Fourth Edition - Frat Stock · Chapter 5 Strength of Materials • Tension, Compression and shear stress pg 5.1 – 5.2 • Deformation and strain pg 5.3 • Elasticity, strength

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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


__________________________________________________________________________________ Copyright © 2012 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, 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 Onouye, Statics and Strength of Materials for Architecture and Building Construction, Fourth Edition, may reproduce material from the instructor’s manual for classroom use.

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: [email protected]


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:


2.5

2.11y x

30˚

40˚F

W=200#

axis

of b

oom

Fy

Fx

Wx

Wy

2.11

axis

of b

oom

30˚

W=200#R =

293#

F = 156#

Tip-to-tail

Scale: 1” = 100#

y x

30˚

40˚F

W=200#

axis

of b

oom

Fy

Fx

Wx

Wy

2.11

axis

of b

oom

30˚

W=200#R =

293#

F = 156#

Tip-to-tail

Scale: 1” = 100#

€

−Wx = −W cos 30° = −0.866W−Wy = −Wsin 30° = −0.50W−Fx = −F cos 40° = −0.766F+Fy = +F sin 40° = +0.642F

€

Ry = ΣFy = 0; − 0.50 200#( ) + 0.642F = 0

∴ F =100#0.642

= 156#

R = Rx = ΣFx = −0.866 200#( ) − 0.766 156#( )R = −173# −120# = −293#

2.12

2.12

25°

y

R = resultant reaction

F1

F2

A

F1 = 7kN

F2 = 6.34kN

R =2.95kN

25˚

2.12

25°

y

R = resultant reaction

F1

F2

A

F1 = 7kN

F2 = 6.34kN

R =2.95kN

25˚

-F2x = -F2cos 25˚-F2y = -F2sin 25˚

Since the resultant must be vertical,Then: Rx = ΣFx = 0

-F2x + F1 = 0

F2cos25˚ = F1From this equation, it is seen that F1Is only a fraction of F2 , therefore, F2 = 7kN.

Then, F1 = F2cos25˚ = (7kN)(0.706)

F1 = 6.34kN and F2 = 7kN

R = F2y = (7kN)(sin25˚) = 2.95kN.

Graphical solution


2.6

2.13

€

−T1x = −T1 cos 30° = −0.866T1

−T1y = −T1 sin 30° = −0.50T1

−T2x = −T2 cos 60° = −0.50T2

−T2y = −T2sin 60° = −0.866T2

€

+Fx = +F cos 45° = +0.707 8k( ) = +5.65k

−Fy = −F sin 45° = −.707 8k( ) = −5.65k

But T1 = T2

For the resultant to be vertical,

€

Rx = ΣFx = 0∴ − 0.866T − 0.50T + 5.65k = 0T = 4.14kR = Ry = ΣFy = −0.50 4.14k( ) − 0.866 4.14k − 5.65k( ) = −11.3k

2.14

W = 25#

F = 20#8’

5’

A

4’

2.14

2.15

A

800N

W=700N

x

2m 1m

2.15

€

ΣM A = 0( )+800N 1m( ) − 700N x( ) = 0

x =800N( ) 1m( )

700N( )= 1.14m

Since x > 1m, the man is OK.

W = 25#

F = 20#8’

5’

A

4’

2.14

2.15

A

800N

W=700N

x

2m 1m

2.13

F = 8k

T1

T2

y

x30°

60°45°

A

A B C

AB = BC so that x-components cancel

T

T

TR

8k

R = 11.3k

xy

Using the parallelogram law

2.13

F = 8k

T1

T2

y

x30°

60°45°

A

A B C

AB = BC so that x-components cancel

T

T

TR

8k

R = 11.3k

xy

Using the parallelogram law

MA = -20#(5’) + 25#(4’) = -100 #-ft + 100 #-ft = 0

The box is just on the verge of tipping over.


2.7

2.16

€

ΣM A = −36# 15"( ) +15# 8"( ) =

− 540# −in( ) + 120# −in( ) = −420# −in.

€

ΣM A = −W 18.8"( ) = −100# 18.8"( ) = −1880# −in. clockwise( )ΣM A = 0[ ] −100# 18.8"( ) + P 45.1"( ) = 0

P =1880# −in.

