IISEE Lecture Note 2009-2010

Structural Testing

By

T. MUKAI

Feb. 19, 2010

International Institute of

Seismology and Earthquake Engineering,

Building Research Institute

STRUCTURAL TESTING

by

Tomohisa MUKAI ( SRI)

February 19, 2010

IISEE Lecture Note 2009-2010

International Institute of Seismoloav and Earthouake Enaineerina

Objective

Structural Test of RC Beams ( 2009-2010 IISEE Earthquake Engineering Course)

by Tomohisa MUKAI ( BRI )

Objectives of the structural test are to experience the test technique and to observe the failure mechanism of RC members. The cracking strengths, flexural strengths, shear strengths, bond splitting strengths and load - deflection relationships of the specimens should be calculated in advance to predict the damaging process, capacities and failure pattern to compared with the test results. It is a clarification of the prediction equations written in the codes.

Test Specimens The test specimens are RC beams with rectangular cross section of 20x30 cm.

Fig. 1 shows the dimensions of the specimens. The specimen-A is designed to fail by shear before yielding of longitudinal steel bar. The specimen-B is designed not to fail by shear. However, the flexural strength of specimen-B caused by yielding of longitudinal steel bar is almost the same as bond splitting strength. The predicted failure pattern of specimen-A is brittle shear failure and that of specimen-B is bond splitting after yielding of longitudinal steel bar.

Variable of this test is shear reinforcement ratio in the test area. The out of the test area of both specimens are reinforced with 010-@65. 016 steel bars are arranged as longitudinal reinforcement.

ISpecimen-lAJ: 06 steel bar are arranged in dir~ction perpendicular to longitudinal reinforcement in the test area as shear reinforcement. Spacing of the shear reinforcements is 195 mm. Shear reinforcement ratio (p w) in the test area is 0.164 %.

ISpecimen-ij : 010 steel bar are arranged in direction perpendicular to longitudinal reinforcement in the test area as shear reinforcement. Spacing of the shear reinforcements is 65 mm. Shear reinforcement ratio (p w) in the test area is 1.10 %.

o Dimensions and strength of materials used in the design of specimens 016 : Cross section area (As) = 1.99cm2, Circumference (ljJ) = 5cm 06 : Cross section area (As) = 0.32cm2 010 : Cross section area (As) = 0.71cm2 Expected yield strength of steel bars (cry) is 3500 kgf/cm 2 . ... C,,,,,,,,,,,,f.,,,.,1 _ ....... """"_I1" ....... ,..~i"o ~ .. I1"~ ..... ,.,. ... h I""\.~ ,.'u'" ..... _ ... o .. o I". \ it!' -1 on 1,,....fJ_1"'V'\2 - L...At-J\J""'LvU \JVIIIt-'lv~~IYv ~U"""II~UI VI vVllvl\;o\V" \V s ' I~ IVV "~IIVIII

Loading The 50 tonf capacity loading machine for flexural test will be used. Anti-

symmetrical moment distribution will be represented by Ohno-type loading set up.

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Measurement Load, displacement, strains of steel bar and crack pattern will be measured. Channel information is as follows,

CH.1 : Load (tf) CH.2 & 3 : Displacement (mm) CH.4 to 13 : Strain of longitudinal steel bar (micro=1 0. 6) CH.14 to 18 : Strain of shear reinforcement (micro=106)

Material test Yield strengths, tensile strengths, Young's moduli and stress - strain

relationships of steel bars and compressive strength, Young's modulus and stress -strain relationship of concrete will be measured by material test, which will be conducted before the structural test.

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~

o o M,_

o o

ML

Specimen-A D1 0 @6 5 06 @1 95 01 0 @6 5

/ I / I

I i- ~ ~

I f

~ I--V 1/ i , ,

I .

I r-- r--1 r- I-- r- !- I--\

6. :::..

