Statics and Strength of Materials For Architecture and Building Construction

Transcription

1 Insstructor s Manual to accompan Statics and Strength of Materials For rchitecture and Building Construction Fourth Edition Barr S. Onoue Upper Saddle River, New Jerse Columbus, Ohio

2 Copright 2012 b Pearson Education, Inc., Upper Saddle River, New Jerse 078. Pearson Prentice Hall. ll rights reserved. Printed in the United States of merica. This publication is protected b Copright and permission should be obtained from the publisher prior to an prohibited reproduction, storage in a retrieval sstem, or transmission in an form or b an means, electronic, mechanical, photocoping, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice Hall is a trademark of Pearson Education, Inc. Pearson is a registered trademark of Pearson plc Prentice Hall is a registered trademark of Pearson Education, Inc. Instructors of classes using Onoue, Statics and Strength of Materials for rchitecture and Building Construction, Fourth Edition, ma reproduce material from the instructor s manual for classroom use ISBN-1: ISBN-10:

3 Instructor s Manual to ccompan Statics and Strength of Materials For rchitecture and Building Construction Fourth Edition Barr Onoue Pearson/Prentice Hall Seattle Public Librar rchitect: Rem Koolhaus Upper Saddle River, New Jerse Columbus, Ohio

4 Preface This Instructor s Manual is intended to accompan Statics and Strength of Materials for rchitecture and Building Construction. It was initiall developed as a stud guide for students to practice on a variet of problems to enhance their understanding of the principles covered in the tet. Solutions were developed in sufficient detail to allow students to use these problems as additional eample problems. lthough the problem solutions contained in this Instructor s Manual have been worked, re-worked, checked and scrutinized b m man students over the ears, there are inevitabl errors that remain to be discovered b others using the book. If ou detect discrepancies, omissions and errors as ou work through these problems, I would appreciate hearing from ou so that I can incorporate the changes for an future editions of the Instructor s Manual or book. I realize that man instructors do not allow student s access to the Instructor s Manual but I have personall found that m students appreciated having it as a stud guide. Fall, 2010 Barr Onoue, Senior Lecturer Dept. of rchitecture College of Built Environments Universit of Washington barro@u.washington.edu

5 Table of Contents Chapter 2 Statics Graphical addition of vectors pg Resolution of forces: and components pg Vector addition b components pg Moment of a force pg Varignon s theorem pg Moment couples pg 2.9 Equilibrium of concurrent forces pg Equilibrium of rigid bodies pg Supplementar problems pg Chapter nalsis of Determinate Sstems Cables with concentrated loads pg.1. Equilibrium of rigid bodies with distributed loads pg.. Planar trusses method of joints pg.6.8 Truss analsis method of sections pg.8.10 Diagonal tension counters pg Zero-force members pg.12 Pinned frames multi-force members pg.1.1 Supplementar problems pg Retaining walls pg.29.2 Chapter Load Tracing Gravit load trace pg.1.8 Lateral load trace pg.8.11 Chapter Strength of Materials Tension, Compression and shear stress pg.1.2 Deformation and strain pg. Elasticit, strength and deformation pg.. Thermal stress and deformation pg... Staticall indeterminate, aiall loaded members pg..6 Chapter 6 Cross-Sectional Properties Centroids pg Moment of inertia pg Moment of inertia for composite sections pg Chapter 7 Bending and Shear Diagrams Equilibrium method for shear and moment diagrams pg Semi-graphical method for shear and moment diagrams pg Chapter 8 Bending and Shear Stress in Beams Bending stress pg Bending and shear stresses pg Deflection in Beams pg

6 Chapter 9 Column nalsis and Design Euler buckling loads and stresses pg iall loaded steel columns - analsis pg Design of steel columns pg iall load wood columns pg Chapter 10 Structural Connections Bolted steel connections pg Framed connections pg 10. Welded connections pg

