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Exam (elaborations)

Solution Manual for Reinforced Concrete Design: A Practical Approach (3rd Edition) by Svetlana Brzev and James K. Wight

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Enhance your concrete design skills with the Solution Manual for Reinforced Concrete Design: A Practical Approach (3rd Edition) by Svetlana Brzev and James K. Wight. This comprehensive manual offers complete, step-by-step solutions and clear explanations for all textbook problems, covering design principles, flexural and shear behavior, columns, slabs, and seismic detailing based on current code standards. Ideal for civil and structural engineering students, it bridges theory and practical application in reinforced concrete design.

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Institution
Reinforced Concrete Design
Course
Reinforced Concrete Design

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Uploaded on
October 21, 2025
Number of pages
435
Written in
2025/2026
Type
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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




@seismicisolation
1-1Copyright © 2006 Pearson Education Canada Inc.@seismicisolation

, 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


@seismicisolation
1-1Copyright © 2006 Pearson Education Canada Inc.@seismicisolation

, = (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




@seismicisolation
1-3Copyright © 2006 Pearson Education Canada Inc.@seismicisolation

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