post-tension flat slab design example

17
Design of Post-Tension Slab Hand Calculation for Typical Flat Slab Design Based on: BS 8110 1997 TR 43 – 1st Edition Prepared by: Nov. 2008 Engineers Training Center www.jea.org.jo [email protected]

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Post-tension Flat Slab Design Example

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Page 1: Post-tension Flat Slab Design Example

Design of Post-Tension Slab

Hand Calculation for Typical Flat Slab Design

Based on: BS 8110 1997

TR 43 – 1st Edition

Prepared by:

Nov. 2008

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Page 2: Post-tension Flat Slab Design Example

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Two Way Post-tension Flat Slabs: A post-tensioned prestressed banded flat slab floor system for an office complex is shown in figure.1 together with atypical sub-frame section. The structure is checked both at serviceability and ultimate limit states. These checks are carried out at serviceability and ultimate limit states. These checks are carried out at transfer and under working loads conditions.

1. Input design data: 1.1 Material: Figure 1: structural plan

- Concrete:

Unit weight = 24 KN/m3

fck = 40 MPa (cylinder strength at 28 day) fcu = 50 MPa (cube strength at 28 day) fcui = 25 MPa (strength at transfer) Ec = 30 GPa (elastic modulus at 28 day) Eci =25 GPa (elastic modulus at transfer)

- Bonded reinforcement:

fy = 460 Mpa

- Prestressing steel:

12.9 mm diameter super strand placed in metal ducts. Pk = 186 KN (characteristic strength of strand) Aps = 100 mm2 (area of strand) fpu = Pk/Aps = 1860 MPa (characteristic strength of prestress steel) Eps= 195 GPa (elastic modulus)

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Page 3: Post-tension Flat Slab Design Example

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1.2 Loading:

- Imposed loading: (according to architectural drawing)

Partition = 2.0 KN/m2

Finishing = 1.5 KN/m2 Services and false ceiling = 0.5 KN/m2 Total S.I.D.L = 4.0 KN/m2

- Live load:

Typical offices building = 3.0 KN/m2

- Total imposed loading = 4.0 + 3.0 = 7.0 KN/m2

1.3 Serviceability classifications:

- BS 8110 code, TR 43 - 1st Eedition. - Class 3.

2. Design step: 2.1 Preliminary designs:

For normal slab load, and initial span, we can take span/depth = 42

mm24042000,10

Punching shear affected by size of column and slab thickness. However to reduce shear reinforcement requirement a depth of 250 mm is chosen.

- Self weight = 6.0 KN/m2

- Total dead load = 4.0 + 6.0 =10.0 KN/m2

- Total live load = 3.0 KN/m2 - Balanced load:

In these example a balanced load consisting 60% of all dead load is chosen 0.6 X 10 = 6 KN/m2.

2.2 Tendon profile:

Nominal cover requirement in accordance with BS 8110-1, sec.3.4:

For 2 hour fire resistance take cover = 25 mm.

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Page 4: Post-tension Flat Slab Design Example

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Take nominal cover to be 25 mm refer to figure.2.

(a) Transverse direction (b) Longitudinal direction

Figure 2:Tendon and reinforcing steel postion

Note: The positioning of reinforcement must be considered at this stage, so as to obtain

practical arrangement of the steel at support. Based on tendon eccentricities show in figure.3 and the position of inflection point (0.1 times the span from the center of the supports) the tendon profile can be calculated.

Figure 3:Transfer tendon profile

At these stage losses are assumed as follows:

- At transfer 10% of the jacking load. - At service 20% of the jacking load.

A through check will be carried out after the stress calculates to check that these initial assumptions are reasonable.

2.3 Initial Prestressing force:

The initial prestress force i.e jacking force, has been taken to be 75% of the characteristic strand strength (BS8110-1:1197, sec 4.7.1).

Calculation of Pavg: - Jacking force (Pj) = 0.75 X 186 = 139.5 KN/strand. - Prestress force at transfer (Po) = 125.55 KN/strand. (10% losses) - Prestress force at service (Pe) =111.6 KN/strand. (20% losses)

Next the value of prestress force required in each span is calculate, this is done using the chosen balanced load of 6.0 KN/m2 (60% of dead load) the distance between points of inflection, s, and the drape, a, as shown in figure.4, (refer to appendix A for profile calculation)

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Page 5: Post-tension Flat Slab Design Example

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Figure 4 : Draps for load balanced

The prestress force is obtained from the following equation which assumes a parabolic profile.

