α±θ)

]k1 − [ρei(α±θ)

]k= ρk cos k(α ± θ) + i sin k(α ± θ)[

1 − ρk cos k(α ± θ)]− iρk sin k(α ± θ)

= ρk cos k(α ± θ) − ρk + i sin k(α ± θ)

1 + ρ2k − 2ρk cos k(α ± θ). (32)


Based on Eq. (32),

Im∞∑n=1

[ρei(α±θ)

]kn = ρk sin k(α ± θ)

1 + ρ2k − 2ρk cos k(α ± θ), (33)

Re∞∑n=1

[ρei(α±θ)

]kn = ρk cos k(α ± θ) − ρ2k

1 + ρ2k − 2ρk cos k(α ± θ). (34)

Substituting Eqs. (33) and (34) into Eqs. (29)–(31) produces

σr = − p

π

⎧⎪⎨⎪⎩2α + k

2

(ρ−2 − 1

)ρk[

sin k(α+θ)

1+ρ2k−2ρk cos k(α+θ)+ sin k(α−θ)

1+ρ2k−2ρk cos k(α−θ)

]+Im

∞∑n=1

1n

[ρei(α+θ)

]kn + Im∞∑n=1

1n

[ρei(α−θ)

]kn⎫⎪⎬⎪⎭ , (35)

σθ = − p

π

⎧⎪⎨⎪⎩2α − k

2

(ρ−2 − 1

)ρk[

sin k(α+θ)



]+Im

∞∑n=1

1n

[ρei(α+θ)

]kn + Im∞∑n=1

1n

[ρei(α−θ)

]kn⎫⎪⎬⎪⎭ , (36)

τrθ = kp

2π(ρ−2 − 1)ρk

{cos k(α − θ) − ρk

1 + ρ2k − 2ρk cos k(α − θ)− cos k(α + θ) − ρk

1 + ρ2k − 2ρk cos k(α + θ)

}. (37)

Because 1nρkneikn(α±θ) = i

∫ρkneikn(α±θ)d (k(α ± θ)),

Im∞∑n=1

1

n

[ρei(α±θ)

]kn =∫

Im(i)∞∑n=1

[ρei(α±θ)

]knd(k(α ± θ))

=∫

Im(i)

[ρk cos k(α ± θ) − ρk + i sin k(α ± θ)

1 + ρ2k − 2ρk cos k(α ± θ)

]d(k(α ± θ))

= ρk∫ [

cos k(α ± θ) − ρk

1 + ρ2k − 2ρk cos k(α ± θ)

]d(k(α ± θ)). (38)

We use the equation developed by Gradshteyn and Ryzhik [18] for the integration in Eq. (38), as follows:

∫A + B cos X

a + b cos XdX = B

bX + Ab − aB

b

2√a2 − b2

tan−1

[√a2 − b2

a + btan

X

2

], (39)

where a2 > b2. Let

A = −ρk, (40)

B = 1, (41)

a = 1 + ρ2k, (42)

b = −2ρk, (43)

X = k(α ± θ). (44)

a2 > b2 because 0 ≤ ρ ≤ 1 and a2 − b2 = 1 + 2ρ2k + ρ4k − 4ρ2k = 1 − 2ρ2k + ρ4k = (1 − ρ2k)2 > 0.Substituting Eqs. (40)–(44) into Eq. (39) gives the following:

∫ [cos k(α ± θ) − ρk

1 + ρ2k − 2ρk cos k(α ± θ)

]d(k(α ± θ))

= 1

ρk

{−k

2(α ± θ) + tan−1

[1 + ρk

1 − ρktan

k

2(α ± θ)

]}+ C±. (45)


Substituting Eq. (45) into Eqs. (35) and (36) gives the following:

σr = − p

π

⎧⎨⎩

(2 − k)α + k2

(ρ−2 − 1

)ρk[

sin k(α+θ)



]+ tan−1

[1+ρk

1−ρk tank2 (α + θ)

