SAMPLE LABORATORY SESSION FOR JAVA MODULE B Calulations fo Sample Coss-Setion. Use Input. Setion Popeties The popeties of Sample Coss-Setion ae shown in Figue and ae summaized below. Figue : Popeties of Sample Coss-Setion.
Conete Popeties Cuently, the entie oss-setion is assumed to be unonfined. The ompessive stess-stain elationship of the unonfined onete is detemined using a method developed by Mande et al. []. The following use speified popeties ae needed: o Conete ompessive stength, f ' = 7.5 MPa o Conete stain oesponding to peak stess ( f ), = 0.00 o The default value used by the module fo o is 0.00. Steel Popeties o Steel yield stength, f y = 400 MPa o Steel Young s Modulus, E s = 00000 MPa The default value used by the module fo Setion Dimensions E s is 00000 MPa. Cuently, only etangula oss-setions ae allowed by the module. Reinfoement o Setion height, h = 60 m o Setion width, b = 36 m o Stiup diamete, d = 0. 95 m s o Numbe of layes of longitudinal bas, n = o Fist laye : bottom laye Numbe of longitudinal bas = 4 Diamete of longitudinal bas, d =. 9 m b l
Distane to ompession (top) fae, d = 54 m o Seond laye: top laye Numbe of longitudinal bas = Diamete of longitudinal bas, d =. 9 m Distane to ompession (top) fae, d = 6 m b The use input fo the fist laye of bas is shown in Figue. Figue : Reinfoement fo Laye. Seleted use-speified popeties of the setion ae displayed in Window as shown in Figue 3. 3
Figue 3: Window epesentation of Sample Coss-Setion.. Axial Foes The use input fo the axial foes is shown in Figue 4. These axial foes (up to five foes) ae used to geneate moment-uvatue elationships fo Figue 4: Axial foes fo Sample Coss-Setion. the setion. The lagest ompessive axial foe that an be speified fo a oss-setion is equal to 0.85 f A + A st f s, whee f is the onete stength, 4
A is the onete aea, A st is the total steel aea, and f s is the steel stess oesponding to onete ushing. The smallest ompessive axial foe that an be speified is zeo. The module is not designed to onside tensile foes, i.e., negative axial loads. The module uses zeo axial load as the default value. As shown in Figue 4, the numbe of axial foes speified fo Sample Coss-Setion is thee, namely, 0, 50 and 00% of the balaned failue load. Only 50% of the balaned load will be onsideed in the sample alulations below..3 Stain Condition fo P-M Inteation The module equies a use-speified maximum ompession stain value to geneate the P-M inteation diagams. A stain value less than o equal to the onete ushing stain may be used. The module uses the onete ushing stain as the default value. A use-speified stain value of 0.003 is speified to geneate the P-M inteation diagams fo Sample Coss-Setion as shown in Figue 5. Figue 5: Stain ondition fo P-M inteation. 5
. Calulations and Equations The following sample alulations ae based on the method employed by Java Module B.. Conete Stess-Stain Relationship The equation fo the unonfined onete stess-stain elationship is []: f = f x + x, whee () x = () =, 0 0.00 o The tangent modulus of elastiity, E, is alulated using: E = 4,74 f MPa = 4,86.0 MPa (3) E se, the seant modulus of elastiity, is the slope of the line onneting the oigin and peak stess on the ompessive stess-stain uve (see Fig. 6). f E se = =3,750.0 MPa (4) o Then, E = =.4 (5) E E se and, the onete stess-stain ( f ) elationship is given as: f 7.5( )(.4) = 0.00.4 + ( ) 0.00.4, (6) The above f elationship is plotted in Figue 6. It is assumed that ushing of onete ous at a stain of = = 0.004. u o 6
Figue 6: Conete stess-stain elationship fo Sample Coss-Setion. In Fig. 6, Cile make: assumed onete linea-elasti limit at Squae make: peak stess at f = f and = o. f f f = fel = and = el =. E Diamond make: assumed ultimate stain at = =. u o. Moment-Cuvatue Relationships Window geneates the moment-uvatue elationships of the setion fo the usespeified axial foes. This is an iteative poess, in whih the basi equilibium equiement (e.g., P= C + Fs Fs Ct fo a setion with two layes of einfoement) and 7
