Page 161 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.25

From Eq. (4.43)

−
φ M = ⎡ ⎣ ′ 2 ω(1 0 625 ω)⎤ ⎦
φ f bd
.
n m
)
2
= 09 20.[ . (763. )(20) 012 1 0625( . )( − . x 012)]
.
= 609 8 k-in . = 50 82 k-ft
.
k
.
Alternatively, we could have determined fk from Table A13. For r = 0.004 and
n
′ f = 2000 psi, fk = 200 psi.
m n
From Eq. (4.47),
⎛ bd 2 ⎞ ⎛ (. ) ( 2
763 20) ⎞
φM = φk n ⎜ ⎟ = 200( ) ⎜ ⎟ = 50 87 k-ft
.
n ⎠ ⎝ 12 0000 ⎠
⎝12 000, ,
The above result is the same as obtained in Example 4.3.
The above calculation assumes the steel has yielded, and e ≥ 1.5e as shown in
y
s
Example 4.3 (calculations not repeated here).
b. Change in flexural strength when Grade 60 reinforcement is changed to Grade 40
reinforcement:
When the grade of reinforcing steel is changed from Grade 60 to 40, then the
yield strength f changes from 60 to 40 ksi. However, the value of the reinforce-
y
ment ratio, r (= 0.004) remains the same as calculated above. From Eq. (4.40),
f ⎛ 40 ⎞
ω = ρ y = 0 004 = 008.
.
′ f ⎝ 20. ⎠
m
From Eq. (4.43)
′
1
(
.
φM = 0 9.[ f bd 2 ω − 0 625 ω)]
n m
−
×
2 2
(
)
= 09 20 763 20 )( 008 1 0625 008.)]
.
.[
.
.)
(
(
.
= 417 5 k-in . = 34 8 k-fft
.
.
The above result is the same as obtained in Example 4.4.
The above calculation assumes the steel has yielded, and e ≥ 1.5e as shown
s
y
previous examples (calculations not repeated here).
The reduction in the design flexural strength of the beam when the reinforcement
grade is changed from Grade 60 to Grade 40:
(φM ) − (φM ) = 50 .8 34− .8 16= . k-ft0
n Gr 60 n Gr 40
% reduction = 16 0 . × 100 = 31 5 .%
u
50 8 .
Or, % increase when Grade 60 steel is used instead of Grade 40 steel:
% increase = 16 .0 × 100 = 46 %
34 .8
Example 4.8 For the beam described in Example 4.3, (a) determine the uniform
service load the beam can carry over a span of 10 ft 8 in. if the dead-to-live load ratio is
0.7, (b) uniform live load the beam can carry safely if the only dead load on the beam
is its own weight.