EXAMPLE CALCULATIONS DTF / DTS

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1 MEMO 830 DTF / DTS EXAMPLE CALCULATIONS DESIGN Dato: Siste rev.: Dok. nr.: K4-10/30E Sign.: Sign.: Control: sss ps CONTENT EXAMPLE CALCULATIONS DTF / DTS EXAMPLE CALCULATIONS DTF / DTS... 1 PART 1 DTF10 USED IN DT GENERAL QUALITIES LOAD GEOMETRY CALCULATIONS... 3 PART DTF00 USED IN A HIGH DT GENERAL QUALITIES LOAD GEOMETRY CALCULATIONS Page 1 of 6

2 PART 1 DTF10 USED IN DT GENERAL QUALITIES Concrete C45/55: f ck 45,0 MPa EC, Table 3.1 f cd α cc f ck/γ c 0,85 45/1,5 5,5 MPa EC, Clause 3.15 f ctd α ct f ctk,0,05/γ c 0,85,70/1,5 1,53 MPa EC, Clause 3.16 f bd,5 η1 η f ctd,5 0,7 1,0 1,53,41 MPa EC, Clause 8.4. Reinforcement 500C: f yd f yk/γ s 500/1, MPa EC, Clause 3..7 Tendons: Diameter: Ø1,7mm. (Nominal diameter. Real diameter11,3mm) Assumed tension after elastic loss: P10kN LOAD Design load F v10kn Page of 6

3 1.1.3 GEOMETRY 1. CALCULATIONS 1) Equilibrium: Page 3 of 6

4 FV (75mm + 4,5mm) 10kN (75mm + 4,5mm) R 61kN 34,5mm 34,5mm R F + R 10kN + 61kN 181kN 1 v ) Reinforcement: 181kN AR 1 416mm 435MPa 61kN AR 140mm 435mm ø1stirrups 45mm 3) Bending of anchoring reinforcement: Minimum mandrel diameter, Ø m,min: Allowable concrete stress in node, EC, clause 6.5.: f ck f cd 0,6 (1 ) f cd ,6 (1 ) 5,5 50 1,5MPa Page 4 of 6

5 Actual concrete stress in node: R1 σ c b Ø sinθ cosθ m b7mm Ø m Mandrel diameter of front reinforcement θassume concrete strut in 45degrees. Solving for Ø m: R1 Øm b σ c sinθ cosθ N Ø m, min 18mm 7mm 1,5MPa sin(45) cos(45) Select mandrel diameter: Ø m160mm 4) Anchoring of Ø1 stirrups in front, EC clause and 8.4.4: l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min l b,reqd Ø σ 4 f sd bd / 4 Stress in stirrup: σ sd 400MPa π l b,reqd 497mm 4, 41 l b,min max(0,3 l b,reqd; 10 Ø; 100mm) 150 mm Page 5 of 6

6 Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. α α 3 α 51,0 1,0 1,01,0 > 0,7 OK l bd 1,0 1,0 1,0 1,0 1,0 497mm 497 mm 5) Lap of Ø1 stirrups, EC clause 8.7.3: l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length, Ø1: Ø σ sd l b,reqd 497mm f 4,41 4 bd l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm)4mm Table 8.: α 1,α,α 3,og α 51,0 as calculated in clause ). Page 6 of 6

7 Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 497mm746mm 750mm 6) Transmission length tendons, EC clause : Bond stress: f bptη p1 η 1 f ctd(t) η p13, (assume 3 or7-wire tendons) η 11,0 (assume good bond conditions ) f ctd(t) α ct 0,7 f ctm(t)/γ c α ct0,85 f ctm(t)(β cc(t)) α f ctm β cc(t)exp(s(1-8/t) 1/ ) Assume: Release after t 1day Assume: s0, β cc(t)exp[0, {1-(8/1) 1/ }]0,43 α1 (t<8days) f ctm3,8mpa f ctm(t)0,43 1 3,8MPa1,60MPa f ctd(t) 0,85 0,7 1,60MPa/1,50,635MPa f bpt3, 1,0 0,635MPa,03MPa Transmission length: l ptα 1 α Ø σ pmo/ f bpt α 11,0 (assume gradual release) α 0,19 (assume 3 or7-wire tendons) Ø1,7mm (Nominal diameter of tendon. Real diameter11,3mm) Page 7 of 6