]+ tan−1

[1+ρk

1−ρk tank2 (α − θ)

]+ C±

⎫⎬⎭ , (46)

σθ = − p

π

⎧⎨⎩

(2 − k)α − k2

(ρ−2 − 1

)ρk[

sin k(α+θ)



]+ tan−1

[1+ρk


]+ tan−1

[1+ρk


]+ C±

⎫⎬⎭ . (47)

Finally, let p = P2αRt , where P is a concentrated force, and t denotes the thickness of the disk. Also,

when the arbitrary number k of uniformly-distributed loads on the circumference of the disk is converted intoconcentrated forces by approaching α → 0 , Eqs. (46), (47), and (37) can be simplified as follows:

limα→0

σr = σr |α→0 = limα→0

(− P

2πRt

)1

α

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

(2 − k)α

+ k2

(ρ−2 − 1

)ρk

[ sin k(α+θ)

1+ρ2k−2ρk cos k(α+θ)

+ sin k(α−θ)


]

+ tan−1[1+ρk


]+ tan−1

[1+ρk


]+ C±

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭

=(

− P

2πRt

){(2 − k) + k2(1 − ρ2)ρk−2[−2ρk + (1 + ρ2k) cos kθ ][

1 + ρ2k − 2ρk cos kθ]2 + k(1 − ρ2k)

1 + ρ2k − 2ρk cos kθ

},(48)

limα→0

σθ = σθ |α→0 = limα→0

(− P

2πRt

)1

α

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

(2 − k)α

− k2

(ρ−2 − 1

)ρk

[ sin k(α+θ)


+ sin k(α−θ)


]

+ tan−1[1+ρk


]+ tan−1

[1+ρk


]+ C±

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭

=(

− P

2πRt

){(2 − k) − k2(1 − ρ2)ρk−2[−2ρk + (1 + ρ2k) cos kθ ][

1 + ρ2k − 2ρk cos kθ]2 + k(1 − ρ2k)


},

(49)

limα→0

τrθ = τrθ |α→0 =(

P

4πRt

)k(ρ−2 − 1)ρk lim

α→0

1

α

⎧⎨⎩

cos k(α−θ)−ρk


− cos k(α+θ)−ρk


⎫⎬⎭

=(

P

2πRt

){k2(ρ−2 − 1)(1 − ρ2k)ρk sin kθ

(1 + ρ2k − 2ρk cos kθ)2

}. (50)

Equations (48)–(50) are full-field stress analytical solutions of k concentrated forceswith evenly spaced appliedaround the circumference of the disk. When k = 2, the problem becomes a disk subjected to concentratedforces at the top and bottom, the full-field stress analytical solution of which is

σr |α→0 = P

πRt

{(1 − ρ2

)2 (ρ4 + 2ρ2 − 1 − 2 cos 2θ

)(ρ4 + 1 − 2ρ2 cos 2θ

)2}

, (51)

σθ |α→0 = P

πRt

{ρ8 + 4ρ4 − 4ρ2 − 1 + 2

(−2ρ6 + ρ4 + 1)cos 2θ(

ρ4 + 1 − 2ρ2 cos 2θ)2

}, (52)

τrθ |α→0 = P

πRt

{2(1 − ρ4

) (1 − ρ2

)sin 2θ(

ρ4 + 1 − 2ρ2 cos 2θ)2}

. (53)

The results in Eqs. (51)–(53) are almost identical to the solutions obtained by Sokolnikoff [19], except for thefact that θ is measured in the opposite direction, resulting in a negative sign for τrθ .


