Insstructor’s Manual to accompany Statics and Strength of Materials For Architecture and Building Construction Fourth Edition
Onouye Upper Saddle River, New Jersey Columbus, Ohio Statics and Strength of Materials for Architecture and Building Construction 4th Edition Onouye Solutions Manual Full Download: http://testbanktip.com/download/statics-and-strength-of-materials-for-architecture-and-building-construction-4th-edition-onouye-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
Barry S.
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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
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Instructor’s
Statics and Strength of Materials
Fourth Edition
Pearson/Prentice Hall
Manual to Accompany
For Architecture and Building Construction
Barry Onouye
Seattle Public Library Architect: Rem Koolhaus
Upper Saddle River, New Jersey Columbus, Ohio
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 variety 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 students over the years, there are inevitably errors that remain to be discovered 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 Instructor’s Manual but I have personally found that my students appreciated having it as a study guide.
Fall, 2010
Barry Onouye, Senior Lecturer Dept. of Architecture College of Built Environments University of Washington
e-mail:
barryo@u.washington.edu
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
Table of Contents Chapter
3
of Determinate
• 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
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
Analysis
Systems
Chapter
Chapter 9 Column Analysis and Design
Chapter 10 Structural Connections
• 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
• Bolted steel connections pg 10.1 – 10.3 • Framed connections pg 10.3 • Welded connections pg 10.4 – 10.5
Chapter 2 Problem Solutions
2.1
2.1 40˚ 20˚ 50˚ x y R=173# Fa=100# Fb=200# 40˚ 20˚ 50˚ x y R=173# Fa=100# Fb=200# or Prob. 2.2 R = 173# θ = 50˚ from horiz. φ = 40˚ from vert. R = 10.2kN θ = 5˚ 35˚ 10˚ 30˚ F1 = 3kN F2 =6kN F3 =5kN x y 2.2 40˚ 20˚ 50˚ x y R=173# Fa=100# Fb=200# 40˚ 20˚ 50˚ x y R=173# Fa=100# Fb=200# or Prob. 2.2 R = 173# θ = 50˚ from horiz. φ = 40˚ from vert. R = 10.2kN θ = 5˚ 35˚ 10˚ 30˚ F1 = 3kN F2 =6kN F3 =5kN x y 2.3 F1=600# F2=720# R = 930# 40˚ 40˚ x y Prob. 2.3 Prob. 2.4 3 4 3 4 5 12 T1=5000# T 2 = 5910# R = 10,000# T3 = 910# 2.4 F1=600# F2=720# R = 930# 40˚ 40˚ x y Prob. 2.3 Prob. 2.4 3 4 3 4 5 12 T1=5000# T 2 = 5910# R = 10,000# T3 = 910#
2.2 2.5 T1 T2 =240lb. 20° 10° W = 1,200 lb. x y Prob. 2.5 2.6 € sin θ = 3 5 and cos θ = 4 5 ∴ Fx = F cos θ = 1000 # ( ) 4 5 ( ) = 800 # Fy = F sin θ = 1000 # ( ) 3 5 ( ) = 600 # € ∴ Fx = 4 5 F = 4 5 1000 # ( ) = 800 # Fy = 3 5 F = 3 5 1000 # ( ) = 600 # € Fx 4 = Fy 3 = F 5 By similar triangles: Prob. 2.6 3 4 F=1000 lb. A B 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
Problem Solutions: 2.3.11-2.3.15
solution using the tip-to-tail method
2.3 2.7 Prob. 2.7 y x T 10° Tx Ty x y T 10° Tx Ty € Tx = T sin 10° Ty = T cos 10° ∴ T = Ty cos 10° = 250 N 0.985 = 254 N 2.8
3 4 F=1000 lb A B x Prob. 2.3.11 Fy Fx y x T 10° Prob 2 3 12 Tx Ty 4 12 θ P=300 lb. Rafter Purlin Detail Prob. 2.3.13 θ Py Px 9 12 16' 16' P = 600 lb Q = 600 lb. A B Prob. 2.3.14 Py Px Qy Qx 3 4 3 4 € θ = tan 1 4 12 ( ) = 18 43° € Px = P 4 12 65 ( ) = 300 # ( ) 0 316 ( ) = 94 9 # Py = P 12 12 65 ( ) = 300 # ( ) 0 949 ( ) = 285# 2.9 Graphical
30 1 1 y F =10k 1 F =12k 2 F =18k 3 x O x y R=7.03k Rx = +5.73 k R y = -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 1 1 y F =10k 1 F =12k 2 F =18k 3 x O x y R=7.03k Rx = +5.73 k R y = -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 1 1 y F =10k 1 F =12k 2 F =18k 3 x O x y R=7.03k Rx = +5.73 k R y = -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 cos 30° = 10k 0.866( ) = 8.66k F1x = + F1 sin 30° = 10k 0 50 ( ) = 5k F2 = F2 x = 12k F3 x = + 1 2 F3( ) = + 18k 2 F3 y = 1 2 F3( ) = 18k 2 € R x = ΣF x = +5k 12k + 18k 2 = +5.73k R y = ΣF y = +8.66k 18k 2 = 4.07k € tan θ = R y R x = 4.07 5.73 = 0 710 θ = tan 1 0 710 ( ) = 35 4° from horizontal € sin θ = R y R R = R y sin θ = R y sin 35.4° ∴ R = 4.07k 0.579( ) = 7 03k
Statics and Strength of Materials for Architecture and Building Construction 4th Edition Onouye Solutions Manual
Full Download: http://testbanktip.com/download/statics-and-strength-of-materials-for-architecture-and-building-construction-4th-edition-onouye-solutions-manual/
2.10 cont’d
Graphical Solution:
2.4 2.10 2.10 x y TABx TABy TAB=600N TACx TAC=800N TACy 60˚ 40˚ x y 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˚ € TAC x = TAC cos 60° = 0 5TAC TACy = TAC sin 60° = 0.866 TAC + TAB x = + TAB cos 40° = +0.766 TAB TAB y = TAB sin 40° = 0.642 TAB € R x = ΣFx = 0.5 ( ) 800 N ( ) + 0.766 ( ) 600 N ( ) = 59.6 N R y = ΣFy = 0 866 ( ) 800 N ( ) 0 642 ( ) 600 N ( ) = 1078 N € θ = tan 1 R y R x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = tan 1 1078 59 6 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = tan 1 18 1 ( ) = 86 8° φ = tan 1 R x R y ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = tan 1 59.6 1078 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = tan 1 0.055 ( ) = 3.2° R = 59 6 2 + 1078 2 = 1079 N x y TABx TABy TAB=600N TACx TAC=800N TACy 60˚ 40˚ x y 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˚ x y 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˚
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