8 σ pmo 100MPa l pt1,0 0,19 1,7mm 100MPa/,03MPa146mm l pt10,8 l pt 0,8 146mm1141mm. To be used in evaluation of local cross section stresses. l pt1, l pt 1, 146mm1711mm. To be used in evaluation of anchorage. Stress after all losses assumed as 0,9σ pmo. (10% loss) 7) Anchoring: Assuming the horizontal part of the front anchoring bar is 750mm ( equals the minimum calculated lap length). I.e. the bar ends at x mm. Section 1: Force anchored in the tendons (7 tendons) at x13mm: Assume 10% loss of pre-stressing force: F sp17 P 0,9 13mm/1711mm7 10kN 0,9 13mm/1711mm54kN Force anchored in Ø1: F Ø1181kN Total anchored force: FF sp1+f Ø154kN+181kN35kN Tension in reinforcement at x13mm: (clause 6..3(7)) S(x)M(x)/z+0,5V Ed (cot(θ)-cot(α)) M(x)/z+0,5 V Ed (cot(45)-cot(90)) (assume 45degrees concrete struts and vertical links) M(x)/z+0,5 V Ed (1-0) M(x)/z+0,5 V Ed Moment at x13: M (x13)10kn (13+75)mm3,8kNm Assume: z0,9d0,9 364mm38mm (approximately) Page 8 of 6

9 S (x13)3,8knm/0,38m+181kn/163kn The anchoring at x13mm is sufficient. Section : Force anchored in the tendons (7 tendons) at x873mm: Assume 10% loss of pre-stressing force: F sp17 P 0,9 873mm/1711mm7 10kN 0,9 873mm/1711mm386kN Force anchored in Ø1: F Ø10kN Total anchored force: FF sp1+f Ø1386kN+0kN386kN Tension in reinforcement at x873mm: (clause 6..3(7)) S(x)M(x)/z+0,5V Ed (cot(θ)-cot(α)) M(x)/z+0,5 V Ed (cot(45)-cot(90)) (assume 45degrees concrete struts and vertical links) M(x)/z+0,5 V Ed (1-0) M(x)/z+0,5 V Ed Moment at x873: M (x873)10kn (873+75)mm113,8kNm Assume: z0,9d0,9 364mm38mm S (x873)113,8knm/0,38m+10kn/407kn The tendons will not have sufficient anchorage at x873mm. Calculating the point where the tendons are sufficient anchored to carry the load: Tension in reinforcement at x: Force anchored in the tendons at x (if x<1711m): S(x)10kN (x+75mm)/38mm+10kn/ F(x)7 10kN 0,9 x/1711mm Hence: 10kN ( x + 75mm) / 38mm + 10kN / 756kN x /1711 0,366x + 87,4 0,44x 0,076x 87,4 x 87,4 / 0,076 x 1150mm The horizontal part of the Ø1 anchoring bars has to be extended beyond the calculated required lap length (evaluated in clause 5), and end at x>1150. Page 9 of 6

10 8) Reinforcement due to splitting stress: 9) Links: A s0, P u1/f s A s0, N/300MPa 616mm To be located within: 0,5 (l lpt1+h 1)0,5 (1141mm+450mm)796mm h 1450mm 450mm Corresponds to: 616mm /0,450m1369mm /m a) Required shear reinforcement x<77: As VEd R N 169mm s z f cotθ z f cotθ 0,38m 435MPa cot(45) ywd ywd / m Shear compression: α cw bw z v1 f cd V Rd (cotθ + tanθ ) α cw1,0 (neglecting effect of pre-tensioning) b w7mm-50mm177mm z38mm υ 10,6 (1-f ck/50) 0,6 (1-45/50)0,49 θassume 45degrees 1,0 177mm 38mm 0,49 5,5MPa V Rd 364,kN > R OK b) Required shear reinforcement 77<x<606: As VEd FV 10000N s z f cotθ z f cotθ 0,38m 435MPa cot(45) ywd ywd 841mm / m Shear compression: α cw bw z v1 f cd V Rd (cotθ + tanθ ) α cw1,0 (neglecting effect of pre-tensioning) b w7mm z38mm υ 10,6 (1-f ck/50) 0,6 (1-45/50)0,49 Page 10 of 6

