1. EAS 453 Pre-stressed Concrete Design
ULTIMATE MOMENT OF RESISTANCE (Mu)
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

2. Part of the exercise under ultimate limit state (the other is shear)
The requirement must be fulfilled only when the member is experiencing load at ultimate (factored)
As loads on the pre-stressed member increase above SLS, cracking start to occurs and the pre-stressing steel begins to behave as conventional reinforcement except the initial strain must be taken into account
1.4 Dead Load Live Load will produce design ultimate moment (M) – case of simply supported beam
Mu > M at any section
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

3. Example 1 Determine the design ultimate moment for 20 meter a beam with 300 x 800 cross section supporting DL = 15 kN/m and LL = 10 kN/m. Assume concrete density = 24 kN/m3. WDL = 0.3 x 0.8 x 24 = 5.76 kN/m n = 1.4 ( ) (10) = 45.1 kN/m M = 45.1 x 202/8 = 2255 kNm
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

4. As part of requirement under Clause 4.3.7
Why need to check at ULS ?
As part of requirement under Clause 4.3.7
If in any condition for Class 2 & 3 members, Mu < M at any section, normally at critical section, additional reinforcement (normally untensioned ) can be added
Assumptions
The strain distribution in the concrete in compression is derived from the assumption that plane section remain plane
The design stresses in the concrete in the compression are derived either from the stress-strain curve given in Figure 2.1 with γm = 1.5 or taken as 0.45fcu for a depth (from the compression face) equal to 0.9 times the depth of the compression zone.
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

5. Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

6. Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

7. Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

8. The tensile strength of the concrete is ignored
(In both cases the strain at the outermost compression fibre is taken as ) = εcu
The tensile strength of the concrete is ignored
The strains in bonded pre-stressing tendons and in any additional reinforcement, whether in tension or compression, are derived from the assumption that plane sections remain plane
 design stresses in bonded pre-stressing tendons, whether initially tensioned on untensioned, and in any additional reinforcement are derived from the appropriate stress-strain curve
The design stress in unbonded pre-stressing tendons is limited to the values given by equation 52 of Clause unless a higher value can be justified by a more rigorous analysis or on the basis of tests
Note : Bonded post-tensioned beam – space between tendon and the duct is grouted after tensioning, similarly unbonded post-tensioned beam is when the duct is not grouted.
**bonded = grouted =turap
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

9. Analysis using strain compatibility
The tendon strain at ultimate consist of 2 components εpb = εpe + εpa εpe = strain in tendon due pre-stress after all loses have occurred = KPo/(Aps Es) εpa = additional strain due applied load in general = β1εe + β2 εu β1 = bond coefficients = 0.5 for unbonded post-tensioned beam) = 1.0 for pre-tensioned and bonded post-tensioned beam β2 = bond coefficient = between 0.1 – 0.25 for unbonded pt beam
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

10. εe = concrete strain in concrete due effective pre-stress force only
  = K/Ec x (P/A + Pe2/I)
εu = average strain in concrete at tendon level at ultimate
= (d – x) εcu (εcu = ) x
Thus, for bonded post-tensioned beam, combining 2 equations;
εpb = εpe + εe + εcu (d – x)
Thus,
x = dεcu / (εcu + εpb - εpe - εe )
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

11. SIMPLIFIED STRESS BLOCKx d h
Aps C
0.45fcu T
d = 0.45x
= Apsfbp
SIMPLIFIED STRESS BLOCK
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

12. From the simplified stress block (rectangular section) T = C
Aps fpb = 0.45 fcu (0.9x)b
fpb = 0.4 fcu bx/Aps
 substituting x 
fpb = 0.4 fcu bd ___ εcu________
Aps (εcu + εpb - εpe - εe )
It seems that there are two unknowns fpb and εpb . Solve using simultaneous equation using appropriate stress-strain curve or trial and error to solve for εpb. The value of εpb will solve fpb.
Solve for x and finally
Mu = Aps fpb (d – dn) ….Equation 5 Clause
** For a rectangular section, dn = 0.45x
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

13. Example :
A bonded pre-stressed concrete beam having rectangular section 400 mm x mm as shown below. The tendon consist of 3300 mm2 of standard strands with fpu = 1770 N/mm2 , stressed to an effective pre-stress (Pe) 910 N/mm2. fcu = 60 N/mm2 with Ec = 36 kN/mm2. Es = 195 kN/mm2.
a) Develop the stress-strain diagram (taking γm = 1.05)- tendon
b) Calculate the ultimate moment of resistance.
400 mm
h/2
870mm
1200mm
270mm
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

14. b) A = 400 x 1200 = 4.8 x 105 mm2 Aps = 3300 mm2 I = 5.76 x 1010 mm4 Pe = 910 x 3300 /1000 = 3003 kN (KPo) e = 270 mm d = 870 mm εcu = εpe = KPo/(Aps Es) = 3003 x 1000 /(3300 x 195 x 1000) = εe = (K/Ec) x (Po/A + Poe2/I) = (1/[36 x 1000]) x (3003 x x 103 x 2702 ) 4.8x x 1010 =
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

15. x = dεcu / (εcu + εpb - εpe - εe )
= 870 x /( εpb – – )
= / (εpb – )
fpb = 0.4 fcu bd ____ εcu _______
Aps (εcu + εpb - εpe - εe )
= / (εpb – )
Locate two points that can fit the graph
εpb = 10 x fpb = 1036 N/mm2
εpb = 8 x fpb = 1352 N/mm2
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

16. Thus fpb = 1400 N/mm2 and εpb = 0.00785 x = 475 mm
x = 475 mm
Mu = Aps fpb (d – 0.45x)
= 3300 x 1400 (870 – 214) / 106
= 3031 kNm
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

17. Alternatively, BS 8110 allowed a simplified table ie Table 4
Alternatively, BS 8110 allowed a simplified table ie Table 4.4 to solve for the 2 unknowns. (fpu/fcu) x (Aps/bd) = (1770/60) x (3300/400 x870) = 0.28 fpe/fpu = 910/1770 = fpb/0.95fpu ≈ fpb ≈ 1423 x/d ≈ x ≈ 483 mm Mu = Aps fpb (d – 0.45x) = 3300 x 1423 (870 – 217) / 106 = 3066 kNm Comments BS 8110 simplified method should give a conservative value. Unlike this example, the Mu worked out from first principle may use a very rough stress-strain diagram
Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM

18. Dr. NORAZURA MUHAMAD BUNNORI (PhD), USM