7 Chapter 2 Problem Solutions Fa = 100# Fb = 200# R = 17# R = 90# 0 F1 = 600# 0 or F2 = 720# 0 R = 17# 20 Fa = 100# 0 Fb = 200# 2. T = 910# 2.2 T1 = 000# F2 =6kN F1 = kn 10 0 F =kn R = 10,000# T2 = 910# R = 10.2kN θ =

8 T 2 = 20 lb. 20 F=1000 lb. F F B T 1 10 W = 1,200 lb. F=1000 lb. θ B similar triangles: F = F = F F = F = 1000# ( ) = 800# F = F = ( 1000# ) = 600# sinθ = and cosθ = F = F cosθ = ( 1000# ) ( ) = 800# F = F sinθ = ( 1000# )( ) = 600# 2.2

9 T T F =10k 1 0 F 1 = +F 1 cos0 =10k ( ) = 8.66k F 1 = +F 1 sin0 =10k( 0.0) = k F 2 = F 2 = 12k 10 O F =12k 2 F = + 1 ( 2 F ) = + 18k 2 T F = 1 ( 2 F ) = 18k 2 T T 10 T 1 1 F =18k R = ΣF = +k 12k + 18k 2 = +.7k R = ΣF = +8.66k 18k 2 =.07k T = T sin10 T = T cos10 R = +.7 k tanθ = R R =.07.7 = θ = tan ( ) =. from horizontal 2.8 T = T cos10 = 20N 0.98 = 2N R = -.07 k θ =.1 R = 7.0 k Resultant sinθ = R R R R = R sinθ = sin. R =.07k ( 0.79) = 7.0k Rafter θ P θ P=00 lb. 12 Purlin Detail θ = tan 1 12 P ( ) = 18. P = P( 12.6) = ( 00# )( 0.16) = 9.9# P = P( 12.6) = ( 00# )( 0.99) = 28# F2 = 12k 1 F = 18k 1 0 R = 7.0k F1 = 10k θ =.1 Graphical solution using the tip-to-tail method 2.

10 cont d TC TB Graphical Solution: TB TB=600N TC TB=600N TC=800N TC=800N R = 1079N T C = T C cos 60 = 0.T C T C = T C sin 60 = 0.866T C +T B = +T B cos 0 = T B T B = T B sin 0 = 0.62T B θ R = 9.6N φ =.2 R = ΣF = 0. R = ΣF = ( )( 800N) + ( 0.766) ( 600N) = 9.6N ( )( 800N) ( 0.62) ( 600N) = 1078N θ = 86.8 φ =.2 Scale 1mm = 10N θ = tan 1 φ = tan 1 R R R R = tan = tan = 9.6 tan 1 = tan ( ) = 86.8 ( ) =.2 R = 1078N R = 1079N R = = 1079N φ 2.

11 F F 2 F 0 F 0 W W = W cos 0 = 0.866W W = Wsin 0 = 0.0W F = F cos 0 = 0.766F +F = +F sin 0 = +0.62F 2 F 1 R = resultant reaction ais of boom W W=200# -F 2 = -F 2 cos 2 -F 2 = -F 2 sin 2 R = ΣF = 0; 0.0( 200# ) F = 0 F = 100# 0.62 = 16# R = R = ΣF = # R = 17# 120# = 29# ( ) 0.766( 16# ) ais of boom Since the resultant must be vertical, Then: R = ΣF = 0 -F 2 + F 1 = 0 F 2 cos2 = F 1 From this equation, it is seen that F 1 Is onl a fraction of F 2, therefore, F 2 = 7kN. Then, F 1 = F 2 cos2 = (7kN)(0.706) F 1 = 6.kN and F 2 = 7kN R = F 2 = (7kN)(sin2 ) = 2.9kN. R = 29# 0 W=200# F1 = 7kN 2 R =2.9kN F = 16# F2 = 6.kN Tip-to-tail Scale: 1 = 100# Graphical solution 2.