2.3 Principal stress analytical solution

Substituting Eqs. (48)–(50) into the equation below,

σ1, σ2|α→0 = σr |α→0 + σθ |α→0

2±√(

σr |α→0 − σθ |α→0

2

)2+ (τrθ |α→0)

2, (54)

produces the principal stress from the analytical solution:

σ1, σ2|α→0 = P

2πRt

{−2[1 − (k − 1)ρ2k + (k − 2)ρk cos kθ ] ± k2ρ−2+k(ρ2 − 1)

1 + ρ2k − 2ρ2k cos kθ

}. (55)

Because k2ρ−2+k(ρ2 − 1) ≤ 0,

σ1|α→0 = P

2πRt

{−2[1 − (k − 1)ρ2k + (k − 2)ρk cos kθ ] − k2ρ−2+k(ρ2 − 1)


}, (56)

σ2|α→0 = P

2πRt

{−2[1 − (k − 1)ρ2k + (k − 2)ρk cos kθ ] + k2ρ−2+k(ρ2 − 1)


}, (57)

where

σ1|α→0 ≥ σ2|α→0 .

Thus, the maximum shear stress is

τmax|α→0 = σ1|α→0 − σ2|α→0

2= P

2πRt

{k2ρ−2+k(1 − ρ2)


}. (58)

3 Numerical calculation

3.1 Replacing series solutions for uniformly-distributed loads with analytical solutions for stress fromconcentrated force

In this section, we discuss discrepancies, resulting from the replacement of series solutions for three uniformly-distributed loads with analytical solutions for the stress from concentrated forces. In p = P

2αRt , when k = 3,and θ = 0, the three stresses in Eqs. (21)–(23) are dimensionless and normalized as the following:

σr/

(P

πRt

)= −

(1

α

){α +

∞∑n=1

[3

2ρ−2 −

(3

2− 1

n

)]ρ3n sin 3nα

}, (59)

σθ/

(P

πRt

)= −

(1

α

){α −

∞∑n=1

[3

2ρ−2 −

(3

2+ 1

n

)]ρ3n sin 3nα

}, (60)

τrθ /

(P

2πRt

)= 0. (61)

Similarly, the three stresses in Eqs. (48)–(50) are dimensionless and normalized as

σr |α→0 /

(P

πRt

)= −

(1

2

){2 + 11ρ + 11ρ2 + 4ρ3 + 4ρ4 + 4ρ5

(1 − ρ)(1 + ρ + ρ2)2

}, (62)

σθ |α→0 /

(P

πRt

)= −

(1

2

){2 − 5ρ − 12ρ2 − 8ρ3 − 4ρ4

(1 + ρ + ρ2)2

}, (63)

τrθ |α→0 /

(P

2πRt

)= 0. (64)


Fig. 2 Dimensionless and normalized radial stress σr when k = 3 resulting from a uniformly-distributed loads with 2α = 5◦; bconcentrated forces; c comparison of 2α = 5◦ and concentrated forces; the results indicate relative error 2.6% at ρ = −0.99 and5.1% at ρ = 0.85; d comparison of 2α = 10◦ and concentrated forces; the results indicate relative error 2.1% at ρ = −0.99 and5.1% at ρ = 0.71

Equations (59)–(61) present the series equations for three uniformly-distributed loads running along the verticalline passing through the center of the disk (k = 3), and Eqs. (62)–(64) are the analytical solutions to the stressinduced by concentrated forces applied at three points and running along the vertical line passing through thecenter of the disk (k = 3). As can be seen, the shear stress in both groups of equations is zero, which meansthat the stress along this line is the principal stress.