45.1 in.= 41.7#

€

ΣM A = 0[ ] − 200# 12"( ) = F 26"( ) = 0; ∴ F = 92.3#

ΣM B = 0[ ] + F 4"( ) − P 36"( ) = 0; ∴ P = 10.3#

2.17

2.18

5

12

8”

17”

15”

5

12

Fy = 15#

Fx = 36#

A

F = 39#

2.16

W = 100# P18.8” 26.3”

6”20”

28”

20˚

A

12” 14”

4”

36”

W = 200# P

F

FA

B

2.19

24m

5kN

10kN

9kN

8kN

8kN

8kN

4m

4m

4m

4m

4m

4m

A

2.19

€

MA = − 5kN( ) 24m( ) − 10kN( ) 20m( ) − 9kN( ) 16m( ) − 8kN( ) 12m( )− 8kN( ) 8m( ) − 8kN( ) 4m( )MA = − 120kN − m( ) − 200kN − m( ) − 144kN − m( )− 96kN − m( ) − 64kN − m( ) − 32kN − m( )MA = −656kN − m


2.8

€

Fx = 1213 1300#( ) =1200#

Fy = 513 1300#( ) = 500#

€

MB = −Fx 5'( ) + Fy 0( ) = −1200# 5'( ) = −6000#−ft

MC = −Fx 5'( ) + Fy 12'( ) = −1200# 5'( ) + 500# 12'( )MC = − 6000#−ft( ) + 6000#−ft( ) = 0

2.202.20

5’

12’ 12’

F = 1300# Fy

Fx

512

AB

C

D

2.21

2.21

6”

6”

4”

25˚

25˚

A

BF = 30#Fy

Fx

€

Fx = Fcos25° = 30#( ) 0.906( ) = 27.2#

Fy = Fsin25° = 30#( ) 0.423( ) =12.7#

€

MA = Fy 6"( ) = 27.2#( ) 4"( ) =108.8#− in CCW

€

MB = −Fy 6"( ) − Fx 2"( ) = − 12.7#( ) 6"( ) − 27.2#( ) 2"( ) = −130.6#− in.

2.22

2.22

A

Fx

Fy F = 1.5kN

60˚

30˚

dx

dy

120mm

200m

m

€

Fx = Fcos30° = 1.5kN( ) 0.866( ) =1.3kN

Fy = Fsin30° = 1.5kN( ) 0.50( ) = 0.75kN

dx = dcos60° = 200mm( ) 0.50( ) =100mm

dy = dsin60° = 200mm( ) 0.866( ) =173mm

€

MA = −Fx dy( ) + Fy 120mm + dx( )MA = −1.3kN 173mm( ) + 0.75kN 220mm( )MA == −60kN − mm = −0.06kN − m

2.23

2.23

60˚

dx1

dx2

dyT = 2000#

Tx

Ty

W

A

B

3’

7’

30˚

€

Tx = T cos 30° = 2000# .866( ) = 1732#

Ty = T sin 30° = 2000# 0.50( ) = 1000#

dx1 = 7' cos 60° = 7' 0.50( ) = 3.5'

dx2 = 10' 0.50( ) = 5'

dy = 7' sin 60° = 7' .866( ) = 6.06'

€

ΣM C = 0[ ] Tx dy( ) − Ty dx1( ) − W dx2( ) = 0

1732#( ) 6.06'( ) − 1000#( ) 3.5'( ) − W 5'( ) = 0

5W = 10,500#−ft − 3500#−ft

W =7000#−ft

5'= 1400#


2.9

2.24

€

MA = − 10k( ) 11.3'( ) − 4k( ) 14'( ) = −113k−ft − 56k−ft = −169k−ft

MB = − 10k( ) 11.3'( ) − 4k( ) 14'( ) = −169k−ft

2.252.25

RAx = 25#

RAy = 150#

RBx = 25#

B

A

2’ 2’

6’

6’

150#

€

RAx and RBx form a coupleMcouple 1 = 25# 12'( ) = +300#−ft

€

RAy and 150# man form a couple

Mcouple2 = − 150#( ) 2'( ) = −300#−ft

Since the moment due to a couple is constant,

MA = MB = MC = +300#-ft - 300#-ft = 0

2.26

2.26

CL

90kN

125mm 180mm

CL

90kN

M = 90kN(0.305m) = 27.45kN-m

2.27

2.27

30°

55°

C

4"

3" 12"

A B

Fy = 69.6#

Fx = 48.7#

F = 85#

€

Fx = 85 lb.( ) cos 55° = 48.8 lb.