12 T 1 0 T 2 60 F = 8k T 1 = T 1 cos 0 = 0.866T 1 T 1 = T 1 sin 0 = 0.0T 1 T 2 = T2 cos 60 = 0.0T 2 T 2 = T2sin 60 = 0.866T 2 8 W = 2# F = 20# +F = +F cos = k F = F sin =.707 8k ( ) = +.6k ( ) =.6k M = -20#( ) + 2#( ) = -100 #-ft #-ft = 0 The bo is just on the verge of tipping over. But T 1 = T 2 For the resultant to be vertical, R = ΣF = T 0.0T +.6k = 0 T =.1k R = R = ΣF = 0.0.1k ( ) 0.866(.1k.6k) = 11.k 2.1 W=700N T 800N TR T 8k 2m 1m B C R = 11.k B = BC so that -components cancel Using the parallelogram law ( ΣM = 0) ( ) 700N( ) = 0 ( )( 1m) = 1.1m ( 700N) +800N 1m = 800N Since > 1m, the man is OK. 2.6

13 W = 200# P 1 F = 9# F = 1# 12 1 F F B 6 ( ) +1# ( 8" ) = ( ) + ( 120# in) = 20# in. ΣM = 6# 1" 0# in 12 F = 6# [ ΣM = 0] 200# ( 12" ) = F( 26" ) = 0; F = 92.# [ ΣM B = 0] + F( " ) P( 6" ) = 0; P = 10.# kn 10kN 9kN m m W = 100# P 8kN m 2m 20 8kN m kN m m ΣM = W( 18.8" ) = 100# ( 18.8" ) = 1880# in. ( clockwise) [ ΣM = 0] 100# ( 18.8" ) + P(.1" ) = 0 P = 1880# in. = 1.7#.1 in. ( )( 2m) ( 10kN) ( 20m) ( 9kN) ( 16m) ( 8kN) ( 12m) ( )( 8m) ( 8kN) ( m) ( ) ( 200kN m) ( 1kN m) ( ) ( 6kN m) ( 2kN m) M = kn 8kN M = 120kN m 96kN m M = 66kN m 2.7

14 2.20 F = 100# F 12 F D 12 B 12 C ( ) =1200# ( ) = 00# F = # F = 1 100# mm F F = 1.kN 0 F d M B = F ' M C = F ' ( ) + F ( 0) = 1200# ( ' ) = 6000# ft ( ) + F ( 12' ) = 1200# ( ' ) + 00# ( 12' ) ( ) + ( 6000# ft) = 0 M C = 6000# ft 120mm 60 d F = Fcos0 = 1.kN F = Fsin0 = 1.kN d = dcos60 = 200mm d = dsin60 = 200mm ( )( 0.866) =1.kN ( )( 0.0) = 0.7kN ( )( 0.0) =100mm ( )( 0.866) =17mm 2.21 M = F ( d ) + F 120mm + d M = 1.kN 17mm ( ) ( ) + 0.7kN( 220mm) M == 60kN mm = 0.06kN m B 2 F F = 0# F F = Fcos2 = 0# F = Fsin2 = 0# ( )( 0.906) = 27.2# ( )( 0.2) =12.7# d T 0 T = 2000# 7 B T W T = T cos 0 = 2000#.866 T = T sin 0 = 2000# 0.0 d 1 = 7' cos 60 = 7' 0.0 d 2 = 10' 0.0 ( ) = 172# ( ) = 1000# ( ) =.' ( ) = ' ( ) = 6.06' d = 7'sin 60 = 7'.866 M = F ( 6" ) = ( 27.2# )( " ) =108.8 # in CCW M B = F ( 6" ) F ( 2" ) = ( 12.7# )( 6" ) ( 27.2# )( 2" ) = 10.6 # in. 60 d1 d2 [ ΣM C = 0] T d ( ) T d 1 ( 172# )( 6.06' ) 1000# W = 10,00 # ft 00 # ft W = 7000# ft ' ( ) W( d 2 ) = 0 ( )(.' ) W( ' ) = 0 = 100# 2.8