Figure2a displays the radial distribution of dimensionless σr based on Eq. (59) with 2α = 5◦. The verticalaxis measures the dimensionless σr , and the horizontal axis measures the normalized radius ρ in the rangeof [−1, 1]; ρ = −1 and ρ = 1 for the points at the top and bottom of the vertical line passing through thecenter of the disk. Figure2b presents the distribution of dimensionless σr based on the concentrated force (62),wherein the magnitude of the total force P is identical to that in Eq. (59). To enable comparisons, we combinedFig. 2a, b to create Fig. 2c. As can be seen, the two curves are highly consistent but diverge stronger whenapproaching ρ = 1. This is due to the boundary conditions associated with the negative load at ρ = 1. Totalforce P is the same; however, the boundary conditions in Fig. 2a are for uniformly-distributed loads, whereasthe boundary conditions in Fig. 2b are for concentrated forces. Thus, when closer to the boundary conditions,the influence of the stress concentration under boundary load (known as Saint-Venant’s principle) becomesincreasingly apparent. Calculations revealed that the relative error between the two dimensionless σr valuesat ρ = −0.99 was 2.6%, which increased to 5.1% at ρ = 0.85. For σr , equating uniformly-distributed loadswith boundary condition 2α = 5◦ to concentrated forces resulted in calculation error of <5% in the range of[−1, 0.85]. Figure2c, d shows relative error of 2.1% between the two dimensionless σr values at ρ = −0.99when 2α = 10◦. The relative error increased to 5.1% at ρ = 0.71. Thus, for σr , equating uniformly-distributed


Fig. 3 Dimensionless and normalized tangential stress σθ when k = 3 resulting from a uniformly-distributed loads with 2α = 5◦;b concentrated forces; c comparison of 2α = 5◦ and concentrated forces; the results indicate relative error 1.4% at ρ = −0.99and 5.1% at ρ = 0.72; d comparison of 2α = 10◦ and concentrated forces; the results indicate relative error 1.1% at ρ = −0.99and 5.1% at ρ = 0.43

loads with boundary condition 2α = 10◦ to concentrated forces resulted in a calculation error of <5% in therange of [−1, 0.71]. Figure2c, d indicates that a greater angle 2α implies a smaller acceptance range for ρ incalculations of σr .

Figure3a presents the distribution of dimensionless σθ based on Eq. (60) with the boundary condition2α = 5◦. The vertical axis measures the dimensionless σθ , and the horizontal axis measures normalized radiusρ in the range of [−1, 1]. Figure3b presents the distribution of dimensionless σθ based on the concentratedforce (63), in which the magnitude of total force P is identical to that in (60). To enable comparisons, wecombined Fig. 3a, b to create Fig. 3c. Beginning at ρ = −1, the two curves are initially consistent; however,the concentration of stress associated with the boundary load under Saint-Venant’s principle becomes moreapparent when approaching ρ = 1. Calculations revealed relative error of 1.4% between the two dimensionlessσθ values at ρ = −0.99, which increased to 5.1% at ρ = 0.72. For σθ , equating uniformly-distributed loadswith boundary condition 2α = 5◦ to concentrated forces resulted in calculation error of <5% in the range of[−1, 0.72]. Thus, the analysis results for σθ are less applicable than those for σr in range ρ. Figures2c and 3dshow a relative error of 1.1% between the two dimensionless σθ values at ρ = −0.99 when 2α = 10◦, whichincreased to 5.1% at ρ = 0.43. Thus, for σr , equating uniformly-distributed loads with boundary condition2α = 10◦ to concentrated forces resulted in a calculation error of <5% in the range of [−1, 0.43].

The above discussion shows that when 2α is <5◦, Eqs. (62)–(64) for concentrated forces can be used toreplace Eqs. (59)–(61) for series solutions in calculations of σr .


Fig. 4 a σr/ (2p/π), b σθ/ (2p/π), and c τrθ / (2p/π) contours based on full-field stress series solution of uniformly-distributedloads applied around the circumference of a disk with k = 3 and 2α = 2◦


3.2 Distribution of full-field stress by series solutions in a disk under uniformly-distributed radial loads withk = 3 − 6

The computation programMathematica was used to calculate the sums of the first 100 terms in Eqs. (21)–(23).To obtain full-field stress series solutions for uniformly-distributed loads around the circumference of thedisk, we derived three full-field stress contour plots: σr/(2p/π), σθ/(2p/π), and τrθ /(2p/π). Figures4 and 5present the three full-field stress contour plots when k = 3 and 2α equals 2◦ and 30◦, respectively. Figures6 and7 present the three full-field stress contour plots when k = 4 and 2α equals 2◦ and 14◦, respectively. Figures8and 9 present the three full-field stress contour plots when k = 5 and 2α equals 2◦ and 14◦, respectively.Figures9 and 10 display the three full-field stress contour plots when k = 6 and 2α equals 2◦ and 14◦,respectively. The thick solid lines in the figures indicate the boundaries of the disk and zero stress contours,whereas the thin dashed and bold lines present negative and positive stress, respectively.

