Flexural Strength of Reinforced Masonry

Hardmasonryflexurereinforced

Use this simplified reinforced-masonry flexure model: a = A_s f_y/(0.8 f′_m b), M_n = A_s f_y (d − a/2), φ = 0.90, with yielding of the steel assumed. Section data: b = 200 mm, d = 400 mm, A_s = 400 mm², f_y = 400 MPa, f′_m = 14 MPa. What is the design flexural strength φM_n?

A58.29 kN·m
B47.31 kN·m
C52.46 kN·m
D53.49 kN·m

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Answer: C. 52.46 kN·m

a = (400)(400)/(0.8 · 14 · 200) = 160,000/2,240 = 71.43 mm. M_n = 160,000 (400 − 71.43/2) = 160,000 (364.29) = 58.29 kN·m. φM_n = 0.90 × 58.29 = 52.46 kN·m. 58.29 kN·m is the nominal strength with φ omitted. 47.31 kN·m uses (d − a) as if a were the full lever-arm drop, i.e. confusing a with the neutral-axis depth (0.90 · 160,000 · 328.57). 53.49 kN·m drops the 0.8 factor in the denominator (a = 57.14 mm).

a

= \frac{A _{s} f _{y}}{0.8 f _{m}^{'} b}

= 71.43

; ϕ M_{n}

= 0.90 (400) (400) (400 - \frac{71.43}{2})

= 52.46

kN·m

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Flexural Strength of Reinforced Masonry

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