15 mm 180mm 90kN M = 90kN(0.0m) = 27.kN-m 8 8 B 7.07 k 10k 7.07 k 7.07 k k 90kN 7.07 k k C L C L M = 10k M B = 10k ( )( 11.' ) ( k) ( 1' ) = 11 k ft 6 k ft = 169 k ft ( )( 11.' ) ( k) ( 1' ) = 169 k ft # B RB = 2# 6 R and R B form a couple M couple 1 = 2# 12' ( ) = +00 # ft R and 10# man form a couple M couple2 = ( 10# )( 2' ) = 00 # ft F = 69.6# " 12" C B 0 " F = 8# F = 8.7# R = 2# 6 F = ( 8 lb. ) cos = 8.8 lb. F = ( 8 lb. )sin = 69.6 lb. M = F ( 1" ) + F ( " ) = ( 69.6 lb. )( 1" ) + ( 8.8 lb. )( " ) = 89 lb. in. R = 10# M B = F ( 12" ) + F ( " ) = ( 69.6 lb. )( 12" ) + ( 8.8 lb. )( " ) = 60 lb. in. Since the moment due to a couple is constant, M = M B = M C = +00 #-ft - 00 #-ft = 0 2.9

16 B 20 C 1000# F = 00N º º C C BC R C RC Free-bod diagram of joint C Force Magnitude F F = cos60 = 0.0 = sin60 = C = Ccos = 0.707C C = Csin = 0.707C [ ΣF = 0] C + = C = 0 C = = [ ΣF = 0] + C # = 0 ( ) =1000# =1000# = 72# C = # ( ) = 18# F 00N -00cos20º = -70N -00sin20º = -171N C? +Csin10º = +0.17C +Ccos10º = +0.98C BC? +BCsin0º = +0.0BC -BCcos0º = BC [ ΣF = 0] 70N C + 0.0BC = 0... Eq ( 1) ΣF = 0 ( ) [ ] 171N C 0.866BC = 0... Eq 2 Solving equations (1) and (2) simultaneousl, [ ]... Eq ( 1) [ ]... Eq ( 2) BC C = 70N BC 0.98C =171N Therefore, ( ) ( ) +0.BC + 0.1C = 07N... Eq 1 0.BC C = 86N... Eq 2 dding the two equations; C = +768N (compression) Substitute and solve for BC; BC = 672N (tension) 2.10

17 2.0 F1 = 0# F2 = 10# P P =.7# cos 61.1 =.7# 0.8 = 92.# Note that the negative sign for P indicates that it was initiall assumed in the wrong direction. 2 α F2 = 10# F1 = 0# W = 60# Force Magnitude F F F1 0 # -0 # cos2 =. # +0#sin2 = # 2 α =61.1 F2 10 # +10 # (/) = +90 # +10 # (/) = +120 # W 60 # 0-60 # P? +Pcosα +Psinα [ ΣF = 0].# +90# +P cosα = 0... ( 1) [ ΣF = 0] # +120# 60# +P sinα = P =.7# cosα P = 81.1# sinα... ( 1)... ( 2) ( ) W = 60# P = 92# Final Free Bod Diagram F2 = 10# W = 60# Equating :.7# cosα = 81.1# sinα P = 92# sinα 81.1# = tanα = cosα.7# = F1 = 0# 2 α =61.1 α = tan 1 ( 1.81) = 61.1 Graphical check 2.11

18 W = 2.kN T T = Tsin = 0.087T T = Tcos = 0.996T P = Pcos20 = 0.90P P = Psin20 = 0.2P B FBD of the sphere W = 2000# P [ ΣF = 0] cos7 Bsin60 = 0 ( ) = B =.6B [ ΣF = 0] sin7 + Bcos60 2.kN = 0.6B ( ) + 0.0B = 2.kN.72B = 2.kN B = 0.67kN = 2.2kN [ ΣF = 0] 0.087T + 0.9P = 0 T = 0.9P =10.8P [ ΣF = 0] T 0.2P 2000# = 0 substituting; ( ) 0.2P = 2000# P P =192# B = T = # ( ) = 207# = 2.2kN B = 0.67kN Forces eerted bt the sphere onto the smooth surface. 2.12