8a ws P

2

rqd =

For span A-B and C-D:

- Prqd = (6.0 X 7.0 X 64002) / (8.0 X 67.5 X 1000) = 3185 KN - Therefore number of the tendons = Prqd / effective force = 3185 / 111.6 = 28.54 - Try 30 tendons per panel “10 X 3S” or “6 X 5S” (more economical).

For span B-C:

- Prqd = (6.0 X 7.0 X 80002) / (8.0 X 120 X 1000) = 2800 KN - Therefore number of the tendons = 2800 / 111.6 = 25 - Try 30 tendons per panel as before “6 X 5S”.

The effect of the tendon in the slab is modeled by mean of equivalent as shown below. It should be noted that the portions of cable from the edges of the slab to grid lines A and D are horizontal and so do not contribute to the equivalent load. The equivalent load w, between any two points of inflection for the chosen number of tendons is given by:

2avPn a 8 W

s=

Where:

n: number of strand. a: drab at the point considered (up +, down -). s: distance between point of inflection.

Figure 5 : Calculation of equivalent load due to tendon forces

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Page 6: Post-tension Flat Slab Design Example

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At transfer: Pav = 125.55 KN n = 30

Table 1 : Calculations of equivalent loads due to transverse tendons A span B B span C C span D

n. Pav (KN) 3766.5 3766.5 3766.5 3766.5 3766.5 3766.5 3766.5 3766.5 3766.5a (mm) 18.6 -67.5 26.3 30 -120 30 26.3 -67.5 18.6 s (mm) 1600 6400 1600 2000 8000 2000 1600 6400 1600

w (KN/m) 218.9 -49.7 309.6 226 -56.5 226 309.6 -49.7 218.9

At service: Pav = 111.6 KN n = 30

Table 2 : Calculations of equivalent loads due to transverse tendons

A span B B span C C span D n. Pav (KN) 3348 3348 3348 3348 3348 3348 3348 3348 3348

a (mm) 18.6 -67.5 26.3 30 -120 30 26.3 -67.5 18.6 s (mm) 1600 6400 1600 2000 8000 2000 1600 6400 1600

w (KN/m) 194.6 -44.1 275.2 200.9 -50.2 200.9 275.2 -44.1 194.6 2.4 Stresses Calculations:

t

S

t

A

tct Z

M Z

M ZP.e

AP f ++−=

b

S

b

A

bcb Z

M - ZM-

ZP.e

AP f +=

Where:

MA = moment due live and dead load. MS = moment due secondary effect. Zt = top section modulus. Zb = bottom section modulus. Ac = 7.0 X 0.25 X 106 =1.75 X 106 mm2

As the section being considered is rectangular and symmetrical abut about the centroid, Zt

and Zb are equal. Zt =Zb = Z = bh2/6 = 7.29 X 107mm3

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Page 7: Post-tension Flat Slab Design Example

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2.4.1 Allowable stresses:

Maximum allowable concrete compressive and tensile stresses for floor with banded tendon are given in BS 8110, part 1, section 4.3.4.2 and 4.3.4.3. - Allowable Stresses at transfer:

Compression = MPafci 5.12255.05.0 =×=

Tension = MPafci 8.12536.036.0 =×=

- Allowable Stresses at service:

Compression = MPafcu 20504.04.0 =×= Tension: tensile stresses calculated to limit the crack width see table 3.

Table 3 : Tensile stresses for class 3 Design stresses for concrete grade

Limiting crack width (mm) 30 N/mm2 40 N/mm2 50 and over N/mm2

0.1 3.2 4.1 4.8

Grouted post-tension tendons

0.2 3.8 5.0 5.8 2.4.2 Applied loads and moment:

Own weight only:

wo,wt = 6 X 7 = 42.0 KN/m

Service load:

wservice = (6 + 4) X 7+ 3 X 7 = 70 KN/m + 80 KN/m

Figure 6 : moment digram for own weight

Figure 7 : moment diagram for service load

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Page 8: Post-tension Flat Slab Design Example

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2.4.3 Stresses at transfer: Po = 30 X 125.55 = 3766.5 KN

Table 4 : Stresses at transver for the transverse direction

ZONE

FIBER

e(mm)

STRESS DUE PRESTRESS

(MPa)

STRESS DUE TO

SELF WEIGHT

(MPa)

TOTAL STRESS

(MPa)

ALLAWABLE

STRESS (MPa)