11 V Rd θ assume 45degrees 1,0 7mm 38mm 0,49 5,5MPa 467kN 1+ 1 > F v OK Summary reinforcement in end of the DT: X< 450mm: Links due to splitting stresses (8) A s 1369 mm / m s Select Ø8 stirrups c/c70 (lapped with u-shaped Ø1 bars) A s ( 4mm) 1436 π mm / m s 0,07m 450<X< 606mm: Links due to shear force (9). A s 841 mm / m s Select Ø8 stirrups c/c100 A s ( 4mm) 1005 π mm / m s 0,1m 606<X: Required shear reinforcement to be evaluated according to shear force distribution in the DT. At x77mm: Required reinforcement: 140mm 1 extra Ø8 link (100mm ) at this location. The remaining 40mm can be carried by the specified distributed links. Page 11 of 6

12 Suggested reinforcement in end of the DT: PART DTF00 USED IN A HIGH DT.1 GENERAL.1.1 QUALITIES Concrete C45/55: f ck 45,0 MPa EC, Table 3.1 f cd α cc f ck/γ c 0,85 45/1,5 5,5 MPa EC, Clause 3.15 f ctd α ct f ctk,0,05/γ c 0,85,70/1,5 1,53 MPa EC, Clause 3.16 f bd,5 η1 η f ctd,5 0,7 1,0 1,53,41 MPa EC, Clause 8.4. Reinforcement B500C: f yd f yk/γ s 500/1, MPa EC, Clause 3..7 Tendons: Diameter: Ø1,7mm. (Nominal diameter. Real diameter11,3mm) Assumed tension after elastic loss: P10kN. Page 1 of 6

13 .1. LOAD Design load F v180kn (NB: The design load in the example is less than the ultimate limit load of the unit.).1.3 GEOMETRY Page 13 of 6

14 Page 14 of 6. CALCULATIONS 1) Equilibrium: kn kn kn R F R kn mm mm mm kn mm mm mm F R v V ) 45 ( ) 45 ( ) Reinforcement: mm mm kn A mm stirrups ø mm MPa kn A R R

15 3) Bending of anchoring reinforcement: Minimum mandrel diameter, Ø m,min: Allowable concrete stress in node, EC, clause 6.5.: f ck f cd 0,6 (1 ) f cd ,6 (1 ) 5,5 50 1,5MPa Actual concrete stress in node: R1 σ c b Ø sinθ cosθ m b10mm Ø m Mandrel diameter of front reinforcement θassume concrete strut in 45degrees. Solving for Ø m: R1 Øm b σ c sinθ cosθ 71000N Ø m, min 361mm 10mm 1,5MPa sin(45) cos(45) Select mandrel diameter: Ø m450mm Page 15 of 6

16 4) Anchoring of Ø16 stirrups in front, EC clause and 8.4.4: l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min l b,reqd Ø σ 4 f sd bd / 4 Stress in stirrup: σ sd 337MPa π l b,reqd 560mm 4, 41 l b,min max(0,3 l b,reqd; 10 Ø; 100mm) 168 mm Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Page 16 of 6

17 Not relevant. α α 3 α 51,0 1,0 1,01,0 > 0,7 OK l bd 1,0 1,0 1,0 1,0 1,0 560mm 560 mm 5) Lap of Ø16 stirrups, EC clause 8.7.3: l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length, Ø16: Ø σ sd l b,reqd 560mm f 4,41 4 bd l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm)51mm Table 8.: α 1,α,α 3,og α 51,0 as calculated in clause ). Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 560mm 840mm Page 17 of 6