Observation of τrθ /(2p/π) contours in Figs. 4c, 5c, 6c, 7c, 8c, 9c, 10c, and 11c reveals that all of the zerostress contours pass through the center of the disks, forming k straight lines radially symmetric to the centerwith lengths equal to diameter 2R. This shows that the stress states in the disks along these straight lines areall principal stress values; therefore, these lines are generally referred to as principal lines.

Figures4a, b, 5a, b, 6a, b, 7a, b, 8a, b, 9a, b, 10a, b, and 11a, b show that the principal lines serve as axesof symmetry for all of the σr/(2p/π) and σθ/(2p/π) stress contours; however, the distribution of shear stresspresented antisymmetric coupling, as shown in Figs. 4c, 5c, 6c, 7c, 8c, 9c, 10c, and 11c. Thus, the principallines serve as axes of antisymmetry for all of the τrθ /(2p/π) contours.

Further observation of the principal lines in Figs. 6 and 7 when k = 4, Figs. 8 and 9 when k = 5,and Figs. 10, and 11 when k = 6, as shown as the bold lines in Figs. 4c, 5c, 6c, 7c, 8c, 9c, 10c, and 11c,



Fig. 7 a σr/ (2p/π), b σθ/ (2p/π), and c τrθ / (2p/π)contours based on full-field stress series solution of uniformly-distributedloads applied around the circumference of a disk with k = 4 and 2α = 14◦

Fig. 8 a σr/(2p/π), b σθ/ (2p/π), and c τrθ / (2p/π)contours based on full-field stress series solution of uniformly-distributedloads applied around the circumference of a disk with k = 5 and 2α = 2◦

revealed perturbation near the center of the disk. The figures were plotted using the series in Eqs. (21)–(23);therefore, the perturbation is likely the result of distortion, resulting from an inadequate number of samplepoints. Increasing the number of sample points could eliminate perturbation and distortion; however, it would


Fig. 9 a σr/(2p/π), b σθ/(2p/π), and c τrθ /(2p/π) contours based on full-field stress series solution of uniformly-distributedloads applied around the circumference of a disk with k = 5 and 2α = 14◦



increase the computation time. Using just 20 sample points and a personal computer with an Intel Core i7-47903.60GHzCPU, the original graph took several seconds to be completed, whereas using 50 sample points wouldlengthen the computation time to 15–20min.


Fig. 12 a σr/(P/πRt), b σθ/(P/πRt), and c τrθ /(P/πRt) contours based on full-field stress analytical uniformly-distributedloads applied around the circumference of a disk with k = 3


3.3 Distributions of full-field stress by analytical solutions in disk under concentrated forces with k = 3 − 6

As outlined in Sect. 3.2, we usedMathematica for the calculation of dimensionless full-field stress contour plotsof σr/(P/πRt), σθ/(P/πRt), and τrθ /(P/πRt) using the full-field stress analytical solutions in Eqs. (48)–(50). The thick solid lines in the figures indicate the boundaries of the disk and the zero stress contours, whereasthe thin bold and dashed lines indicate positive and negative stress, respectively. Figures12, 13, 14 and 15present the three types of full-field stress contour plots for k = 3 to k = 6. Due to the fact that Eqs. (48)–(50)are analytical functions, Mathematica can complete these plots in just a few seconds, even when using 50sample points. With an adequate number of sample points, no perturbation appeared in the principal lines nearthe center of the disk.