STATE

A top 0 -2.152 3.00 0.85 1.8 OK bottom 0 -2.152 -3.00 -5.152 -12.5 OK

AB(sgging) top 80 1.98 -1.54 0.44 -12.5 OK bottom 80 -6.29 1.54 -4.75 1.8 OK

B top -70 -5.77 3.13 -2.64 1.8 OK bottom -70 1.47 -3.13 -1.66 -12.5 OK

B top -70 -5.77 4.79 -0.98 1.8 OK bottom -70 1.47 -4.79 -3.32 -12.5 OK

BC(sgging) top 80 1.98 -2.41 -0.43 -12.5 OK bottom 80 -6.29 2.41 -3.88 1.8 OK

C top -70 -5.77 2.41 -3.66 1.8 OK bottom -70 1.47 -2.41 -0.94 -12.5 OK

C top -70 -5.77 3.13 -2.64 1.8 OK bottom -70 1.47 -3.13 -1.66 -12.5 OK

CD(sgging) top 80 1.98 -1.54 0.44 -12.5 OK bottom 80 -6.29 1.54 -4.75 1.8 OK

D top 0 -2.152 3.00 0.85 1.8 OK bottom 0 -2.152 -3.00 -5.15 -12.5 OK

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Page 9: Post-tension Flat Slab Design Example

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2.4.4 Stresses at service: Po = 30 X 111.6 = 3348 KN

Table 5 : Stresses after all stresses for the transverse direction

ZONE

FIBER

e(mm)

STRESS DUE PRESTRESS

(MPa)

STRESS DUE TO

SELF WEIGHT

(MPa)

TOTAL STRESS

(MPa)

ALLAWABLE

STRESS (MPa)

STATE

A top 0 -1.913 6.55 4.64 5.8 OK bottom 0 -1.913 -6.55 -8.46 -20.0 OK

AB(sgging) top 80 1.76 -3.35 -1.59 -20.0 OK bottom 80 -5.59 3.35 -2.24 4.8 OK

B top -70 -5.12 6.76 1.64 5.8 OK bottom -70 1.30 -6.76 -5.46 -20.0 OK

B top -70 -5.12 10.38 5.8 OK bottom -70 1.30 -10.38 -9.08 -20.0 OK

BC(sgging) top 80 1.76 -5.23 -3.47 -20.0 OK bottom 80 -5.59 5.23 -0.36 4.8 OK

C top -70 -5.12 10.38 5.26 5.8 OK bottom -70 1.30 -10.38 -9.08 -20.0 OK

C top -70 -5.12 6.76 1.64 5.8 OK bottom -70 1.3 -6.76 -5.46 -20.0 OK

CD(sgging) top 80 1.76 -3.35 -1.59 -20.0 OK bottom 80 -5.59 3.35 -2.24 4.8 OK

D top 0 -1.913 6.55 4.64 5.8 OK bottom 0 -1.913 -6.55 -8.46 -20.0 OK

2.5 Ultimate limit state: 2.5.1 Determination of hyperstatic action:

Hyperstatic moments can be calculated either directly or indirectly (Alami, 1998b) for skeletal members, such as beams and floor systems that are modeled as strips of isolated slab frames , hyperstatic actions can be successfully calculated using both methods.

In this example we will use the indirect method (covenantal method) in calculation of hyperstatic action. Indirect method is based on the following relationship:

Mhyp = Mbal – P.e

Where: e = eccentricity of post-tensioning with respect to the neutral axis of the section (positive if

CGS is above the neutral axes otherwise negative). Mhyp =hyperstatic moment. Mbal = balanced moment due to balanced load.

P = post-tensioning force “positive”

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Page 10: Post-tension Flat Slab Design Example

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Balanced loads:

As described before:

s

Pn a 8 W 2av

bal =

- At support A: -

Mbal= 232 KN.m , Pe = 3348 KN , e = 0.0 mm Mhyp= (232) + (3348 X 0 / 1000) = 232 KN.m

- At center of span AB:

Mbal= -117.7 KN.m , Pe = 3348 KN , e = 45-125 = -80 mm Mhyp= (-117.7) + (3348 X 80 / 1000) = 150.14 KN.m

- At left of support B:

Mbal= 238.1 KN.m , Pe = 3348 KN , e = -70 mm Mhyp= (238.1) - (3348 X 70 / 1000) = 3.74 KN.m

- At right of support B:

Mhyp= (417.8) - (3348 X 70 / 1000) = 183.44 KN.m - At center of span B-C:

Mhyp= (-209.7) + (3348 X 80 / 1000) = 58.14 KN.m - At left of span C:

Mhyp= (417.8) - (3348 X 70 / 1000) = 183.44 KN.m - At right of support C:

Mhyp= 3.74 KN.m

- At center of support C-D:

Mhyp= 150.14 KN.m

- At support D:

Mhyp = 232 KN.m

Figure 8 : moment diagram for balance load

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Page 11: Post-tension Flat Slab Design Example

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2.5.2 Total ultimate moment:

The load combination for ultimate strength design:

MU = 1.4 MD.L + 1.6 ML.L + 1.0 Mhyp

Figure 9 : Moment diagram for hiperstatic

Figure 10 : Moment diagram for dead load

Figure 11 : Moment diagram for live load

Figure 12 : Moment diagram for ultimate load pear panel

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Page 12: Post-tension Flat Slab Design Example

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2.5.3 Design section: Section analyses may be carried out in accordance with clause 3.7 of BS 8110, part 1, the moment capacity of the section can calculated using following equation. Mu = fPb Aps (d- dn) Where: Mu = design moment of resistance of the section. fPb = design tensile strength in the tendon. Aps = area of prestressing tendon in the tension zone. d = effective depth to the centroid of the steel area. dn = depth of centroid of the compression zone = 0.45x. x = depth of neutral axis. b = effective width of the section. - When d = 125 mm.

127.0125700050100301860

=××××

=bdfAf

cu

pspu

6.018601116

==pk

pe

ff

From table 4.4 BS 8110. Sec 4.3.7.3 finds value of fPb and x. fPb = 1731.7 MPa x = 36.25 mm dn = 0.45 x 36.25 = 16.3 mm Mu = -1731.7 x 30 x 100 x (125-16.3) = -564 KN.m

When d = 205 mm.

078.0205700050100301860

=××××

=bdfAf

cu

pspu

6.018601116

==pk

pe

ff

fPb = 1767 MPa x = 35.9 mm dn = 0.45 x 35.9 = 16.14 mm Mu = 1767 x 30 x 100 x (205-16.14) = 1001 KN.m - When d = 195 mm.

08.0195700050100301860

=××××

=bdfAf

cu

pspu

6.018601116

==pk

pe

ff

fPb = -1767 MPa x = 35.1 mm dn = 0.45 x 35.1 = 15.8 mm Mu = -1767 x 30 x 100 x (195-15.8) = -950 KN.m

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Page 13: Post-tension Flat Slab Design Example

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Table 6 : Comparison between applied moment and moment capacity at ultimate limit state ZONE M (KN.m)/panel

(applied) Mu (KN.m)/panel

(capacity) A -411 -564

AB 503.42 1001 B -708.62 -950 B -911.0 -950

BC 609.8 1001 C -911.0 -950 C -708.62 -950

CD 503.42 1001 D -411 -564

Because the moment capacity at ultimate limit state > applied moment, no un-tensioned reinforcement is required.

2.5.4 Minimum un-tensioned reinforcement requirement:

Reinforcement required = 0.075% Ac Ac = hb = 250 X 7000 = 1.75 X 106 mm2 As = 0.00075 X 1.75 X 106 = 1312.5 mm2

Use 7 T 16 = 1407 mm2

The reinforcement should extend into the span by 0.2 × span measured from the centerline of the column and the width of strip is the column breadth plus 3 times the slab depth as shown in figure 13.

Figure 13 : Un-tensioned reinforcement details

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Page 14: Post-tension Flat Slab Design Example

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2.6 Post-tension losses: 2.6.1 Short-term losses:

2.6.1.1 Friction losses:

)x(-

ox e P P ϖαμ +×=

Where:

Px = Force at distance x from stressed end. Po = 139.5 KN (Stressing force (at anchor)). μ = 0.2 (friction coefficient). α = angle change in tendon from anchor to point considered (radians). ω = 0.0085 (“wobble” factor (radians/m))

Total drape from figure.3 (refer to appendix A for profile calculation)

- Total drab for span A-B and span C-D: (18.6+26.3)/2 + 67.5 = 89.95 mm - Total drab for span B-C: (30+30)/2 + 120 = 150 mm

×16

= 2Ldrapetotalα

mrad /31 0.023 = 8

)10 × 89.95 × (16= = 2

−3

αα

mrad /2 0.024 = 10

)10 × 150 × (16= 2

−3

α

PA = 139.5 KN PB = (139.5) e-(0.2)(8)(0.023+0.0085) = 132.64 KN PC = (132.64) e-(0.2)(10)(0.024+0.0085) = 124.3 KN PD = (124.3) e-(0.2)(8)(0.023+0.0085) = 118.2 KN

KN 128.66 4

118.2) 124.3 132.64 (139.5 Pavg =+++

=

Po - Pavg = 139.5 – 128.66 = 10.84 KN

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Page 15: Post-tension Flat Slab Design Example

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2.6.1.2 Wedge set losses:

Because Length of the tendon 26 m < 30 m so use one live end.