a linea stain diagam ae used to find the neutal axis fo a patiula maximum onete ompessive stain, m, seleted (see Figue 7). Figue 7: Setion stains, stesses, and stess esultants. The alulation of the following fou points on a moment-uvatue uve will be shown in this example: 0. 5 m = u = 0.5 m u 0. 75 m = u m u = (onete ushing) Axial Load, P The axial load onsideed fo these sample alulations is 50% of the balaned failue load. The balaned load, P b, is omputed as follows: The neutal axis, f u y = b = d, whee y = (7) u + y Es With u = 0.004 and d = 54 m, this gives a value of b = 36.0 m. The onete ompessive esultant, onete stess distibution uve. C, is detemined by numeially integating unde the 8
b u C = f bdx = f b d =,795.6 kn, whee 0 0 m f is fom Eq. (6) (8) The top and bottom steel foes, F s and F s, espetively, ae alulated using similaity to find the stains in the layes. Balaned failue ondition, by definition, has stain values m = = 0.004 fo onete and = = 0.00 fo bottom steel. Fo the top steel, u s y d' s = s =0.0033 (9) d whih implies that the top steel is also at yield stess. Hene, Fs = sesas = fyas = 458.4 kn (tension) (0) Fs = sesas fyas = 9. kn (ompession) () whee, As and A s ae the total einfoing steel aeas in eah laye. The module onsides the onete tensile stength in the tension egion. ACI-38 [] eommends the modulus of uptue to be taken as f = 0.6 f MPa () fo nomal weight onete. Thus, fo f = 7.5 MPa, f = 3.3 MPa. The onete tension foe At = 6.96 kn (3) whee, At is the aea of onete in tension alulated based on the linea stain diagam. Then, the balaned axial load is found fom equilibium as Pb = C + Fs Fs Ct =,559.4 kn. (4) Theefoe, 50% of the balaned load used in the example is P = 79.7 kn. 9
Instant Centoid The module assumes that the axial load ats at an instant entoid loation fo the alulation of the moment-uvatue elationship. The loation of the instant entoid is detemined by assuming an initial ondition whee only the use-seleted axial load ats on the oss-setion without moment. This loading ondition podues a unifom ompession stain distibution thoughout the oss-setion. Let the unifom ompession stain be equal to i. Then s = s = i and fsi = fs = fs = Esi fy (5) Fom equilibium, f A + f A + f A = P (6) i s s s s i 7.5( )(.4) 0.00 ( A) + ( fsi)( As + As) = P= 79.7 i.4.4 + ( ) 0.00 kn (7) whee A = bh ( As + As) = 43.0 m (8) A tial-and-eo solution on Eq. (7) is needed sine it is not known in advane if the bas ae yielding; fom whih the stain i an be alulated as: = 0.0007 f = 5.6 MPa and f = 45.4 MPa (bas not yielding). i i si Then, the loation of the instant entoid, x, fom the top ompession fae is detemined as x f Ah/ + A f d + A f d' i s si s si = = fi A + As fsi + As fsi 30.48 m (9) 0
Point The alulation of the fist sample point on the moment-uvatue elationship of the setion an be summaized as follows:. 0.5 = 0. 00 m = u. Assume the neutal axis depth, a distane = 5 m. 3. Fom the linea stain diagam geomety m s = ( d ) = 0.006 (at yield stess) and (0) m s = ( d ) = 0.0006 (below yield stess) () 4. The steel stess esultants ae Fs = fyas = 458.4 kn (tension) and () Fs = sesas = 68.8 kn (ompession). (3) 5. Detemine C by integating numeially unde the onete stess distibution uve. m C = f bdx = f b d = 644.3 kn. (4) 0 0 m whee f is given by Eq. (6). The onete that has not aked below the neutal axis ontibutes to the tension foe. At =. kn with f = 3.3 MPa fom Eq. () (5) 6. Chek to see if P= C + Fs Fs Ct (6) But P = 7 kn 43.5 kn = C + Fs Fs Ct
So, the neutal axis must be adjusted downwad, fo the patiula maximum onete stain that was seleted in Step, until equilibium is satisfied. This iteative poess detemines the oet value of. Tying neutal axis depth = 3.9 m gives: s = 0.00064 (below yield stess) and s = 0.0008 (below yield stess). F s = 46.7 kn (tension) and F s = 93.7 kn (ompession). m C = f bdx = f b d =,369. kn. (7) 0 0 m At = 6.0 kn. (8) P = 79 kn 85 kn = C + Fs Fs Ct O.K. Setion uvatue an then be found fom: ψ m 0.00 5 = = = 3.04 0 3.9 m (9) The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z = 9.5 m and z t = 4.39 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 38.5 kn-m. (30) Point The alulation of sample point two on the moment-uvatue elationship of the setion an be summaized as follows:. = 0.5 = 0.00 m u. Assume the neutal axis depth, a distane = 8 m. 3. Fom the linea stain diagam geomety
