Insstructor’s Manual to accompany

Statics and Strength of Materials For

Architecture and Building Construction

Fourth Edition

Barry S. Onouye

Upper Saddle River, New Jersey Columbus, Ohio

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10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-511455-1 ISBN-10: 0-13-511455-1

Instructor’s Manual to Accompany

For Architecture and Building Construction

Fourth Edition

Barry Onouye

Seattle Public Library Architect: Rem Koolhaus

Pearson/Prentice HallUpper Saddle River, New Jersey Columbus, Ohio

Statics and Strength of Materials

Preface

This Instructor’s Manual is intended to accompany Statics and Strength of Materials for Architecture and Building Construction. It was initially developed as a study guide for students to practice on a va-riety of problems to enhance their understanding of the principles covered in the text. Solutions were developed in sufficient detail to allow students to use these problems as additional example problems.

Although the problem solutions contained in this Instructor’s Manual have been worked, re-worked, checked and scrutinized by my many stu-dents over the years, there are inevitably errors that remain to be discov-ered by others using the book. If you detect discrepancies, omissions and errors as you work through these problems, I would appreciate hearing from you so that I can incorporate the changes for any future editions of the Instructor’s Manual or book.

I realize that many instructors do not allow student’s access to the Instruc-tor’s Manual but I have personally found that my students appreciated having it as a study guide.

Fall, 2010

Barry Onouye, Senior LecturerDept. of ArchitectureCollege of Built EnvironmentsUniversity of Washingtone-mail: barryo@u.washington.edu

Table of Contents

Chapter 2 Statics • Graphical addition of vectors pg 2.1 - 2.2 • Resolution of forces: x and y components pg 2.2 - 2.3 • Vector addition by components pg 2.3 - 2.6 • Moment of a force pg 2.6 - 2.7 • Varignon’s theorem pg 2.7 – 2.8 • Moment couples pg 2.9 • Equilibrium of concurrent forces pg 2.10 – 2.13 • Equilibrium of rigid bodies pg. 2.13 – 2.16 • Supplementary problems pg 2.16 – 2.26

Chapter 3 Analysis of Determinate Systems • Cables with concentrated loads pg 3.1 – 3.3 • Equilibrium of rigid bodies with distributed loads pg 3.4 – 3.5 • Planar trusses – method of joints pg 3.6 – 3.8 • Truss analysis – method of sections pg 3.8 – 3.10 • Diagonal tension counters pg 3.10 – 3.12 • Zero-force members pg 3.12 • Pinned frames – multi-force members pg 3.13 – 3.15 • Supplementary problems pg 3.16 – 3.28 • Retaining walls pg 3.29 – 3.32

Chapter 4 Load Tracing • Gravity load trace pg 4.1 – 4.8 • Lateral load trace pg 4.8 – 4.11

Chapter 5 Strength of Materials • Tension, Compression and shear stress pg 5.1 – 5.2 • Deformation and strain pg 5.3 • Elasticity, strength and deformation pg 5.3 – 5.4 • Thermal stress and deformation pg. 5.4 – 5.5 • Statically indeterminate, axially loaded members pg 5.5 – 5.6

Chapter 6 Cross-Sectional Properties • Centroids pg 6.1 – 6.3 • Moment of inertia pg 6.3 – 6.7 • Moment of inertia for composite sections pg 6.7 – 6.9

Chapter 7 Bending and Shear Diagrams • Equilibrium method for shear and moment diagrams pg 7.1 – 7.4 • Semi-graphical method for shear and moment diagrams pg 7.5 – 7.10

Chapter 8 Bending and Shear Stress in Beams • Bending stress pg 8.1 – 8.5 • Bending and shear stresses pg 8.6 – 8.12 • Deflection in Beams pg 8.13 – 8.15

