Solutions Manual for Reinforced Concrete Design: A Practical Approach (3rd Edition) by Svetlana Brzev

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 350mm
b=

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

Load analysis:

- self-weight
m
b h w= 0.35m 0.7m -------
kN
24 =

,kN m
----
5.9




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, kN
DL 6.0 m----

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




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

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




1.4.

a) DL = 10.0kPa

LL = 5.0kPa

Load on beam B1:

Tributary width = 3.0m
s

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
= 1.25 10.0 + 1.5 kN
160 ----
5.0 8.0= m




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,b) Tributary area for the column C1




2
A = 8m 9m = 72m

Factored axial load

Pf = 1.25DL + 1.5LL A

= 1.25 10.0 + 1.5 5.0 72.0 = 1 440kN




1.5.




Cross sectional area
A= 1100 100 + 300 600= 290 000mm 2 = 0.29m 2




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,Unit weight 3
= 24.0kN m
w

Beam self-weight:
2 kN kN
kN
w = A w = 0.29m 24.0 ---- = 6.96 ---- 7.0 ----
3 m m
m
Hence,
kN
w = ------
7.0 m



1.6.

a) LL = 2.4kPa (Garage for passenger cars, NBC 2005 Table 4.1.5.3)

b) A typical T-beam

- Tributary s = 4500mm = 4.5m
width

- Cross sectional area
6 2 2
A = 4500 200 + 700 450 = 1.215 10 mm = 1.2m
kN
- Unit weight = 24 ------
w m3




- Beam self-weight
2 kN kN
kN
w = A w = 1.2m 24.0 ---- = 28.8 ---- 29.0 -----
m3 m m
c) Load analysis for a typical T-beam


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,- Self kN
= 29.0 -----
weight m
- Superimposed dead 0.3kPa s = 0.3kPa 4.5m = kN
1.35 ----
load m


kN
DL 30.4 -------
m
- Live load kN kN
2.4kPa s = 2.4kPa 4.5m = 10.8 11
LL=
----- -----
m m
Total factored load (NBC 2005 Table 4.1.3.2):
= = = --kN
--
Case 1:
wf 1.4DL 1.4 30.4 42.6
m
Case 2: wf = 1.25DL + 1.5LL

= 1.25 30.4 + 1.5 11.0= 54.5kN m

It can be concluded that Case 2 load combination gives larger value and it governs, that is,
= -kN
---
wf 54.5 .
m


1.7.

PD = 1000kN
PL =
PS = 800kN

500kN
Consider all possible load combinations from NBC 2005 Table 4.1.3.2.

Case 1: P f = 1.4PD = 1.4 1000 = 1400kN
Pf = 1.25PD + 1.5PL + 0.5PS
Case 2:
= 1.25 1000 + 1.5 800 + 0.5 500

= 2700kN

Case 3: P f = 1.25PD + 1.5PS + 0.5PL




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, = 1.25 1000 + 1.5 500 + 0.5 800

= 2400kN

It can be concluded that Case 2 governs, that is,

Pf = 2700kN .




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