250+ TOP MCQs on Design of Pretensioned Beams and Answers

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

1. Design a pretensioned roof pull-in to suit the data F_cu, concrete cube strength = 50n/mm², effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, Dc = 2400, and determine ultimate moment and shear?
a) 42 and 27.75
b) 54 and 27.75
c) 34 and 27.75
d) 20 and 28
Answer: a
Clarification: W_min/W_ud = KD_cgβ(L/h)L/f_cu(d/h)² = 7.5×2400 x 9.81 x 0.125×25 x 6/50x 10⁶x(0.85)² = 0.094
F_cu, concrete cube strength = 50n/mm², effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, D_c = 2400, f_cu = 50n/mm², w_min = (0.094)(9.25) = 0.86kn/m, m_u = (0.125×9.25×6²) = 42knm, v_u = (0.5×9.25×6) = 27.75kn.

2. Design cross sectional dimensions of pretensioned roof pull given that b is 0.5d?
a) 250
b) 260
c) 270
d) 280
Answer: c
Clarification: M_u = 0.10f_cubd² and if b = 0.5d
D = (42×10⁶x2/0.10×50)^1/3 = 270mm.

3. Find the approximate thickness of web if b is 0.5d, d is 270mm, d/h ratio is 0.85, h is 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent?
a) 45mm
b) 43mm
c) 41mm
d) 42mm
Answer: b
Clarification: b = 0.5d, d = 270mm, d/h = 0.85, h = 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent:
Thickness of flange = (0.2×275) =55mm Approximate thickness of web = (0.85v_u/f_th) = (0.85×27.75×10³/1.7×320) = 43mm.

4. Find minimum range of stresses if f_ct is 15n/mm², f_cw is 17, f_tw is zero, f_u is -1n/mm², ɳ is 0.8?
a) 12 and 18n/mm²
b) 13 and 14n/mm²
c) 12 and 15n/mm²
d) 10 and 16n/mm²
Answer: a
Clarification: Range of stress f_br = (ɳf_ct-f_cw) = (0.85×15-0) = 12n/mm², f_tr = (f_cw – ɳ_fu) = (17-0.8x(-1)) = 17.8n/mm², f_ct = 15n/mm², f_cw = 17, f_tw = 0, f_u = -1n/mm², ɳ = 0.8.

5. Find minimum section modulus given data is mg is 3.86×10⁶, m_q is 22.50×10⁶, f_br is given as 12 and the loss ratio is 0.8?
a) 134×10⁴
b) 182×10⁴
c) 123×10⁴
d) 120×10⁴
Answer: b
Clarification: m_g = 3.86×10⁶, m_q = 22.50×10⁶, f_br = 12, loss ratio = 0.8
Z_b > or equal (m_q+(1-ɳ)m_g/f_br) > or equal ((22.50×10⁶)+(1-0.8)3.86×10⁶)/12)
Greater than equal to 182x104mm³.

6. Find the supporting force if given characteristic strength is -1, moment of gravity is 3.86×10⁶, z_t = 230×10⁴?
a) -2.68n/mm²
b) -3.45n/mm²
c) -1.23n/mm²
d) 13.56n/mm²
Answer: a
Clarification: p = (A(f_infZ_b+f_subZ_t)/Z_t+Z_b)
F_inf = ((f_tw/ɳ+(m_q+m_g)/ɳz_b)) = (0+ (26.36×10⁶/0.8x230x10⁴))
F_sup = (f_u – m_g/z_t) = (-1 – (3.86×10⁶)/(230×10⁴)) = -2.68n/mm².

7. Check for ultimate flexural strength if given A_ps is 154mm², f_pu is 1600n/mm², b is 160mm, f_cu is 50n/mm²and diameter is 265mm?
a) 9.65
b) 0.116
c) 3.442
d) 2.345
Answer: b
Clarification: A_ps = (38.5xy) = 154mm², f_pu = 1600n/mm², b = 160mm, f_cu 50n/mm², d = 265mm
(A_psf_pu/bdf_cu) = (154×1600/160x265x50) = 0.116.

8. Find ultimate shear strength (check it for safe against shear failure) if v_u is 27.75kn, Loss ratio is 0.8, prestressing force is 182000, area is 31400, breadth is 50 where height is 320, prestressing force is 1.7, f_cp = 4.65, f_t is 1.7?
a) Safe
b) Unsafe
c) Zero
d) Collapse
Answer: a
Clarification: F_cp = (ɳp/A) = (0.8×182000/31400) = 4.65n/mm²
V_cw = 0.67bh(f₁²+0.8f_cpf_t)^1/2 = (0.67x50x320(1.72+0.8×4.65×1.7)1/2/10³) = 33.2kn
V_cw > V_u hence safe against shear failure.

9. Check for deflection due to prestressing force if given data is Prestressing force is 182×10³ eccentricity of cable is 10⁵, Length of the cable is 1000, elastic modulus of concrete is 34×10³, Moment of inertia is 3200×10⁵?
a) 9.4
b) 4.5
c) 6.8
d) 9.8
Answer: c
Clarification: P = 182×10³ e = 10⁵, L = 1000, elastic modulus of concrete = 34×10³, I = 3200×10⁵
PeL²/8E_cI = (182×10³x10⁵x6²x1000²/8x34x10³x3200x10⁵) = 6.8mm.

10. Find the deflection due to self weight given that ϕ = 1.6, E_e = 2.6E_ce, elastic modulus of concrete is 34×10³, gravity is given as 6, self weight is 0.76, Length of the cable is 1000, elastic modulus of concrete is 34×10³ , Moment of inertia is 3200×10⁵?
a) 1.66mm
b) 5.3mm
c) 23.4mm
d) 1.02mm
Answer: d
Clarification: E_ce = E_c/1+ϕ, ϕ = 1.6, E_e = 2.6E_ce
Deflection due to self weight g = (5gL⁴/384E_cI) = (5×0.76×6⁴x1000⁴/384x34x10³x3700x10⁵) = 1.02mm.