J t r ai n g a u q e Specimen-B

010 @6 5 01 0 @6 5 D10 @6 5 / 7 I / ~

I

~ - "-- ' , I - r - - tf- -i I ~L I , I I ~ - - ir- - - --- f- ----' ,

'" "

11 2 0 I 5 2 0 7 8 0 5 2 0 \120 I . 2 060 I

~ Test area J :- ;

0,-V'-

O !

C\J 1 C\J I O~

v

4 0 4 0 n 11 ~ I 01 6

06

~

4 0 4 0 n n

Or

~l ll~)1 0 1_

D16

01 a v

! 2 0 0 I

390 910 r

Load ( P) CII. I

Q Displacement CII.2&3

520 7 8 0 \S 2 0

Test area ~

Fig. 2 Loading method and measuring method

5 6 7 8 CH . 4 \ \7 ~~~-------------------------------------

f-- t--- f-- t--- ,

14 15 16 17 18

f--- r-- r- r--~

9 1 0 1 1 1 2 ~\

L 780 -J ~f--_.=-_- _- _ --T e-s-t -a r-e-a--~

Fig. 3 Location of strain gauges

4

1 3

Report (Requirement)

Submit a report on structural testing of RC beams. Items of the following contents should be included in the report. This report is an examination for your assessment. Submit a report by March 19th to T. MUKAI, who is the lecturer of RC Structure, or IISEE staff by papers.

Tentative Contents of Report

1. Objective

2. Test Specimens ~Fig. 1

3. Loading Setup and Measuring Methods ~Fig. 2

4. Material Test 4.1 Concrete 4.2 Steel bars

~ Strength, Young's modulus, a-f. concrete -> ~'N)00 1 ~ Strengths, Young's modulus, a-f. 06,010,016 bar ---3> f,'-1 1"""0 I(dSwov

1'4J..J J

5. Test Results 5.1 Failure pattern and Damage state 5.2 Shear force - deflection relationship 5.3 Strain distribution of longitudinal steel bar 5.4 Strain distribution of shear reinforcement bars

~Photos & drawings -> ~I",.J ) ~V-o . ~f.- location relationship ~f.- location relationship

6. Comparison between Test Results and Prediction -.j) C~O(.V\ (Ptv \5)

7. Personal Impression Participants should describe your personal impression individually.

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Calculation of Strengths

1. Dimensions Dimensions and strengths of materials used in the calculation are as follows.

b = beam width ( = 20 cm) o = beam depth ( = 30 cm) d = distance from extreme compression fiber to centroid of longitudinal

reinforcement ( = 26 cm) = distance from the centroid of the compressive resultant to the centroid

of the tensile resultant ( = (7/8) d = 22.25 cm) j l = distance between top and bottom bars in a beam ( = 22 cm) L = clear span of a beam ( = 78 cm) LID = length-depth ratio ( = 2.6 ) M = moment ( = 15.6 P = 15.6x2.5V = 39 V (kgf . cm)) P == Load (kgf) V = shear force (kgf) MNd = shear span ratio ( = 1.5 ) cr B = concrete compressive strength ( = 180 kgf/cm 2)

Longitudinal reinforcement (016) : Ag = nominal cross section area ( = 1.99 cm 2) d b = diameter of longitudinal reinforcement ( = 1.6 cm) lj.I = nominal circumference ( = 5 cm) cr y = nominal yield strength ( = 3500 kgf/cm 2) A l = gross cross section area of tensile longitudinal steel ( = 5.97 cm 2) P l = ratio of tension reinforcement ( = (A g t) lb d) = 0.015 = 1.15 % )

Shear reinforcement (06) : A ~ = nominal cross section area ( = 0.320 cm 2) cr w y = nominal yield strength ( = 3500 kgf/cm 2) s = spacing of shear reinforcement ( = 19.5 cm) p w = shear reinforcement ratio (= 2 A vI (b' s) = 0.00164 = 0.164 %)

( P w 0' Wy= 5.74 kgf/cm 2)

Shear reinforcement (010) : A v = nominal cross section area ( = 0.71 cm 2) cr w y = nominal yield strength ( = 3500 kgf/cm 2) s = spacing of shear reinforcement ( = 6.5 cm) p w = shear reinforcement ratio ( = 2 A v I (b . s) = 0.0110 = 1.10 % )

