250+ TOP MCQs on Design of Post Tensioned Beams and Answers

Prestressed Concrete Structures Multiple Choice Questions on “Design of Post Tensioned Beams”.

1. Calculate ultimate moment and shear of effective span is 30m, live load is 9kn/m, dead load excluding self weight is 2kn/m, load factors for dead load is 1.4 for live load is 1.6 cube strength of concrete f_cu is 50n/mm² cube strength at transfer is f_ci is 35n/mm², tensile strength of concrete E_c is 34kn/mm² loss ratio ɳ is 0.85 and 8mm diameter high tensile strength f_pu is 1500n/mm² are available for use and the modulus of elasticity of high tensile wires is 200kn/mm²?
a) 340 and 450kn
b) 240 and 340kn
c) 140 and 240kn
d) 100 and 200kn
Answer: a
Clarification: W_min/W_ud = (50x2400x9.81×0.125x25x30/50×10⁶x0.85²) = 0.31
Ultimate load excluding the factored selfweight = (1.4×2)+(1.6×9) = 17.2kn/m, W_ud = 17.2/1-1.4×0.31 = 30KN/M, W_min = (0.31×30) = 9.3kn/m, Ultimate moment, M_u = (0.125x30x30²) = 3400knm, Ultimate shear, V_u = (0.5x30x30) = 450kn.

2. Find cross-sectional dimensions thickness of web if h_f/d ratio is 0.23 and b_w/b ratio is 0.25 and b is 0.5d?
a) 100mm
b) 110mm
c) 120mm
d) 30mm
Answer: c
Clarification: h_f/d =0.23 and b_w/b = 0.25 and b = 0.5d,
M_u = 0.10f_cubd² d = (3400×10⁶/0.10x50x0.5)^1/3 = 1130mm, h = (1130/0.85) = 1300, b = 600mm, h_f = (0.2×1130) = 250mm, adopt an effective depth, d = 1150mm, thickness of web, b_w = (0.6vu/fth) = (0.6x450x10³/1.7×1300) = 120mm.

3. Calculate working moment if design working load is 19.8kn/m covered over a span of 30m (actual self weight of girder is 8.8kn/m)?
a) 3000
b) 2000
c) 4340
d) 2230
Answer: d
Clarification: Actual self weight of the beam and the girder = 8.8kn/m, span = 30m
Minimum moment M_min = 990knm, Design working load = 19.8kn/m,
Working moment M_d = (0.125×19.8×30²) = 2230knm.

4. Find the Permissible stresses and range of stresses for class 1 structure f_cu = 50n/mm², f_ck = 35n/mm² according to BS: 8110 recommendations for f_cu = 50n/mm² and f_ci = 35n/mm²,f_ct = 0.5f_ci = 17.5n/mm²?
a) 16.5n/mm²
b) 12.56n/mm²
c) 13.56n/mm²
d) 12.00n/mm²
Answer: a
Clarification: f_cu = 50n/mm²,f_ck = 35n/mm² according to BS: 8110 recommendations for f_cu = 50n/mm² and f_ci = 35n/mm², f_ct = 0.5f_ci = 17.5n/mm²For class 1 structure f_u = h_tw = 0, f_br = (ɳf_ct-ftw) = (0.85×17.5) = 15n/mm², f_cw = 0.33f_cu = (0.33×50) = 16.5n/mm², f_cu = (fcw-ɳfu) = 16.5n/mm².

5. Find prestressing force if area is 36.75mm² of eccentricity 580 given f_inf is 26.5kn/m and z_b is 99×10⁶?
a) 405
b) 308
c) 453
d) 206
Answer: b
Clarification: Area = 36.75mm², e = 580, f_inf = 26.5kn/m, z_b = 99×10⁶
p = (Af_infZ_b/Z_b+A_e) =(367500×26.5x99x10⁶/(99×10⁶)+(367500×580)) = 308x104kn/m².

6. Find force in cable using Freyssinet cables 12-8mm diameter and stressed to 1100n/mm² of eccentricity 50 and the prestressing force is given as 1000n/mm²?
a) 660kn
b) 234kn
c) 300kn
d) 230kn
Answer: a
Clarification: 12 diameter, stress = 1100n/mm², e = 50, prestressing force = 1000n/mm²
Force in each cable = ((12x50x1100)/1000)) = 660kn.

7. Find ratio for ultimate flexural strength at the centre – span section given that A_ps = 3000mm², d = 1150mm, f_cu = 50n/mm², b_w = 150mm, f_pu = 1500n/mm², b = 600mm, h_t = 250mm, design ultimate moment m_ud = 3400knm?
a) 9.5
b) 0.23
c) 6.7
d) 3.4
Answer: b
Clarification: A_ps = 3000mm², d = 1150mm, f_cu = 50n/mm², b_w = 150mm, f_pu = 1500n/mm², b = 600mm, h_t = 250mm, design ultimate moment m_ud = 3400knm, according to BS: 8110-1985, _Aps = (A_pw+A_pf) = A_pf = 0.45×50(600-150)(250/1500) = 0.45xf_cu(b-b_w)(h_f/f_pu) = 1680mm², A_pw = (1300-1680) = 1320mm², ratio(f_puA_pw/f_cub_wd) = (1500×1320/50x150x1150) = 0.23.

8. Calculate the slope of cable section at support uncracked in flexure given that eccentricity is 410, length is 30m and stress induced is 1000?
a) 0.0547
b) 2.456
c) 0.0234
d) 0.0123
Answer: a
Clarification: e = 410, length = 30m, stress induced = 1000
Slope of cable θ = (4e/l) = ((4×410)/(30×1000)) = 0.0547.

9. Calculate the span section cracked in flexure (M=M0) F_cp = 23.4n/mm², z_b is 99×106 and stress induced is 1000?
a) 1200kn
b) 1850kn
c) 2300kn
d) 4300kn
Answer: b
Clarification: F_cp = 23.4n/mm², zb is 99×10⁶, stress is 1000
m₀ = (0.8f_cpZ_b) = (0.8 x 23.4 x (99×10⁶/1000)) = 1850knm.

10. Find resultant maximum long term deflection if ϕ is 2.6, α_y is 38.5mm, α_g is 46mm, α_p is 74.7mm?
a) 95mm
b) 35mm
c) 55mm
d) 20mm
Answer: a
Clarification: E_ce = (E_c/1+ϕ) = (E_c/2.6), ϕ = 2.6, α_y = 38.5mm, α_g = 46mm, α_p = 74.7mm, resultant maximum long term deflection = (2.6×46)+38.5-(0.85×74.7) = 95mm which is less than the code limit (span/250) = 120mm, ɳ = 0.85.