The symmetry of the full-field stress contour plots obtained using the analytical solutions in Figs. 12,13, 14, and 15 is entirely consistent with that of the plots discussed in Sect. 3.2. Further comparison of theseries and analytical solutions revealed that because 2α equals 2◦, the uniformly-distributed loads under smallcontact angle are very similar to concentrated forces. For example, Fig. 4 is very similar to Fig. 12, and otherexample pairs include Figs. 6 and 13, Figs. 8 and 14, and Figs. 10 and 15. The only greater difference is thatthe σθ/(P/πRt) plots from analytical solutions displayed k zero stress curves near the rim of the disk, whichdid not appear in the σθ/(2p/π) plots from series solutions.

3.4 The principal stresses comparison between series solutions and analytical solutions

Recall Eqs. (21), (22),(23), and (54), the series solutions of principal stresses were deduced as follows:


Fig. 14 σr/(P/πRt), σθ/(P/πRt), and τrθ /(P/πRt) contours based on full-field stress analytical uniformly-distributed loadsapplied around the circumference of a disk with k = 5


σ1 =(

P

πRt

)(1

α

){−α −

∞∑n=1

1

nρkn sin knα cos knθ + k

2(ρ−2 − 1)

}, (65)

σ2 =(

P

πRt

)(1

α

){−α −

∞∑n=1

1

nρkn sin knα cos knθ − k

2(ρ−2 − 1)

}, (66)

where

=√√√√[ ∞∑

n=1

ρkn sin knα cos knθ

]2+[ ∞∑n=1

ρkn sin knα sin knθ

]2(67)

and σ1 ≥ σ2.We take k = 3 to discuss the difference of principal stresses between and Eqs. (56) and (57). For convenient

description, we declare that Eqs. (65) and (66) are series solutions of principal stresses, and Eqs. (56) and (57)are analytical solutions of principal stresses in this section.

Figure16 shows the full-field stress distribution of σ1/(P/πRt). The right-half disk with dashed lines is thestress distribution by series solution used Eq. (56). The left-half disk with solid lines is the stress distribution bythe analytical solution (65). The value of non-dimensional principal stress values is presented near the line ofcontour. Figure16a, b shows2α = 5◦ and2α = 10◦, respectively, as the contours of two solutions.Approachingthe disk center, the agreement is more accurate between the analytical and series solutions. However, whenthe non-dimensional principal stress approaches almost 0.7, the analytical and numerical solutions mismatchin the case 2α = 5◦, as shown in Fig. 16a. When the non-dimensional principal stress approaches almost to


Fig. 16 σ1/(P/πRt) contours in series solution (dashed line) and σ1|α→0 /(P/πRt) contour in analytical solution (solid line).a 2α = 5◦, k = 3, b 2α = 10◦, k = 3

0.5, the analytical and numerical solutions mismatch in the case 2α = 10◦, as shown in Fig. 16b. As the 2α issmall enough (We suggested <5◦), the analytical solutions can replace on series solutions.

Figure17 shows the full-field stress distribution of σ2/(P/πRt). The right-half disk with dashed lines is thestress distribution by series solution used Eq. (57). The left-half disk with solid lines is the stress distributionby the analytical solution (66). The value of non-dimensional principal stress values is presented near theline of contour. Figure16a, b shows 2α = 5◦ and 2α = 10◦, respectively, as the contours of two solutions.Approaching the disk center, the agreement is more accurate between the analytical and series solutions.However, when the non-dimensional principal stress approaches almost −8, the analytical and numericalsolutions mismatch in the case 2α = 5◦, as shown in Fig. 16a. When the non-dimensional principal stressapproaches almost −2.5, the analytical and numerical solutions mismatch in the case 2α = 10◦, as shownin Fig. 16b. As the 2α is small enough (we suggested <5◦), the analytical solutions can replace the seriessolutions.