€

Fy = 85 lb.( ) sin 55° = 69.6 lb.

€

M A = −Fy 15"( ) + Fx 4"( ) = − 69.6 lb.( ) 15"( ) + 48.8 lb.( ) 4"( ) = −849 lb.− in.

€

M B = −Fy 12"( ) + Fx 4"( ) = − 69.6 lb.( ) 12"( ) + 48.8 lb.( ) 4"( ) = −640 lb.− in.

2.24

A

B

7.07 k

7.07 k4 k

7.07 k

7.07 k

6’ 8’ 8’

8’8’ 10k

10k11.3’


2.10

2.28

2.28

B

Ax

AyRA RC

Cx

Cy

1000#

60º 45º

y

x

1000#

RC RA

Cx

Cy Ay

Ax

B

2.29

F = 500N

ACBC

20˚

30˚10˚

x

y

Free-body diagram of joint C

C

2.29

€

ΣFx = 0[ ] − Cx + Ax = 0−0.707C + 0.50A = 0

C =0.50A0.707

= 0.707A

€

ΣFy = 0[ ] + Cy + Ay −1000#= 0

0.707 0.707A( ) + 0.866A =1000#∴ 1.37A =1000#A = 732#C = 0.707 732#( ) = 518#

€

ΣFx = 0[ ] − 470N + 0.174AC + 0.50BC = 0 ... Eq 1( )ΣFy = 0[ ] −171N + 0.984AC − 0.866BC = 0 ... Eq 2( )

Solving equations (1) and (2) simultaneously,

€

0.866 × 0.50BC + 0.174AC = 470N[ ] ... Eq 1( )0.50 × −0.866BC − 0.984AC =171N[ ] ... Eq 2( )

Therefore,

€

+0.433BC + 0.15AC = 407N ... Eq 1( )−0.433BC + 0.492AC = 86N ... Eq 2( )

Adding the two equations;

AC = +768N (compression)

Substitute and solve for BC;

BC = 672N (tension)

€

Ax = Acos60° = 0.50AAy = Asin60° = 0.866ACx = Ccos45° = 0.707CCy = Csin45° = 0.707C

Force Magnitude Fx__________________ Fy____________________

F 500N -500cos20º = -470N -500sin20º = -171N

AC ? +ACsin10º = +0.174AC +ACcos10º = +0.984AC

BC ? +BCsin30º = +0.50BC -BCcos30º = -0.866BC


2.11

2.30

2.30

F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ α x

y

P F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ x

y

α =61.1˚

P = 925#

3

4

F1 = 50#

F2 = 150#W = 60#

P = 925#

α =61.1˚25˚

Graphical check

2.30

F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ α x

y

P F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ x

y

α =61.1˚

P = 925#

3

4

F1 = 50#

F2 = 150#W = 60#

P = 925#

α =61.1˚25˚

Graphical check

2.30

F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ α x

y

P F1 = 50#

F2 = 150#

W = 60#

3

4

25˚ x

y

α =61.1˚

P = 925#

3

4

F1 = 50#

F2 = 150#W = 60#

P = 925#

α =61.1˚25˚

Graphical check

Force Magnitude Fx__________________ Fy__________________

F1 50# -50#cos25° = −45.3# +50#sin25° = +21.1#

F2 150# +150#(3/5) = +90# +150#(4/5) = +120#

W 60# 0 -60#

P ? +Pcosα +Psinα

€

ΣFx = 0[ ] − 45.3# +90# +P cosα = 0 ..... 1( )ΣFy = 0[ ] + 21.1# +120# −60# +P sinα = 0 .... 2( )

€

P =−44.7#cosα

.... 1( )

P =−81.1#sinα

.... 2( )

€

Equating : −44.7#cosα

=−81.1#sinα

sinαcosα

= tanα =−81.1#−44.7#

= +1.81

α = tan−1 1.81( ) = 61.1°

€

P =−44.7#

cos 61.1°=

−44.7#0.483

= −92.5#

Note that the negative sign for Pindicates that it was initially assumedin the wrong direction.