19 C 00# B C CD DC DE 6 [ ΣM = 0] +100 # ( ' ) 00 # ( ' ) + B ( 10' ) = 0 B = +0 # BC 1 FBD(b) D FBD(a) W = 200 lb. Solving FBD(a) first: Force F F 100# [ ΣF = 0] # B = # 0 # = 0 = +10 # [ ΣF = 0] + 00 # = 0 = +00 # DC DC = 0.80DC + DC = +0.60DC DE +DE cos1 = DE +DE sin1 = +0.29DE W lb. 2. 0kN 0kN F = 0.80DC DE = 0 ; DC = 1.21DE F = (1.21DE) DE 200lb. = 0 ( ) DC 0.98DE = 200lb. ; DE = +20lb.; DC = 1.21( 20lb. ) = 26lb. 2.m 2.m 2.m C D B Writing equations of equilibrium for FBD(b); Force F F CD + ( 26lb. ) = +197lb. ( 26lb. ) = 18lb. C C cos 60 = 0.0C Csin 60 = 0.866C BC 0 +BC F = 0.0C +197lb. = 0 ; C = 9lb. ( T) [ ΣM = 0] 0kN( 2.m) 0kN(.0m) + B ( 7.m) = 0 B = 6.7kN [ ΣF = 0] + 0kN 0kN + 6.7kN = 0 =.KN No horizontal reaction is necessar for this load case. F = 0.866( 9lb. ) + BC 18lb. = 0 ; BC = +89lb. ( C) 2.1

20 k k k 1kN [ ΣF = 0] No force to balance, = 0 B 1kN m 12 6m 6m B B B ( 1) ( 1kN) = 0.8kN ( )( 1kN) = 0.92kN 12 1 [ ΣM = 0] 2k( 20' ) k( 0' ) k( 60' ) + B ( 80' ) = 0 B = 0k ft +120 k ft + 20 k ft 80' = +k [ ΣM B = 0] + k( 20' ) + k( 0' ) + 2k( 60' ) ( 80' ) = 0 = +k [ ΣM = 0] + B( 12m) ( 0.92kN) ( 2.m) + ( 0.8kN) ( 2.m) ( 0.8kN kN) ( 6m) = # 100# 000# D D Solving for B; B = 1 B = 12 1 B = 0.767kN ( )( 0.767kN) = 0.29kN ( )( 0.767kN) = 0.708kN Reverting back to the unresolved forces; [ ΣF = 0] + +1kN ( 0.29kN) = 0 = +0.70kN ( B ) [ ΣF = 0] + 1kN + ( 0.708kN) = 0 ( B ) = kN [ ΣM = 0] 100 # ( 17.' ) 000 # ( 8.67' ) + D ( 0' ) = 0 D = +17 # [ ΣF = 0] 100 # cos0 000 # cos0 100 # cos # = 0 = +6 # [ ΣF = 0] +100sin sin0 +100sin 0 D = 0 D = +000 # 2.1

21 k # 00# 20k 2 6 C k k D 16k 6 6 k 00# 2 10 C D D Upper beam: B E F B 10 B C C +D =8k; C = D = k Upper beam: Left beam: [ ΣM = 0] 20k( 2' ) k( 60' ) + B ( 8' ) = 0 B = +1k [ ΣF = 0] + 20k +1k k = 0 = +9k Right beam: [ ΣM E = 0] + k( 12' ) 16k( 2' ) + F ( 8' ) k( 60' ) = 0 F = +12k [ ΣF = 0] k + E 16k +12k k = 0 E = +12k [ ΣM D = 0] + 00 # ( 12' ) C ( 10' ) = 0 C = +80 # [ ΣF = 0] 00 # + C + D = 0 D = 80 # + 00 # = 80 # [ ΣF = 0] + 00 # + D = 0 D = 00 # ( ) Lower beam: ( ) [ ΣM B = 0] C ( ' ) ( 10' ) = 0 = 20 # ( ) [ ΣF = 0] 20 # + B 80 # = 0 B = +720 # ( ) [ Σ = 0] B = 0 2.1