Now,we discuss the full-field absolute error in percentage between analytical solutions and series solutions.Let the absolute error be


Fig. 17 σ2/(P/πRt) contours in series solution (dashed line) and σ2|α→0 /(P/πRt)contour in analytical solution (solid line).a 2α = 5◦, k = 3, b 2α = 10◦, k = 3

∣∣∣∣ σi |α→0 /(PRt) − σi/(PRt)

σi/(PRt)

∣∣∣∣× 100%, (68)

where i = 1 or 2.Figure18 shows the full-field contours of the absolute error in percentage between the analytical solution,

σ1|α→0 /(P/πRt), and the series solution, σ1/(P/πRt). Figure18a, b shows 2α = 5◦ and 2α = 10◦,respectively, as the full-field contours of absolute error. In Fig. 18a, the errors are indicated on the thick solidline (0.1%), thin solid line (0.5%), and dashed line (3%), respectively. In Fig. 18b, the errors are indicated onthe thick solid line (0.2%), thin solid line (5%), and dashed line (15%), respectively. Obviously, the principalstresses in the analytical solution are similar to those in the series solution as 2α = 5◦, because the contourdistribution of absolute error as 2α = 5◦ is less intensive and the error value is also smaller.

Figure19 shows the full-field contours of absolute error in percentage between the analytical solution,σ2|α→0 /(P/πRt), and the series solution, σ2/(P/πRt). Figure19a, b shows 2α = 5◦ and 2α = 10◦,respectively, as the full-field contours of absolute error. In Fig. 19a, the errors are indicated on the thick solidline (0.1%), thin solid line (0.2%), and dashed line (0.3%), respectively. In Fig. 19b, the errors are indicated


Fig. 18 Full-field contours of absolute error in percentage between σ1/(P/πRt) and σ1|α→0 /(P/πRt). a 2α = 5◦ and k = 3.The absolute errors indicate on thick solid line (0.1%), thin solid line (0.5%), and dashed line (3.0%), b 2α = 10◦ and k = 3.The absolute errors indicate on thick solid line (0.2%), thin solid line (5%), and dashed line (15%)

on the thick solid line (0.1%), thin solid line (0.5%), and dashed line (1%), respectively. Because the contourdistribution of absolute error as 2α = 5◦ is less intensive and the error value is also smaller, we reach the sameconclusion as with σ1/(P/πRt) as that the analytical solutions can replace on series solutions, when 2α issmall enough.

4 Photoelastic experiment

Based on Eq. (58), we can derive that the relationship between the interference fringes and parameters of thespecimens are used in the photoelastic experiment as follows:

N fσt

= σ1|σ→0 − σ2|σ→0 = P

πRt

k2ρ−2+k(1 − ρ2)

1 + ρ2k − 2ρk cos kθ, (69)


Fig. 19 Full-field contours of absolute error in percentage between σ2/(P/πRt) and σ2|α→0 /(P/πRt). a 2α = 5◦ and k = 3.The absolute errors indicate on thick solid line (0.1%), thin solid line (0.2%), and dashed line (0.3%), b 2α = 10◦ and k = 3.The absolute errors indicate on thick solid line (0.1%), thin solid line (0.5%), and dashed line (1%)

where N denotes the fringe order, fσ is the material fringe constant, and t indicates thickness of the specimen.The fringe order N can be obtained from images obtained during the photoelastic experiment, by eliminating

the effects of specimen thickness using Eq. (69):

N fσ = P

πR

k2ρ−2+k(1 − ρ2)

1 + ρ2k − 2ρk cos kθ. (70)

Equation (70) is a simplified equations that can be applied in experiments to evaluate the difference of principalstress. Obtaining any two of the three unknowns P , N , or fσ makes it possible to obtain the third in practicalapplication. The fringe order N can be derived using the differences between the dark fields and light fieldsobtained using a conventional polariscope. UsingMATLAB,we obtained