18 6) Transmission length tendons, EC clause : Bond stress: f bptη p1 η 1 f ctd(t) η p13, (assume 3 or7-wire tendons) η 11,0 (assume good bond conditions ) f ctd(t) α ct 0,7 f ctm(t)/γ c α ct0,85 f ctm(t)(β cc(t)) α f ctm β cc(t)exp(s(1-8/t) 1/ ) Assume: Release after t 1day Assume: s0, β cc(t)exp[0, {1-(8/1) 1/ }]0,43 α1 (t<8days) f ctm3,8mpa f ctm(t)0,43 1 3,8MPa1,60MPa f ctd(t) 0,85 0,7 1,60MPa/1,50,635MPa f bpt3, 1,0 0,635MPa,03MPa Transmission length: l ptα 1 α Ø σ pmo/ f bpt α 11,0 (assume gradual release) α 0,19 (assume 3 or7-wire tendons) Ø1,7mm (Nominal diameter of tendon. Real diameter11,3mm) Page 18 of 6

19 σ pmo 100MPa l pt1,0 0,19 1,7mm 100MPa/,03MPa146mm l pt10,8 l pt 0,8 146mm1141mm. To be used in evaluation of local cross section stresses. l pt1, l pt 1, 146mm1711mm. To be used in evaluation of anchorage. Stress after all losses assumed as 0,9σ pmo. (10% loss) 7) Anchoring: Assuming the horizontal part of the front anchoring bar is 840mm ( equals the minimum calculated lap length). I.e. the bar ends at x mm. Section 1: Force anchored in the tendons (5 tendons) at x70mm: Assume 10% loss of pre-stressing force: F sp15 P 0,9 70mm/1711mm5 10kN 0,9 70mm/1711mm85kN Force anchored in Ø16: F Ø1671kN Total anchored force: FF sp1+f Ø1685kN+71356kN Tension in reinforcement at x70mm: (clause 6..3(7)) S(x)M(x)/z+0,5V Ed (cot(θ)-cot(α)) M(x)/z+0,5 V Ed (cot(45)-cot(90)) (assume 45degrees concrete struts and vertical links) M(x)/z+0,5 V Ed (1-0) Page 19 of 6

20 M(x)/z+0,5 V Ed Moment at x70: M (x70)180kn (70+75)mm6,1kNm Assume z0,9d0,9 733mm660mm (approximately) S (x70)6,1knm/0,660m+71kn/30kn The anchoring at x70mm is sufficient. Section : Force anchored in the tendons (5 tendons) at x1110mm: F sp15 P 0,9 105mm/1711mm5 10kN 0,9 1110mm/1711mm350kN Force anchored in Ø16: F Ø160kN Total anchored force: FF sp1+f Ø16350kN+0kN350kN Tension in reinforcement at x1110mm: (clause 6..3(7)) S(x)M(x)/z+0,5V Ed (cot(θ)-cot(α)) M(x)/z+0,5 V Ed (cot(45)-cot(90)) (assume 45degrees concrete struts and vertical links) M(x)/z+0,5 V Ed (1-0) M(x)/z+0,5 V Ed Moment at x1110: M (x1110)180kn ( )mm13,3kNm Assume z0,9d0,9 733mm660mm (approximately) S (x1110)13,3knm/0,660m+180kn/413kn The tendons will not have sufficient anchorage at x1110mm. Calculating the point where the tendons are sufficient anchored to carry the load: Tension in reinforcement at x: S(x)180kN (x+75mm)/660mm+180kn/ Force anchored in the tendons at x: If x<1711m: F(x)5 10kN 0,9 x/1711mm If x>1711m: F(x)5 10kN 0,9+ σ sp(x) A sp 5 σ sp(x)f bpd/(α Ø) (x-1711) 10-3 F(x) 5 10kN 0,9+f bpd/(α Ø) (x-1711) 10-3 A sp 5 F(x) 540kN+,03MPa/(0,19 1,7) (x-1711) mm 5 F(x) 540+0,40x-700,40x-180 Page 0 of 6