Final Free Body Diagram


2.12

2.31

2.31

A

B

W = 2.5kN

75˚

30˚

FBD of the sphere

15˚ 60˚

A = 2.24kN

B = 0.67kN

Forces exerted bt the sphere onto the smooth surface.

90˚

90˚

2.31

A

B

W = 2.5kN

75˚

30˚

FBD of the sphere

15˚ 60˚

A = 2.24kN

B = 0.67kN

Forces exerted bt the sphere onto the smooth surface.

90˚

90˚

€

Tx = Tsin5° = 0.087TTy = Tcos5° = 0.996TPx = Pcos20° = 0.940PPy = Psin20° = 0.342P

€

ΣFx = 0[ ] − 0.087T + 0.94P = 0

T =0.94P0.087

=10.8P

ΣFy = 0[ ] + 0.996T − 0.342P − 2000#= 0

substituting;0.996 10.8P( ) − 0.342P = 2000#P =192#AB = T =10.8 192#( ) = 2074#

2.32

€

ΣFx = 0[ ] Acos75° − Bsin60° = 0

A =B 0.866( )

0.259= 3.346B

€

ΣFy = 0[ ] Asin75° + Bcos60° − 2.5kN = 0

3.346B 0.966( ) + 0.50B = 2.5kN3.732B = 2.5kNB = 0.67kNA = 2.24kN

2.32

y

x

W = 2000#

P

T5˚

20˚

2.33

34

60°

y

x

CA

CD

BC

34

DC

y

W = 200 lb.

DE

15°FBD(b)

FBD(a)

C

D


2.13

2.33

Solving FBD(a) first:Force _________Fx_________ __________Fy__________

DC

€

−45

DC = −0.80DC

€

+35

DC = +0.60DC

DE

€

+DE cos15° = +0.966DE

€

+DE sin15° = +0.259DE

W 0 -200 lb. ___________________ _____________________

€

Fx∑ = − 0.80DC + 0.966DE = 0 ; DC = 1.21DE

€

Fy∑ = + 0.60(1.21DE)DCx( )

+ 0.259DE − 200lb. = 0

€

0.985DE = 200lb. ; DE = +203lb.; DC = 1.21 203lb.( ) = 246lb.

Writing equations of equilibrium for FBD(b);

Force Fx Fy

CD

€

+45

246lb.( ) = +197lb.

€

−35

246lb.( ) = −148lb.

CA

€

−CA cos 60° = −0.50CA

€

−CAsin 60° = −0.866CA

BC 0 +BC

€

Fx∑ = − 0.50CA +197lb. = 0 ; ∴ CA = 394lb. T( )

€

Fy∑ = − 0.866 394lb.( ) + BC −148lb. = 0 ; BC = +489lb. C( )

2.34

€

ΣMA = 0[ ] +100# 4'( ) − 300# 3'( ) + Bx 10'( ) = 0

∴ Bx = +50#

ΣFx = 0[ ] + Ax −100# − Bx = 0

+Ax −100# − 50# = 0∴ Ax = +150#

ΣFy = 0[ ] + Ay − 300# = 0

∴ Ay = +300#

2.34

Ax

Ay

Bx

300#

100#

3’ 2’

6’

4’

2.35

€

ΣMA = 0[ ] − 40kN 2.5m( ) − 50kN 5.0m( ) + By 7.5m( ) = 0∴ By = 46.7kN

ΣFy = 0[ ] + A − 40kN − 50kN + 46.7kN = 0

∴ = 43.3KN

No horizontal reaction is necessary for this load case.

2.35

40kN 50kN

A BC D

2.5m2.5m 2.5m

2.36

20’ 20’ 20’20’

2k 3k 4k

15’15’ 20’

Ax

Ay By

2.32

y

x

W = 2000#

P

T5˚

20˚

2.33

34

60°

y

x

CA

CD

BC

34

DC

y

W = 200 lb.