m s = ( d ) = 0.004 (at yield stess) and m s = ( d ) = 0.003 (below yield stess) 4. The steel stess esultants ae Fs = fyas = 458.4 kn (tension) and Fs = sesas = 5.8 kn (ompession). 5. Detemine C by integating numeially unde the onete stess distibution uve. m C = f bdx = f b d = 00.7 kn. 0 0 m whee f is given by Eq. (6). The onete that has not aked below the neutal axis ontibutes to the tension foe. At = 7 kn with f = 3.3 MPa fom Eq. () 6. Chek to see if P= C + Fs Fs Ct But P = 7 kn 894.4 kn = C + Fs Fs Ct So, the neutal axis must be adjusted downwad, fo the patiula maximum onete stain that was seleted in Step, until equilibium is satisfied. This iteative poess detemines the oet value of. Tying neutal axis depth = 3.76 m gives: s = 0.005 (at yield stess) and s = 0.005 (below yield stess). F s = 458.4 kn (tension) and F s = 7.3 kn (ompession). m C = f bdx = f b d =,585. kn. 0 0 m 3
At = 9.4 kn. P = 79 kn 88 kn = C + Fs Fs Ct O.K. Setion uvatue an then be found fom: ψ m 0.00 5 = = = 8.4 0 3.76 m The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z =.6 m and z t = 5.7 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 49.6 kn-m. Point 3 The alulation of sample point thee on the moment-uvatue elationship of the setion an be summaized as follows:. = 0.75 = 0.003 m u. Assume the neutal axis depth, a distane = 0 m. 3. Fom the linea stain diagam geomety m s = ( d ) = 0.005 (at yield stess) and m s = ( d ) = 0.00 (at yield stess) 4. The steel stess esultants ae Fs = fyas = 458.4 kn (tension) and Fs = fyas = 9. kn (ompession). 5. Detemine C by integating numeially unde the onete stess distibution uve. 4
m C = f bdx = f b d = 55.3 kn. 0 0 m whee f is given by Eq. (6). The onete that has not aked below the neutal axis ontibutes to the tension foe. At = 5.3 kn with f = 3.3 MPa fom Eq. () 6. Chek to see if P= C + Fs Fs Ct But P = 7 kn 90.8 kn = C + Fs Fs Ct So, the neutal axis must be adjusted upwad, fo the patiula maximum onete stain that was seleted in Step, until equilibium is satisfied. This iteative poess detemines the oet value of. Tying neutal axis depth = 9.86 m gives: s = 0.005 (at yield stess) and s = 0.00 (at yield stess). F s = 458.4 kn (tension) and F s = 9. kn (ompession). m C = f bdx = f b d =,54.6 kn. 0 0 m At = 5. kn. P = 79 kn 80 kn = C + Fs Fs Ct O.K. Setion uvatue an then be found fom: ψ 0.003.5 0 9.86 m 4 = = = m 5
The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z =.33 m and z t = 0.0 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 50.6 kn-m. Point 4 The alulation of sample point fou on the moment-uvatue elationship of the setion an be summaized as follows:. =.0 = 0.004 m u. Assume the neutal axis depth, a distane = m. 3. Fom the linea stain diagam geomety m s = ( d ) = 0.0063 (at yield stess) and m s = ( d ) = 0.009 (at yield stess) 4. The steel stess esultants ae Fs = fyas = 458.4 kn (tension) and Fs = fyas = 9. kn (ompession). 5. Detemine C by integating numeially unde the onete stess distibution uve. m C = f bdx = f b d = 630.4 kn. 0 0 m whee f is given by Eq. (6). The onete that has not aked below the neutal axis ontibutes to the tension foe. At = 4. kn with f = 3.3 MPa fom Eq. () 6
6. Chek to see if P= C + Fs Fs Ct But P = 7 kn 396.7 kn = C + Fs Fs Ct So, the neutal axis must be adjusted upwad, fo the patiula maximum onete stain that was seleted in Step, until equilibium is satisfied. This iteative poess detemines the oet value of. Tying neutal axis depth = 9.48 m gives: s = 0.007 (at yield stess) and s = 0.008 (at yield stess). F s = 458.4 kn (tension) and F s = 9. kn (ompession). m C = f bdx = f b d =,5.4 kn. 0 0 m At = 3.8 kn. P = 79 kn 79.4 kn = C + Fs Fs Ct O.K. Setion uvatue an then be found fom: ψ m 0.004 4 = = = 9.48.05 0 m The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z =.8 m and z t = 0.6 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 493.5 kn-m. 7