22 FD FD FD FD = ( )FD FD = ( )FD [ ΣM = 0] + FD ( ' ) + FD ( 20' ) 2k( 16' ) 8k( 2' ) = 0 ( )FD( ' ) + ( )FD( 20' ) 2 k ft 26 k ft = 0 FD = 18.9k FD = FD = ( )( 18.9k) = 1.2k ( )( 18.9k) = 11.k [ ΣF = 0] + FD = 0 ( ) = +1.2k 2k [ ΣF = 0] + 2k 8k +11.k = 0 ( ) = 1.k 8k F1 = 00# R = 720# θr = 72. θ1 = 7 θ2 = 0 Parallelogram Method F2 = 00# DC BD DF = 1.2k DF = 18.9k DF = 11.k F1 = 00# F2 = 00# θ2 = 0 R = 720# [ ΣF = 0] DC + BD +1.2k = ( DF ) ( ) ( ) DC BD [ ΣF = 0] DC + BD 11.k.16.1 ( DF ) = 0 ( DC ) ( BD ) Solving the two equations simultaneousl; BD = 17.9k θ1 = 7 θr = 72. Tip-to-Tail Method DC = 19.7k 2.16

23 C = 1kN R = ΣF = 6k R = k + 2.6k = 0.k ( ) cos 60 + ( k) cos 0 R =.kn 0 B = 1.8kN k R = ΣF = ( 6k)sin 60 ( k)sin 0 R =.2k 1.k = 6.7k R = -0.k θ = 2 O = 2kN 12 6k θ = 86.6 Tip-to-Tail Method θ = tan 1 R R = tan = tan ( ) = 86.6 R = R 2 + R 2 = ( 0.0) 2 + ( 6.7) 2 = 6.71k R = -6.7k 2. R = 6.71k S = 20.k 0 θ = k 60 Q = 22k R = 2.k (vertical) 7 R = 6.71k 6k P = 16k Tip-to-tail O 2.17

24 D F1 F1 = 800# F1 F2 D = 90kN k BD = kn CD = 110kN P = 00# Component F F F2 F2 = 1200# D = 90kN 1 90kN 2 ( ) = 90kN 2 = 6.6kN 1 90kN 2 ( ) = 6.6kN Force F F P=00# 0-00# F 1 +F 1 cos0 =(800#)(0.866)=692.8# + F 1 sin0 =(800#)(0.0)=00# F 2 R = 700# + F 2 cos0 =(1200#)(0.866)=109.2# - F 2 sin0 =-(1200#)(0.0)=-600# R = ΣF = #+109.2#= +172# ( ) R = ΣF = 00#+00# 600#= 700# ( ) R = ( 172# ) 2 + ( 700# ) 2 =1868# θ = 22 R = 172# R = 1868# lternate wa to find the resultant R: sinθ = R R ; R = R sinθ = 700 sin22 = =1867# BD = kn CD = 110kN R = -1.6kN ( kn) = 27kN ( kn) = 6kN + ( 110kN) cos 0 = +9.kN ( 110kN)sin 0 = kn R = +.7kN R = F = +.7kN R = F = 1.6kN θ = 88. R = R 2 + R 2 = (.7) 2 + ( 1.6) 2 = 1.7kN θ = tan 1 R R = tan θ = tan 1 ( 2.9) = 88. tanθ = R R ; 700 θ = tan 1 = tan ( ) = 22 R = 1.7kN Resultant 2.18