( P w cr wy= 38.5 kgf/cm 2)

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2. Calculation of Moment - Rotation relationship of RC beam specimens 2.1) Initial stiffness

I = b031 12 = 20x303/12 = 45000 cm4

Ie = b031 12 = 1.44x20x3031 12 = 64800 cm4

E c = 2.1x105 x(2.3/2.3) 1.5X f (1801200) = 1.99x105 kgf/cm 2 S = 6 E c II L = 6x1.99x10 5 x45000 178 = 6.89x10 8 kgf-cm S' = 6 E c Ie I L = 6x1 .99x10 5 x64800 178 = .9,..9.2 x 1 0 8 kgf-cm

2.2) Flexural cracking moment (steel are ignored in this case for simplicity) Z c = 45000 I 15 = 3000 cm 3 M c = 1.8 f a B . ' Z c = 1.8 f 180x3000 = 72449 kgf-cm = 72.4 tf'cm V c = M c I 39 = 1858 kgf = 86 P c = V c x2.5 = 4645 kgf = 4.65 tf

Following calculation is a case when the steel is taken into account.

3. Calculation of shear strength and bond splitting strength

3.1) Shear Cracking Strength (Arakawa's equation) (V. c)

V s c = b . j . k c (500 +0 B) . M 1 (V ~.~~5+ 1.7 = 20 . 22.75 . 0.77 . (500+180) . 0.0851 (1.5+1.7) = 6328 kgf = 6.33 tf where, k c: coefficient depend on d (see Fig.4 in the lecture note for RC structure)

Pc1 =6.33*2.5=15.8tf

3.2) Shear strength (Arakawa's equation) (V u)

Vu =Vc +Vs where, V c: shear strength provided by concrete

V s: shear strength provided by shear reinforcement

V c = b . j {k u . k p (180 +0 B) . M 1 (V .0 d~ 2+ O. 12 = 20x22.75 [0.833 . 0.847 . (180+180) . 0.12 1 (1.5+0.12) ] = 8560 kgf = 8.56 tf where, k u : coefficient depend on d

k p : coefficient depend on P t (= 0:82 p l 0.23= 0.847) (see Fig.4 and Fig.5 in the lecture note for RC structure by Mukai)

V s = b . j {2. 7 ,,(p \V 0 W Y)} . = 20x22.75 x2.7 ,,(p wOw Y) = {2943 kgf (Specimen-A)

7623 kgf (Specimen-B) ]

Vu =Vc +Vs = { 8560 + 2943 = 11.50 tf (Specimen-A) 8560 + 7623 = 16.28 tf (Specimen-B)

Pu 1 =11.50*2.5=28. 75tf (Specimen-A) Pu 1 =16.18*2.5=40.45tf (Specimen-B)

(Specimen-A) Vu (11.5tf) < V y (12.5tf)

~ Specimen-A will fail by shear before yielding of member.

(Specimen-B) V u ( 16.3 tf) > V y ( 12.5 tf )

~ Specimen-B will not fail by shear before yielding of member.

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3.3) Shear strength (AIJ Guidelines * 1, pp.77-89) (V u) *1: AIJ Design Guidelines for Earthquake Resistant Reinforced Concrete Buildings Based on Ultimate Strength Concept (1990), Chapter 6 Design for shear and bond

v u = b . j t . p w a wy cot

The shear force of specimen-B at yielding of member is 12.5 tf. Then shear failure will not occur in specimen-B after yielding.

(Specimen-A) Vu (9.6tf) < V y (12.5tf)

~ Specimen-A will fail by shear before yielding of member.

(Specimen-B) V u ( 23.1 tf) > V y ( 12.5 tf ) V u (Rp ~ O. 0 25 ) (16. ~ tf) > V y( 12.5tf)

~ Specimen-B will not fail by shear before and after yielding of member.