Chapter 9 Column Analysis and Design • Euler buckling loads and stresses pg 9.1 – 9.2 • Axially loaded steel columns - analysis pg 9.3 – 9.4 • Design of steel columns pg 9.5 – 9.6 • Axially load wood columns pg 9.6 – 9.9

Chapter 10 Structural Connections • Bolted steel connections pg 10.1 – 10.3 • Framed connections pg 10.3 • Welded connections pg 10.4 – 10.5

2.1

Chapter 2 Problem Solutions

2.1

40˚

20˚

50˚ x

y

R = 17

3#Fa = 100#

Fb = 200#

40˚

20˚50˚

x

y

R = 17

3#Fa = 100#

Fb = 200#

or

Prob. 2.2

R = 173#θ = 50˚ from horiz.φ = 40˚ from vert.

R = 10.2kNθ = 5˚

35˚10˚

30˚F1 = 3kN

F2 =6kN

F3 =5kN

x

y

2.2

40˚

20˚

50˚ x

y

R = 17

3#Fa = 100#

Fb = 200#

40˚

20˚50˚

x

y

R = 17

3#

Fa = 100#

Fb = 200#

or

Prob. 2.2

R = 173#θ = 50˚ from horiz.φ = 40˚ from vert.

R = 10.2kNθ = 5˚

35˚10˚

30˚F1 = 3kN

F2 =6kN

F3 =5kN

x

y

2.3

F1 = 600#

F2 = 72

0#

R =

930

#

40˚

40˚

x

yProb. 2.3

Prob. 2.43

4

3

4

5

12

T1 = 50

00#

T2 = 5910#

R =

10,

000#

T3 = 910#

2.4

F1 = 600#

F2 = 72

0#

R =

930

#

40˚

40˚

x

yProb. 2.3

Prob. 2.43

4

3

4

5

12

T1 = 50

00#

T2 = 5910#

R =

10,

000#

T3 = 910#

2.2

2.5

T1

T2 = 240 lb.

20°

10°

W =

1,200 lb.

x

y

Prob. 2.5

2.6

€

sinθ =35

and cosθ =45

∴ Fx = F cosθ = 1000#( ) 45( ) = 800#

Fy = F sinθ = 1000#( ) 35( ) = 600#

€

∴ Fx = 45 F = 4

5 1000#( ) = 800#

Fy = 35 F = 3

5 1000#( ) = 600#€

Fx4

=Fy

3=

F5

By similar triangles:

Prob. 2.6

3

4

F=1000 lb.

AB

x

Fy

Fx

Prob. 2.7

3

4

F=1000 lb.

θ

5

5

y

x

T

10°

Tx

Ty

x

y

T

10°

Tx

Ty

2.3

2.7

Prob. 2.6

3

4

F=1000 lb.

AB

x

Fy

Fx

Prob. 2.7

3

4

F=1000 lb.

θ

5

5

y

x

T

10°

Tx

Ty

x

y

T

10°

Tx

Ty

€

Tx = T sin10°

Ty = T cos10°

∴ T =Ty

cos10°=

250N0.985

= 254N

2.8

Problem Solutions: 2.3.11-2.3.15

3

4

F=1000 lb.

AB

x

Prob. 2.3.11

Fy

Fx

y

x

T

10°

Prob. 2.3.12

Tx

Ty

412

θ

P=300 lb.

Rafter

Purlin Detail

Prob. 2.3.13

θPy

Px

9

12

16' 16'

P = 600 lb.

Q = 600 lb.

