Solutions to Reinforced Concrete Design: A Practical Approach (3rd Edition) by (Brzev, 2017) | All 13 Chapters Covered

ALL 13 CHAPTERS COVERED




SOLUTIONS MANUAL

,Chapter 1 - Solutions




1.1.

a) 2.4 kPa (NBC 2005 Table 4.1.5.3)

b) 2.4 kPa (NBC 2005 Table 4.1.5.3)

c) 1.9 kPa (NBC 2005 Table 4.1.5.3)

d) 7.2 kPa (stack rooms, NBC 2005 Table 4.1.5.3)



1.2.

NBC 2005 Table 4.1.5.3 prescribes occupancy live load of 2.4 kPa for classrooms with or without
fixed seats.

To determine the actual occupancy live load, add together the weight for all students in the class-
room and divide by the classroom plan area.



1.3.

a) Beam properties:

width b= 350mm

depth h= 700mm
3
unit weight w = 24kN  m

Load analysis:

- self-weight
kN kN
b  h  w= 0.35m  0.7m  24 ---- = 5.9--- -
m3 m




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, kN
DL  6.0----
m
- Live load LL= 15kN  m
kN
Total load w= DL + LL= 6.0 + 15.0= 21.0--- -
m




2
wl
Ms= --------- (simply supported beam, see Table A.16)
8
kN 2
21.0 --- (8.0 m )
m
= = 168kNm
8
Ms= 168kNm

b) wf = 1.25DL + 1.5LL (NBC 2005 Table 4.1.3.2)
kN
= 1.25  6.0 + 1.5  15.0 = 30.0----
m
kN 2
2 30 ---- (8.0m)
wf  l m
Mf = ----------- = = 240kNm
8 8
Mf = 240kNm



1.4.

a) DL = 10.0kPa

LL = 5.0kPa

Load on beam B1:

Tributary width s = 3.0m

wf = (1.25DL + 1.5LL)  s


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, = (1.25  10.0 + 1.5  5.0)  3.0

= 60kN  m




Load on girder G1:

Tributary width s= 8m

wf = (1.25DL + 1.5LL)  s
kN
= (1.25  10.0 + 1.5  5.0)  8.0= 160 ----
m




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