Page 188 - Design of Reinforced Masonry Structures

4.52 CHAPTER FOUR

4.8.2 Determination of A When the Beam Width b
S
and Depth d are Known: Direct Solution
Instead of using the iterative procedure described in the preceding section, it is possible to
develop a procedure for direct determination of the required area of reinforcement if the
beam width b and the depth d are known. This can be done by combining Eqs. (4.9) and
(4.13) (repeated as follows):
Af
a = sy (4.9 repeated)
.
080 fb ′
m
⎛ a⎞
M = A f d − ⎟ (4.13 repeated)
s y ⎜
n ⎝ ⎠ 2
Substitution of Eq. (4.9) into (4.13) and multiplying both sides with φ yields
⎡ ⎛ Af ⎞ ⎤
φ
M = ⎢ A f d − s y ⎟ ⎥ (4.79)
s y ⎜
u ⎝ fb ′ ⎠
.
⎣ 16 m ⎦
2

Equation (4.79) can be rearranged as a quadratic in A (of the form Ax + Bx + C = 0):
s
⎛ φ f 2 ⎞
y
2
)
⎜ ⎝ 16 fb ′ ⎠ ⎟ A − ( φ f d A + M = 0 (4.80)
y
s
s
u
.
m
Equation (4.80) can be simplified and expressed as Eq. (4.81):
⎛ f ′ ⎞ ⎛ 16 . bf ′ ⎞ ⎛ M ⎞
2
s ⎜
16
m
A − . bd ⎜ m ⎟ A + 2 ⎟ ⎜ u ⎟ = 0 (4.81)
s
⎝ f ⎠ ⎝ f y ⎠ ⎝ ⎝ φ ⎠
y
With the known values of ′ f , f , b, d, and M , either Eqs. (4.80) or (4.81) can be
m
u
y
solved as quadratic equations for the required value of A . This procedure is illustrated in
s
Example 4.17. Note that both these equations contain the strength reduction factor f. Its
value can be taken as 0.9 for solving these equations, but must be verified later based on
the value of inelastic strain in the tension reinforcement (e ≥ 1.5e ).
s
y
Example 4.17 Solve Example 4.16 by Eq. (4.81).
Solution
Equation (4.80) is
⎛ f ′ ⎞ ⎛ bf ′ ⎞ ⎛ M ⎞
m
2
A − .( bd) ⎜ m ⎟ A + 16 . 2 ⎟ ⎜ u ⎟ = 0 (4.81 repeated)
s ⎜
16
s
⎝ f ⎠ ⎝ f y ⎠ ⎝ φ φ ⎠
y
Given: b = 7.63 in., d = 20 in., ′ f = 4000 psi, f = 60 ksi, M = 63.05 k-ft = 756.6 k-in.
m
y
u
(calculated in Example 4.16), φ = 09. (assumed, verified in Example 4.16)
Calculate the coefficients in Eq. (4.80).
⎛ f ′ ⎞ ⎛ 40 ⎞
.
.
.(
.
(
16.(bd ) ⎜ m ⎟ = 16 763 20) = 16628
)
⎝ f ⎠ ⎝ 60 ⎠
y
⎛ 16bf ′ ⎞ ⎛ M ⎞ 16 763 4.0 756 6) ( . )
(
.
.
.
(
0
)
m
.
⎜ 2 ⎟ ⎜ u ⎟ = = 11 4
⎝ f y ⎠ ⎝ φ ⎠ ( 60) 2 09
.