AB

Prob. 2.3.14

PyPx

Qy

Qx

3

4

34

x

3

4

y

12

5

F=2500N

Inclinedrafter

Problem 2.3.15

Fy

Fx

α

β

φ

α

90∞

€

θ = tan−1 412( ) = 18.43°

€

Px = P 412.65( ) = 300#( ) 0.316( ) = 94.9#

Py = P 1212.65( ) = 300#( ) 0.949( ) = 285#

2.9

Graphical solution using the tip-to-tail method

30

11

y F =10k1

F =12k2

F =18k3

xO

x

y

R = 7.03 k

Rx = +5.73 k

Ry

= -4

.07

k

Resultant

θ = 35.1°

1

1

30 F1 =

10k

F2 = 12k

F3 = 18k

R = 7.03k

x

y

θ = 35.1°

30

11

y F =10k1

F =12k2

F =18k3

xO

x

y

R = 7.03 k

Rx = +5.73 k

Ry

= -4

.07

k

Resultant

θ = 35.1°

1

1

30 F1 =

10k

F2 = 12k

F3 = 18k

R = 7.03k

x

y

θ = 35.1°

30

11

y F =10k1

F =12k2

F =18k3

xO

x

y

R = 7.03 k

Rx = +5.73 k

Ry

= -4

.07

k

Resultant

θ = 35.1°

1

1

30 F1 =

10k

F2 = 12k

F3 = 18k

R = 7.03k

x

y

θ = 35.1°

€

F1y = +F1 cos30° =10k 0.866( ) = 8.66k

F1x = +F1 sin30° =10k 0.50( ) = 5kF2 = −F2x = −12k

F3x = +12

F3( ) = +18k

2

F3y = −12

F3( ) = −18k

2

€

Rx = ΣFx = +5k −12k +18k

2= +5.73k

Ry = ΣFy = +8.66k −18k

2= −4.07k

€

tanθ =Ry

Rx

=4.075.73

= 0.710

θ = tan−1 0.710( ) = 35.4° from horizontal

€

sinθ =Ry

R

R =Ry

sinθ=

Ry

sin35.4°

∴ R =4.07k0.579( )

= 7.03k

2.4

2.10

2.10

x

y

TABx

TABy

TAB=600N

TACx

TAC=800N

TACy

60˚ 40˚

xy

TAC=

800N

TAB=600N

60˚ 40˚

R = 1079N

φ = 3.2˚

Scale 1mm = 10N

φ

x

y

θ

Rx = 59.6N

Ry = 1078N R = 1079N

φ = 3.2˚

θ = 86.8˚ €

−TACx= −TAC cos 60° = −0.5TAC

−TACy = −TAC sin 60° = −0.866TAC

+TABx= +TAB cos 40° = +0.766TAB

−TABy= −TAB sin 40° = −0.642TAB

€

Rx = ΣFx = − 0.5( ) 800N( ) + 0.766( ) 600N( ) = 59.6N

Ry = ΣFy = − 0.866( ) 800N( ) − 0.642( ) 600N( ) = −1078N

€

θ = tan−1 Ry

Rx

⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 1078

59.6⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 18.1( ) = 86.8°

φ = tan−1 RxRy

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = tan−1 59.6

1078⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 0.055( ) = 3.2°

R = 59.62 +10782 = 1079N

2.10

x

y

TABx

TABy

TAB=600N

TACx

TAC=800N

TACy

60˚ 40˚

xy

TAC=

800N

TAB=600N

60˚ 40˚

R = 1079N

φ = 3.2˚

Scale 1mm = 10N

φ

x

y

θ

Rx = 59.6N

Ry = 1078N R = 1079N

φ = 3.2˚

θ = 86.8˚

2.10

x

y

TABx

TABy

TAB=600N

TACx

TAC=800N

TACy

60˚ 40˚

xy

TAC=

800N

TAB=600N

60˚ 40˚

R = 1079N

φ = 3.2˚

Scale 1mm = 10N

φ

x

y

θ

Rx = 59.6N

Ry = 1078N R = 1079N

φ = 3.2˚

θ = 86.8˚

2.10 cont’d

Graphical Solution:

Statics and Strength of Materials for Architecture and Building Construction 4th Edition Onouye Solutions ManualFull Download: https://alibabadownload.com/product/statics-and-strength-of-materials-for-architecture-and-building-construction-4th-edition-onouye-solutions-manual/

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