21 Try if x<1711: 180kN ( x + 75mm) / 660mm + 180kN 0,73x + 110,5 0,316x 0,043x 110,5 x 110,5 / 0, kN x /1711 x 570mm Calculated x>1711 the second expression shall be used: 180kN ( x + 75mm) / 660mm + 180kN / 0,40x 180 0,73x + 110,5 0,40x 180 0,147x 90,5 x 90,5 / 0,147 x 1976mm / The horizontal part of the Ø16 anchoring bars has to be extended beyond the calculated required lap length (evaluated in clause 5), and end at x> ) Reinforcement due to splitting stress: A s0, P u1/f s A s0, N/300MPa 440mm To be located within: 0,5 (l lpt1+h 1)0,5 (1141mm+863mm)100mm h 1863mm 863mm Corresponds to: 440mm /0,863m510mm /m 9) Links/strut and tie model: For this particular DT, the height z is approximately equal to 3xb. Hence, the bending moment from the cantilevering may be assumed transferred to the main reinforcement of the beam through a local truss with three levels. This is illustrated as a blue truss in the end of the DT, see below Figure. The compression struts in this truss will be angled in approximately 45 degrees. The support reaction force/shear force itself may be carried by a truss with a height equal to z. This is illustrated as a red truss. The two trusses are both in equilibrium, and the forces may be superimposed. Page 1 of 6

22 Evaluation of local strut and tie model: The model implies that only 1/3 of the force R have to be carried by the vertical link at back of the unit. Nevertheless, it is chosen to calculate the required extra reinforcement at this point based on the value of R. (conservative assumption) Considering the two horizontal ties towards the end of the DT as smeared, the horizontal force per unit height of the DT becomes: (a/z) F v/(z/3). Where z/3 b, hence this corresponds to: (a/z) F v/b(a/b) F v/zr /z. Prescribe horizontal bars anchoring this force. Page of 6

23 a) Required shear reinforcement x<45+b: As VEd R N 944mm s z f cotθ z f cotθ 0,660m 435MPa cot(45) ywd Shear compression: α cw bw z v1 f V Rd (cotθ + tanθ ) V Rd cd ywd α cw1,0 (neglecting effect of pre-tensioning) b w10mm z660mm υ 10,6 (1-f ck/50) 0,6 (1-45/50)0,49 θassume 45grader 1,0 10mm 660mm 0,49 5,5MPa 497kN > R OK b) Required shear reinforcement 45+b<x<45+z: As VEd FV N 67mm s z f ywd cotθ z f ywd cotθ 0,660m 435MPa cot(45) Shear compression: α cw bw z v1 f cd V Rd (cotθ + tanθ ) α cw1,0 (neglecting effect of pre-tensioning) b w10mm z660mm υ 10,6 (1-f ck/50) 0,6 (1-45/50)0,49 θassume 45grader 1,0 10mm 660mm 0,49 5,5MPa VRd 497kN > Fv OK c) Horizontal stirrups at the end of the DT: A R s N 317mm / m s z f ywd 0,660m 435MPa Amount of reinforcement below the unit: A s 317mm / m / 3z 317mm / m / 3 0,66m 140mm x U-shaped Ø8 bars (A s00mm ). / m / m Page 3 of 6

24 Summary reinforcement in end of the DT: The required amount of reinforcement due to splitting stresses is less than the required amount of shear reinforcement calculated in 9a and 9b. X< 8mm: Links due to shear force (9a) A s 944 mm / m s Select Ø8 stirrups c/c75 (lapped with u-shaped Ø1 bars) A s ( 4mm) 1340 π mm / m s 0,075m 8<X< 705mm: Links due to shear force (9b) A s 67 mm / m s Select Ø8 stirrups c/c150 ( lapped with u-shaped Ø1 bars) A s ( 4mm) 670 π mm / m s 0,150m 705<X: Required shear reinforcement to be evaluated according to shear force distribution in the DT. At X8: Required reinforcement: 10mm. 1 extra Ø8 (100mm ) at this location. The remaining 110mm can be carried by the specified distributed links. Anchoring (7): Horizontal part of Ø16 anchoring bars in front 1700mm. (Ends at x>1976mm) Page 4 of 6

25 Suggested reinforcement in end of the DT: Page 5 of 6

26 REVISION HISTORY Date: Description: First Edition Included revision history table. Updated reinforcement quality New template Page 6 of 6

Detailing. Lecture 9 16 th November Reinforced Concrete Detailing to Eurocode 2

Detailing. Lecture 9 16 th November Reinforced Concrete Detailing to Eurocode 2 Detailing Lecture 9 16 th November 2017 Reinforced Concrete Detailing to Eurocode 2 EC2 Section 8 - Detailing of Reinforcement - General Rules Bar spacing, Minimum bend diameter Anchorage of reinforcement