∣∣dark field–light field∣∣ images, whichrefers to images of the absolute differences between the light field image and the dark field image. Thus, basedon the theory of photoelasticity, the dark fringes in the dark field, light field, and |dark field–light field| images,respectively, represent the information of three types of fringe orders:


Fig. 20 a Dark field isochromatic pattern, b theoretical analysis of patterns for N = 0 and N = 1 corresponding to dark field,and c superposition with experimental pattern (a) and dark–light inverse pattern (b)

Dark field: N = 0, 1, 2,3,…,Light field: N = 0, 0.5, 1.5, 2.5,…,|Dark field–light field|: N = 0, 0.25, 0.75, 1.75,…

The experimental conditions in this study were k = 3 and P = 60N, which involved three centripetalforces of 60N applied at the edge of the disk specimen with angles of 120◦ separating them. Based on Eq. (70),the fringe order equation for k = 3 is

N =(

9P

fσ πR

)ρ(1 + ρ)

(−1 + ρ)(1 + ρ + ρ2)2. (71)

Sodium light (wavelength=589nm) served as the light source of the polariscope in the experiment. The CCDresolution was 1600 × 1200 pixels, and the disk specimen (radius of 40mm) comprised PSM-1 photoelastic


Fig. 21 a Light field isochromatic pattern, b theoretical analysis of patterns for N = 0.5 and N = 1.5 corresponding to lightfield, and c superposition with experimental pattern (a) and light–dark inverse pattern (b)

material manufactured by VISHAY. According to the manufacturer, the material fringe constant of PSM-1under green light (wavelength=546nm) is fσ = 7.01KPa m/fringe, which can be converted to sodium lightas 7.56KPa m/fringe.

Figures20a, 21a, and 22a present dark field, light field, and |dark field–light field| images obtained during thephotoelastic experiment. Figures20b, 21b, and 22b present the respective fringe distributions, as calculatedusing Eq. (71) in Mathematica. To compare the experimental and theoretical results, we used MATLAB tocombine Fig. 20a and 20b into Figs. 20c, 21a, b into Figs. 21c, and 22a and 22b into Fig. 22c. Thus, theresults showed the influence of stress concentration under boundary load known as Saint-Venant’s principle.Discrepancies between the experiment and theoretical results became increasing obvious closer to the negativeload; however, they still presented good consistency.


Fig. 22 a Isochromatic pattern of absolute values of the subtraction from dark field to light field, b theoretical analysis ofpatterns for N = 0.25, N = 0.75 and N = 1.25 corresponding to absolute value of the subtraction from dark field to light fieldisochromatic pattern, and c superposition with experimental pattern (a) and light–dark inverse pattern (b)

5 Conclusion

This study derived the full-field stress series solution for multi-loads with small contact area uniformly distrib-uted within even space around the circumference of a disk and then formulated a simple and useful full-fieldstress analytical solution for concentrated forces applied within even space around the circumference of a disk.When 2α, the contact angle between the uniformly-distributed loads, is<5◦, the full-field stress series solutioncan be replaced with the full-field stress analytical solution to simplify computation. Full-field stress contourplots drawn using numerical calculations for the series solution and analytical solution revealed symmetry inthe stress contours. Furthermore, all of the stress contours pass along the principal lines through the center ofthe disk.


Finally, the results obtained from the analytical solution of the full-field stress distributions, which can bederived from force P , fringe order N , and material fringe constant fσ , are consistent with stress distributionresults obtained in photoelastic experiments.We used numerical calculations to obtain and plot the fringe orderN of the stress distributions with k = 3 to enable a comparison with the results obtained in the photoelasticexperiment, which showed high consistency between the two methods. This study provides a full-field stressanalytical solution for multi-forces applied within around the circumference of a disk. These findings can alsobe used to derive the internal stress distributions of other disk structures under load.

Acknowledgements Authors gratefully acknowledge the financial support of this research by the Ministry of Science andTechnology (Republic of China) under Grant for Outstanding Young Scholar 104-2628-E-011-002-MY3.

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