DE

15°FBD(b)

FBD(a)

C

D


2.14

2.36

€

ΣFx = 0[ ] No force to balance Ax, ∴ Ax = 0

ΣMA = 0[ ] − 2k 20'( ) − 3k 40'( ) − 4k 60'( ) + By 80'( ) = 0

By =40k−ft +120k−ft + 240k−ft

80'= +5k

ΣMB = 0[ ] + 4k 20'( ) + 3k 40'( ) + 2k 60'( ) − Ay 80'( ) = 0∴ Ay = +4k

2.35

40kN 50kN

A BC D

2.5m2.5m 2.5m

2.36

20’ 20’ 20’20’

2k 3k 4k

15’15’ 20’

Ax

Ay By

2.37

2.37

30° 30°

90° 90°1500#

1500#

3000#

Ay Dy

Dx8.66’

8.66’

30’

60°

2.38

5

126m

6m

2.5m

12

5

5

12

1kN

1kN

Ax

Ay

By

Bx

B

€

ΣMA = 0[ ] −1500# 17.33'( ) − 3000# 8.67'( ) + Dy 30'( ) = 0

∴Dy = +1733#

ΣFy = 0[ ] −1500# cos30° − 3000# cos30° −1500# cos30° + Ay +1733# = 0

∴ Ay = +3463#

ΣFx = 0[ ] +1500sin 30° + 3000sin30° +1500sin30° − Dx = 0

∴ Dx = +3000#

€

513( ) 1kN( ) = 0.385kN1213( ) 1kN( ) = 0.923kN

2.37

30° 30°

90° 90°1500#

1500#

3000#

Ay Dy

Dx8.66’

8.66’

30’

60°

2.38

5

126m

6m

2.5m

12

5

5

12

1kN

1kN

Ax

Ay

By

Bx

B

2.38

€

ΣMA = 0[ ] + B 12m( ) − 0.923kN( ) 2.5m( ) + 0.385kN( ) 2.5m( ) −

0.385kN + 0.923kN( ) 6m( ) = 0

Solving for B; B = 0.767kNBx = 5

13( ) 0.767kN( ) = 0.295kN

By = 1213( ) 0.767kN( ) = 0.708kN

Reverting back to the unresolved forces;

€

ΣFx = 0[ ] + Ax +1kN − 0.295kN( )Bx( )

= 0

∴ Ax = +0.705kN

ΣFy = 0[ ] + Ay −1kN + 0.708kN( )By( )

= 0

∴ Ay = +0.292kN


2.15

2.39 2.40

Left beam:

Right beam:

€

ΣM E = 0[ ] + 4k 12'( ) −16k 24'( ) + Fy 48'( ) − 4k 60'( ) = 0∴ Fy = +12k

ΣFy = 0[ ] − 4k + Ey −16k +12k − 4k = 0

∴ Ey = +12k

Upper beam:Cy +Dy =8k; Cy = Dy = 4k Upper beam:

Lower beam:€

ΣMA = 0[ ] − 20k 24'( ) − 4k 60'( ) + By 48'( ) = 0∴ By = +15k

ΣFy = 0[ ] + Ay − 20k +15k − 4k = 0

∴ Ay = +9k

€

ΣMD = 0[ ] + 400# 12'( ) − Cy 10'( ) = 0

Cy = +480#

ΣFy = 0[ ] − 400# + Cy + Dy = 0

Dy = −480# + 400# = −80# ↓( )

ΣFx = 0[ ] + 300# + Dx = 0

Dx = −300# ←( )

€

ΣMB = 0[ ] − Cy 5'( ) − Ay 10'( ) = 0

Ay = −240# ↓( )

ΣFy = 0[ ] − 240# + By − 480# = 0

By = +720# ↑( )

Σx = 0[ ] Bx = 0

2.39

Ay By

Cy Dy

Ey Fy

8k

16k20k

4k

24’ 36’

48’

12’ 12’

12’ 12’ 12’

36’ 36’

2.40500# 400#

300#Dx

DyCy

2’ 10’

Bx

ByAy

Cy

10’ 5’

4k 4k

2.39

Ay By

Cy Dy

Ey Fy

8k

16k20k

4k

24’ 36’

48’

12’ 12’

12’ 12’ 12’

36’ 36’

2.40500# 400#

300#Dx

DyCy

2’ 10’

Bx

ByAy

Cy

10’ 5’