3.4) Shear strength (ACI Building Code * 2) (V u) *2: Building Code Requirements for Reinforced Concrete (ACI 318-2002), Chapter 11 Shear and Torsion

Notation of the following equations are that of ACI code. Following are conversion factors.

1 psi = 0.07031 kgf/cm 2 1 in = 2.54 cm 1 Ib = 0.453592 kgf

Vn =V c +V 5 (ACI318, Eq.11-2)

V c = ( 1.9..r fc+ 2500p w' (V u . d)1 M u)b w' d (AC1318, Eq.11-5) V 5 = A v ' f y' dIs (AC1318, Eq.11-15)

f c = 180 I 0.07031 = 2560 psi ..r f c = 50.6 psi P w (~ l lJi~A1i lt) = A 51 b w ' d = 3 . 1.991 (20 ' 26) = 0.0115 V u . dIM u = 0.667 b w d

A v

f y

s

= 20 I 2.54 = 7.874 in = 26 I 2.54 = 10.236 in = {2' 0.320/6.45163 = 0.0992 in 2 (Specimen-A)

2 . 0.713/6.45163 = 0.2210 in 2 (Specimen-B) = 3500 I 0.07031 = 49780 psi = {19.5/2.54 = 7.677 in (Specimen-A)

6.5 I 2.54 = 2.559 in (Specimen-B)

V c = ( 1.9' 50.6 + 2500'0.01150.667) 7.874' 10.236 = 9294 Ib = 9294x0.453592 kgf = 4216 kgf = 4.22 tf

Vc ~ 3.5 (..r f c) b w d = 14273.951b = 6.47tf ~ OK

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(Specimen-A) V s = 0.0992 . 49780 . 10.236/7.677

= 6584 Ib = 6584x0.453592 kgf = 2987 kgf = 2.99 tf (Specimen-B)

V 5 = 0.221 . 49780 . 10.236/2.559 = 44006 Ib V s :-::;; 8 (.r f c) b w d = 32626 Ib

Then V s = 32626x0.453592 kgf = 14799 kgf = 14.8 tf

V = {4.22 + 2.99 tf = 7.2 tf (Specimen-A) u 4.22 + 14.8 tf = 19.0 tf (Specimen-B)

Pu3=7.2*2.5=18.0tf (Specimen-A) Pu3=19.0*2.5=47.5tf (Specimen-B)

(Specimen-A) Vu (7.2tf) < V y(12.5tf)

- Specimen-A will fail by shear before yielding of member.

(Specimen-B) Vu( 19.0tf) > V y(12.5tf)

- Specimen-B will not fail by shear before yielding of member.

3.5) Bond Splitting Strength (AIJ Guidelines * 1 I pp.105-112) *1: AIJ Design Guidelines for Earthquake Resistant Reinforced Concrete Buildings Based on Ultimate Strength Concept (1990), Chapter 6 Design for shear and bond

Notation of the following equations are that of AIJ Guidelines.

a) Design bond stress a-1) Bond stress at member yielding point

The longitudinal steel yield in tension. Compressive steel is also assumed yielding in compression as safety assumption.

T (1 = d b fla/ (4 L) (= AtAa/ (ljJL ) ) = 1.6 . 2 . 3500 / (4 . 78) = 35.9 kgf/cm 2

a-2) Bond stress at large deflection after yielding when the hinge length elongate to D

T (2 = d b fla/ [ 4 (L-D) ] = 1.6 . 2 . 3500 / [4 . (78-26) ]

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= 53.8 kgf/cm 2

a-3) Bond stress required for truss mechanism T I 3 = b . P w a w y cot

Since the bond stress requirement at member yielding is satisfied, though the bond stress required for tress mechanism is not satisfied, Specimen-B will exhibit member yielding. However, since the bond stress requirement at large deflection after yielding when the hinge length elongate to D are not satisfied, Specimen-B will fail by bond splitting after yielding of member.

4. Prediction of the behavior and failure pattern of specimens

4.1) Summary of the calculation

Initial stiffness S = 6.89x10 8 kgfcm

Flexural cracking V c =1.9tf (5 c = 0.082 mm

Member yielding V y = 12.5tf (5 y = 2.47 mm

Shear Cracking V 5 C = 6.3 tf .