δPw = 2 L’ P’

Where:

δPw = Force losses due draw-in.

Α Δ

= E L ''

PPSPS (Length of the tendon affected by draw-in).

)L L (L

)P -(P P321

DA'

++= (Slop of the force profile).

Δ = 6 mm (Wedge draw-in). Eps = Modulus of elasticity of the tendon. Aps = Area of the tendon.

KN 0.82 8) 10 (8

118.2)-(139.5 P' =++

=

m 11.95 0.82

100) X 195 X 10 X (6 L-3

' == < Length of the tendon

So δPw = 2 P’ L’ = 2 X 0.82 X 11.95 = 19.6 KN

2.6.1.3 Elastic shortening losses:

δPes = εes Eps Aps Where:

δPes = Force losses due Elastic shortening. εes = 0.5 fco/ Eci fco = n Po /bh (stress in concrete adjacent to the tendon after transfer). Eci = Modulus of elasticity of concrete at time of transfer. fco = (5 X 139.5 X 103)/ (1500 X 250) = 1.86 MPa εes = 0.5 X1.86/ 2500 = 3.72 X 10-5

δPes = 3.72 X 10-5 X 195000 X 100 / 1000 = 0.725 KN

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Page 16: Post-tension Flat Slab Design Example

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2.6.2 Long-term losses: 2.6.2.1 Creep of concrete losses:

δPcr = εcr Eps Aps Where:

δPcr = Force losses due creep. εcr = φ fco/ Eci φ = 2.9 (creep coefficient (BS 8110, part 2, figure.1)) εcr = 2.9 X 1.86 / 25000 = 2.16 X 10-4 δPcr = (2.16 X 10-4 X 195000 X 100) / (1000) = 4.21 KN

2.6.2.2 Shrinkage of concrete losses:

δPsh = εsh Eps Aps

εsh = 300 X 10-6

δPcr = (300 X 10-6X 195000 X 100) / (1000) = 5.85 KN

2.6.2.3 Relaxation of the tendon losses:

δPr = 1000 hour relaxation value x relaxation factor x the pre-stressed force at transfer. Pi =139.5 -10.84 – 19.6 – 0.725 = 108.3 KN δPr = 0.035 X 1.5 X 108.3 = 5.69 KN

2.6.3 Summary for P.T losses:

- Short-term losses:

1- Friction losses = 10.84 KN ………………….…… 7.7% 2- Wedge set losses = 4.48 KN ……………………... 3.2% 3- Elastic shortening losses = 0.725 KN …………..... 0.5%

Total short-term losses = 16.1 KN ………...………….... 11.5% - long-term losses:

1- Creep of concrete losses = 4.21 KN ………….……. 3.0% 2- Shrinkage of concrete losses = 5.85 KN ……….….. 4.2% 3- Relaxation of the tendon losses = 5.69 KN …….….. 4.0%

Total long-term losses = 15.75 KN ………………….….. 11.2% Total losses = 31.85 KN ………………..……………….... 22.8%

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Page 17: Post-tension Flat Slab Design Example

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► APPENDIX A: Calculation of Tendon Geometry:

jX + mX+n = 0 m = (p2-2L)(q1 – q2 ) + p1 (q3- q2) = (800 - 2 X 8000) (125-45) + 800 (195 - 45) = -1096000 n = (q1-q2)(L-P2)L = (125-45)(8000-800)8000 =4608X106

L’ =(-m ± √ [ m2 – 4 j n ] )/ 2 j = (1096000 ± 1577973.384 ) /-140 = -19094 , 3442.67 L’ = 3442.67 mm a1 = [(q1 – q2)P1]/ L’ = [80 X 800] / 3442.67 = 18.6 mm a2 = [(q3 – q2)P2]/ [L-L’]= [150 X 800] / 4557.33 = 26.33 mm - for span B-C: q1 = q3 = 195 mm, P1=1000 , L’= 5000 a4 = [195 - 45] X 1000 / 5000 = 30 mm

Figure 14 : force profile for full-length longitudinal tendons

Figure 15 : Tendon geometry

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