4k 4k


2.16

2.41

2.41

Ax

Ay

FD FDy

FDx

2k 8k

x

y

DCDF = 18.9k

BD

DFx = 15.2k

DFy = 11.3k

2.41

Ax

Ay

FD FDy

FDx

2k 8k

x

y

DCDF = 18.9k

BD

DFx = 15.2k

DFy = 11.3k

€

FDx = 45( )FD FDy = 3

5( )FD

€

ΣM A = 0[ ] + FDx 4'( ) + FDy 20'( ) − 2k 16'( ) − 8k 32'( ) = 04

5( )FD 4'( ) + 35( )FD 20'( ) − 32k−ft − 256k−ft = 0

∴ FD = 18.9kFDx = 4

5( ) 18.9k( ) = 15.2k

FDy = 35( ) 18.9k( ) = 11.3k

€

ΣFx = 0[ ] + Ax − FDx = 0

∴ Ax = +15.2k →( )

ΣFy = 0[ ] + Ay − 2k − 8k +11.3k = 0

∴Ay = −1.3k ↓( )

€

ΣFx = 0[ ] − 3DC3.16

⎛

⎝ ⎜

⎞

⎠ ⎟

DCx( )

+BD5.1

⎛

⎝ ⎜

⎞

⎠ ⎟

BDx( )

+15.2kDFx( )

= 0

ΣFy = 0[ ] − DC3.16

⎛

⎝ ⎜

⎞

⎠ ⎟

DCy( )

+5BD5.1

⎛

⎝ ⎜

⎞

⎠ ⎟

BDy( )

−11.3kDFy( )

= 0

Solving the two equations simultaneously;

BD = 17.9k

DC = 19.7k

2.42

2.42

θ1 = 75° θ2 = 30°

F1 = 500#

F2 = 400#

R =

720#

θR = 72.5°

Parallelogram Method

x

y

F2 = 400#

F1 = 500#

R =

720#

θR = 72.5°θ1 = 75°

θ2 = 30°

Tip-to-Tail Method


2.17

2.432.43

512

11

30°

A = 2kN

B =

1.8k

N

C = 1kN

R = 3.4

kN

x

y

θ = 52°

O

Tip-to-Tail Method

2.442.44

45°

75°

30°

P = 16k

Q = 22k

S = 20.5k

R = 42.5k(vertical)

O x

2.452.45

y

x

6k

3k

A

60° 30°

R = 6.71k

Rx = -0.4k

Ry = -6.7k

θ = 86.6°

30°3k

6k

Tip-to-tail

xy

θ = 86.6°

R = 6.71k

60°

€

Rx = ΣFx = − 6k( ) cos 60° + 3k( ) cos 30°

Rx = −3k + 2.6k = −0.4k

Ry = ΣFy = − 6k( ) sin 60° − 3k( ) sin 30°

Ry = −5.2k −1.5k = −6.7k

€

θ = tan−1 Ry

Rx

⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 6.7

0.4⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 16.75( ) = 86.6°

€

R = Rx2 + Ry

2 = 0.40( )2+ 6.7( )2

= 6.71k

2.45

y

x

6k

3k

A

60° 30°

R = 6.71k

Rx = -0.4k

Ry = -6.7k

θ = 86.6°

30°3k

6k

Tip-to-tail

xy

θ = 86.6°

R = 6.71k

60°

2.45

y

x

6k

3k

A

60° 30°

R = 6.71k

Rx = -0.4k

Ry = -6.7k

θ = 86.6°

30°3k

6k

Tip-to-tail

xy

θ = 86.6°

R = 6.71k

60°


2.18

2.462.46

x

y

F1 = 800#

F2 = 1200#

30°

30°

F1x

F1y

F2x

F2y

P = 500#

Rx = 1732#

Ry = 700#

R = 1868#

θ = 22°

2.46

x

y

F1 = 800#

F2 = 1200#

30°

30°

F1x

F1y

F2x

F2y

P = 500#

Rx = 1732#

Ry = 700#

R = 1868#

θ = 22°

Force Fx______________________ Fy __________________________

P=500# 0 -500#

F1 +F1cos30°=(800#)(0.866)=692.8# + F1sin30°=(800#)(0.50)=400#

F2 + F2cos30°=(1200#)(0.866)=1039.2# - F2sin30°=-(1200#)(0.50)=-600#

Alternate way to find the resultant R:

€

Rx = ΣFx = +692.8#+1039.2#= +1732# (→)Ry = ΣFy = −500#+400#−600#= −700# (↓)

R = 1732#( )2+ −700#( )2

=1868#

€

tanθ =Ry

Rx

;

θ = tan−1 7001732

⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 0.404( ) = 22°

€

sinθ =Ry

R;