25 # 100# T1 d1 = 10 d2 = 1 F d2 T2 T2 = 700# T2 6 T1 T1 = 00# 20 d1 C d ( ) = 9.9' ( ) = 9.9' d 2 = d 2 cos = 1' d 2 = d 2 sin = 1' d 1 = d 1 cos 20 = 10' ( 0.9) = 9.' [ ΣM C = 0] + 20 # ( d 1 ) 100 # d 2 F = 20# 9.' 9.9' ( ) 100 # ( 9.9' ) ( ) F d 2 = 17. # ( ) = 0 Force F F T 1 = 00# (00#)cos1 o = 8# (00#)sin1 o =129.# T 2 = 700@# (700#)cos10 o = 689.# (700#)sin10 o = 121.8# M = +T 2 (0 ) + T 2 (6 ) T 1 ( ) T 1 ( ) =0 MR M = 689.#(0 ) #(6 ) 8#( ) 129.#( ) = +990#-ft 1.m F =.21kN F = kn 0 1.1m 1m 1m F =.8kN [ ΣF = 0] +.8kN = 0 ( F ) =.8kN [ ΣF = 0] +.21kN = 0 ( F ) =.21kN M =.21kN.1m ( ).8kN( 1.m) M = 9.9kN m.09kn m = 1.0kN m 2.19

26 O origin 10# 7# 6# 8 18# O origin 1m 100N 0N/m 2m 10N Weight of wood member: ω = 0N/m ssume the member weight is located at the center of the length. 10N R = ΣF = 10# + 7# + 6# - 18# = +# M O = +(7#)( ) + (6#)(9 ) (18#)(17 = - 22#-in. 100N 1m R() = 22 #-in; = 22 # in # =.8" O origin 1.m W = 90N (beam wt.) R = # R = ΣF = -100N 90N + 10N = -0N M O = - (100N)(1m) (90N)(1.m) + (10N)(m) = +21 N-m O origin.8 R() = MO = 21N m 0N =.m R = 0N =.m O origin m For a 0N force to produce a moment directed counter-clockwise, the R = 0N force will be at an imaginar location where =.m to the left of the origin. 2.20

27 # 100# 00# B B B 12 B origin 20#/ft C C C W = k 200# 100# 00# Total beam weight equals (20 # / ft )(16 ) = 20# at the center of the beam length. B Force F F B 12 1 B + 1 B C C C origin W 0 -k W = 20# For the beam to remain stationar and horizontal, the moments taken about points and B should be balanced b the opposing moments due to B and respectivel, resulting in no resultant moment. [ ΣM = 0] # ( ' ) +100 # ( 8' ) 20 # ( 8' ) 00 # ( 12' ) B( 16' ) = 0 B = 60 # [ ΣM B = 0] + 00 # ' = 60 # ( ) + 20 # ( 8' ) 100 # ( 8' ) 200 # ( 12' ) ( 16' ) = 0 R = ΣF = # +100 # 20 # 00 # + B = 0 = +60 # # +100 # 20 # 00 # + 60 # = 0 [ ΣF = 0] 12 1 B C = 0; 12 1 B = C = 0 B = 1 1 C [ ΣF = 0] + 1 C C k = 0; + C C = k; C =.6k ( compression) B = 1 1.6k ( ) = +.6k ( tension ) C C = +k 2.21

28 F = 1.6kN F = 2kN F = 1.2kN C 60 BC 60 BC = 2.1kN F = 2kN B 12 0 BC DB W = 800# C C BC C = 0.27kN Tip-to-tail Force F F Force F F C C C BC BC cos 60 = 0.0BC +BC sin 60 = BC B 12 1 B 1 B DB +DBsin 0 = +0.0DB +DB cos 0 = DB F=2kN + ( 2kN) = +1.2kN ( 2kN) = 1.6kN W 0-800# [ ΣF = 0] C 0.0BC +1.2kN = 0; C + 0.0BC = 1.2kN [ ΣF = 0] C BC 1.6kN = 0; C BC = 1.6kN Solving simul tan eousl; BC = 2.1kN (compression) C = 0.27kN (tension) [ R = ΣF = 0] 12 2 B + 0.0DB = 0; DB = 1 1 B [ R = ΣF = 0] 1 B DB 800# = 0 2 B B = 800 # B = 68.2 # DB = 2 ( 68.2 # ) = #