Plots The thee axial foes speified to geneate the moment-uvatue elationships fo Sample Coss-Setion ae epesented as P, P and P3 in Window as shown below. The M and ψ pais alulated fo the fou points above ae plotted on the moment-uvatue uve fo P = 0.50P b. Figue 8: Sample Coss-Setion moment-uvatue elationships shown in Window.. Axial-Foe-Bending-Moment Inteation Diagam Window 3 geneates P-M inteation diagams fo the use-defined oss-setion by detemining the axial load and moment pais of the setion fo a use-speified maximum onete ompession stain, m. The P-M inteation diagam fo eah oss-setion is geneated by seleting suessive hoies of the neutal axis distane,, fom an initial small value to a lage one that gives a pue axial loading ondition. The initial neutal axis value, o, oesponds to the pue bending ondition (i.e., no axial foe) of the osssetion. 8
The alulation of the following thee points on the P-M inteation diagam of Sample Coss-Setion will be shown in this example. = o =.5 o = 3 o Figue 9: Setion stains, stesses, and stess esultants fo P-M inteation. Instant Centoid The module alulates an instant entoid fo P-M inteation, simila to the instant entoid used fo the setion moment-uvatue elationship. The axial load is assumed to be applied at the instant entoid, whih is detemined fom a unifom ompession stain distibution ove the setion, equal to the use-speified maximum onete stain, m, fo P-M inteation. Fo Sample Coss-Setion, the use-speified maximum onete ompession stain = 0.003 (see Figue 5). m Thus, 0.003 f = 4. 83 MPa [fom Eq. (6)] and i = i = = 0.003 f = 400 MPa (at yield). si i si 9
With A = bh ( As + As) = 43 m, the loation of the instant entoid, x, fom the top ompession fae is detemined as fbh i / + As fd si + As fd si ' x = = 30.9 fbh+ A f + A f i s si s si m Point The alulation of the fist sample point on the P-M inteation diagam of the setion is desibed below. The initial neutal axis loation, o, oesponding to pue bending is detemined as follows: s d o = m and s o d o = m (3) o whee the use speified m = 0.003 Then, fs = ses fy and fs = ses fy (3) Sine P = 0, fom equilibium: C + f A = f A + C (33) s s s s t Substituting fo the onete and steel stess esultants gives the following equation m f b d f A f A f + = + A (34) 0 o s s s s t m An iteative solution of Eq. (34) is needed (sine the yielding bas ae not known in advane), fom whih the value of the neutal axis an be obtained as o = 6.0 m. The intenal foes an then be alulated as: F s = 458.4 kn (tension, at yield) and F s = 0.57 kn (ompession, below yield). o m o 0 0 m kn C = f bdx = f b d = 458.4 0
At =.6 kn The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z = 8.46 m and z t = 4.7 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 36. kn-m. Point The alulation of the seond sample point, =. 5, on the P-M inteation diagam of o the setion is desibed below. s d = m and s = m d The neutal axis is = 9.0 m. The intenal foes an then be alulated as: Fs = fyas = 458.4 kn (tension, at yield) and Fs = sesas = 5. kn (ompession, below yield). m C = f bdx = f b d = 687.5 kn 0 0 m At =.4 kn Then, the setion axial foe an be alulated as P= C + Fs Fs Ct = 34.8 kn. The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z = 7. m and z t =.6 m, espetively. Then, the setion moment an be alulated as
M = Cz + Fs zs+ Fszs Ctzt = 3. kn-m. Point 3 The alulation of the thid sample point, = 3. 0, on the P-M inteation diagam of the o setion is desibed below. s d = m and s = m d The neutal axis is = 8.03 m. The intenal foes an then be alulated as: Fs = fyas = 458.4 kn (tension, at yield) and Fs = fyas = 9. kn (ompession, at yield). m C = f bdx = f b d =,375. kn 0 0 m At = 4.7 kn P= C + Fs Fs Ct = 4. kn The intenal leve ams fo the esultant ompession and tension foes of the onete measued fom the instant entoid ae z = 3.5 m and z t =.35 m, espetively. Then, the setion moment an be alulated as M = Cz + Fs zs+ Fszs Ctzt = 485.8 kn-m Plots The M and P pais alulated fo the thee points above ae plotted on the inteation diagam shown in Figue 0 below.
Figue 0: Sample Coss-Setion inteation diagam shown in Window 3 3