Shear strength < Specimen-A>

V u (Arakawa's equation) = 11.5 tf V u (AIJ Guidelines) = 9.6 tf (without planning of yield hinge) V u (ACI Building Code) = 7.2 tf

- Specimen-A will fail ~y shear before yielding of member.

< Specimen-B > V u(Arakawa's equation) = 16.2 tf V u (AIJ Guidelines) = 23.1 ,tf (without planning of yield hinge)

= 16.9 tf (with planning of yield hinge: R p > 0.025) V u(ACI Building Code) = 19.0 tf

- Specimen-B will not fail by shear before and after yielding of member.

Bond Splitting Strength Design bond stress at member yielding point

T r 1 = 35.9 kgf/cm 2 Design bond stress at large deflection when the hinge length elongate to D

T r 2 = 53.8 kgf/cm 2 Design bond stress required for truss mechanism

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T r 3 = Bond strength

{ 15.3 kgf/cm 2 69.9 kgf/cm 2

(Specimen-A) (Specimen-B)

{ 26.0 kgf/cm 2 (Specimen-A) 39.1 kgf/cm 2 (Specimen-B)

{ 8.6 tt (Specimen-A) 12.9 tt (Specimen-B)

4.2) Prediction of the behavior and failure pattern of specimens

Specimen-A

At beginning of the loading, initial stiffness (S) will be 6.89x10 8 kgf-cm. First flexural crack will initiate at V= 1.9 tf, (5= 0.082 mm. First shear Crack will initiate at V= 6.3 tt. Before yielding of member, which shear force is predicted as V= 12.5 tt, shear

failure will occur. The predicted shear force when the shear failure occur will V=11.5 tt by Arakawa's equation, V=9.6 tf by AIJ Guidelines and V=7.2 tf by ACI Building Code.

Since the bond stress required for truss mechanism is satisfied, Specimen-A can exhibit it's shear capacity without bond splitting failure.

Then, Specimen-A is predicted as to fail by shear before yielding of member. Bond splitting failure will not be observed.

Specimen-B

At beginning of the loading, initial stiffness (S) will be 6.89>

[p(tfJJ

V(t{)

...... VI I

[37.5J 15.0

[31.3tf! 12.5tf

[25J 10.0

f12.5J 5.0

[4.8tfl 1.9tf

I Shear Cracking I 6.33tf [15.8tfl(Arakawa's Eq.)

~ , .' , ,

I Yielding I

S.bSl~r.Jllill~~~tY . .Qf..s.I>~f.imSlg_~_.(A).

1l.5tf [28.8tflCArakawa's Eq.) 9.6tf [24. Otfl (AJ.J A - method, without hinge) 7.2tf f18.0tflCACI Code)

S.bSl~r.J:lill~~~tY . .Qf..s.I>_~f.imSlg_~_.m) .

16.2tf [40. 5tfl CArakawa's Eq.) 23.ltf [57.8tfl(AJ.J A - method, without hinge) 16.9tf [42.3tfl(AJ.J A -method, with hinge, Rp=0.02)

, , , 19.0tf [47. 5tfl CACI Code)

, , , , , , , , , , , , , , 1?.9.ni.S.p'l}..Hin..K.S.tr..~J).,g,1;b_(._:r_.~.\l.2

, , , , , ,

( . I Specimen - (A) '" ~ Flexural Cracking r b u = 26.0kgflcm 2 > r f for truss mechanism = 15.3kgflcm 2 ~..........-- ,,~

,/ Specimen - (B) S ,/ r b u = 39.1kgflcm 2 > r f at member yielding = 35.9kgflcm 2 w\",OS

0.082 nan 1.0 2.0 2.47 mm 3.0 4.0

Fig.4 Predicted V [pJ- 8 relationship of beam specimens 8 (mm)

Notes) [pJ = 2.5V (tf)

=9.812.5V (kN)