29 Joint B: C C CB CB C 0 CB BC BC BC B 12 B B = 160# BE W Force F F C C cos = 0.707C Csin = C CB +CB cos 0 = CB +CBsin 0 = +0.0CB Force F F B=160# + 12 ( ) 1 160# = 10 # ( ) 1 160# = 600 # BE 0 +BE BC BC BC W 0 -W [ ΣF = 0] 0.707C CB = 0; C = 0.866CB = 1.22CB This relationship indicates the C > CB, therefore, C = 1.8kN Then, CB = (1.8kN)/1.22 = 1.8kN [ ΣF = 0] +10 # BC = 0; BC = 1800# [ ΣF = 0] 600 # + BE ( ) 1800# = 0; BE = 1680 # [ ΣF = 0] 0.707C + 0.0CB W = 0 W = kN ( ) + 0.0( 1.8kN) = 2.0kN 2.2

30 2.8 cont d 2.9 Joint C: CD CD CD CB CB = 1800# CB 10 00# 26 B B 10 B 2 W Force F F CD CD CD CB = 1800 # + ( ) 1800# = 10 # + ( ) 1800# = # W 0 -W [ ΣF = 0] 0.707CD +10 # = 0; CD = 207 # [ ΣF = 0] ( 207 # ) # W = 0; W = 220 # [ ΣM = 0] 00 # ( 10' ) + B ( 1 10' ) + 12B ( 1 2' ) = 0 ( B ) ( B ) 0B B 1 = 000# ft ; B = 192. # B = 7 # ; B = 177. # [ ΣF = 0] + 7 # ( ) = 0; = 7 # B [ ΣF = 0] + 00 # # = 0; = 22. # ( B ) 2.2

31 kN 2.7kN 20 2k k 2k B hinge C C MRC C k B 2m 2.m m m k k 1.8kN 2m 2.m B B 2.7kN [ ΣF = 0] + k = 0; = +k ( ) [ ΣM = 0] 2k( 20' ) k( 0' ) 2k( 60' ) k( 0' ) + k( 20' ) + B ( 80' ) = 0 B = +k ( ) [ ΣF = 0] + 2k k 2k k + k ( B ) = 0 = +6k ( ) B m m MRC B C C Beam B: [ ΣF = 0] B = 0 [ ΣM B = 0].m [ ΣF = 0] +1kN 1.8kN + B = 0; B = 0.8kN Beam BC: [ ΣF = 0] C = 0 ( ) +1.8kN( 2.m) = 0; = 1kN [ ΣF = 0] 0.8kN 2.7kN + C = 0; C =.kn [ ΣM C = 0] M RC + 2.7kN m ( ) + 0.8kN( 6m) = 0 M RC = 8.1kN m+.8kn m = 12.9kN m 2.2

32 # 00# 180# 200# B B D # 60# D C Upper Beam: [ ΣM = 0] 00 # ( ' ) + B ( 8' ) = 0; B = # ( ) [ ΣF = 0] # = 0; = +180 # ( ) ΣF = 0 ( ) [ ] # 20 # + = 0; = +2. # Lower Beam: [ ΣM D = 0] # ( ' ) + C ( 6' ) 187. # ( 9' ) 80 # ( 1' ) = 0 C = +22 # ( ) [ ΣF = 0] + D 60 # = 0; D = +60 # ( ) [ ΣF = 0] 200 # + D + 22 # 187. # 80 # = 0 ( ) D = +1. # 2.26