R =Ry

sinθ=

700sin22°

=700

0.375=1867#

2.47

Component Fx Fy

AD = 90kN

€

−12

⎛

⎝ ⎜

⎞

⎠ ⎟ 90kN( ) = −

90kN2

= −63.6kN

€

−12

⎛

⎝ ⎜

⎞

⎠ ⎟ 90kN( ) = −63.6kN

BD = 45kN

€

− 35 45kN( ) = −27kN

€

− 45 45kN( ) = −36kN

CD = 110kN

€

+ 110kN( ) cos 30° = +95.3kN

€

− 110kN( )sin 30° = −55kN

€

Rx = Fx∑ = +4.7kN

€

Ry = Fy∑ = −154.6kN

€

R = Rx2 + Ry

2 = 4.7( )2+ 154.6( )2

= 154.7kN

€

θ = tan−1 Ry

Rx

⎛

⎝ ⎜

⎞

⎠ ⎟ = tan−1 154.6

4.7⎛

⎝ ⎜

⎞

⎠ ⎟

θ = tan−1 32.9( ) = 88.3°

D

AD = 90kN k BD = 45kN CD = 110kN

34

11

30°

x

y

2.47

x

y

θ = 88.3˚

R = 154.7kN

Ry = -154.6kN

Rx = +4.7kN

Resultant

D

AD = 90kN k BD = 45kN CD = 110kN

34

11

30°

x

y

2.47

x

y

θ = 88.3˚

R = 154.7kN

Ry = -154.6kN

Rx = +4.7kN

Resultant


2.19

€

d2x = d2 cos 45° = 14' 0.707( ) = 9.9'

d2y = d2 sin 45° = 14' 0.707( ) = 9.9'

d1x = d1 cos 20° = 10' 0.94( ) = 9.4'

€

ΣM C = 0[ ] + 250# d1x( ) −100# d2y( ) − F d2x( ) = 0

F =250# 9.4'( ) −100# 9.9'( )

9.9'= 137.4#

€

ΣFx = 0[ ] + Ax − 3.83kNFx( )

= 0 Ax = 3.83kN

ΣFy = 0[ ] + Ay − 3.21kNFy( )

= 0 Ay = 3.21kN

2.48

2.48

C

250#100#

F

20° 45°

d1x d2x

d2y

d2 = 14

’

d1 = 10’

2.49

2.50

T2 = 700#

T2x

T2y6’

4’

5’

30’

T1 = 500#

T1x

T1y

A

MA = +T2x(30’) + T2y(6’) – T1y(35’) – T1y(4’) =0

MA = 689.5#(30’) + 121.8#(6’) – 483#(35’) – 129.5#(4’) = +3990#-ft

Force Fx_________________ Fy_________________

T1 = 500# (500#)cos15o = 483# (500#)sin15o =129.5#

T2 = 700@# (700#)cos10o = 689.5# (700#)sin10o = 121.8#2.49

4

3

Ax

Ay

MRA

1.33m

1.1m 1m 1m

F = 5kNFy = 3.21kN

Fx = 3.83kN

40°

€

MA = −3.21kN 3.1m( ) − 3.83kN 1.33m( )MA = −9.95kN − m − 5.09kN − m = −15.04kN − m


2.20

2.51

R = ΣFy = 10# + 7# + 6# - 18# = +5#

MO = +(7#)(4”) + (6#)(9”) – (18#)(17” = - 224#-in.

R(x) = 224 #-in;

€

x =224#−in

5# = 44.8"

ω = 30N/m

Assume the member weight is located at the center of the length.

Weight of wood member:

R = ΣFy = -100N – 90N + 150N = -40N

MO = - (100N)(1m) – (90N)(1.5m) + (150N)(3m) = +215 N-m

R(x) = MO

€

∴ x =215N − m

40N= 5.4m

For a 40N force to produce a moment directed counter-clockwise, the R = 40N force will be at an imaginary location where x = 5.4m to the left of the origin.

2.52

2.51

y

x

10#7# 6#

18#

4” 5” 8”origin

O

origin

y

x

R = 5#

44.8”O

x

2.51

y

x

10#7# 6#

18#

4” 5” 8”origin

O

origin

y

x

R = 5#

44.8”O

x

2.52

x

y

origin1m 2m

100N

150N

30N/m

1m

x

y

origin

100N

150N

W = 90N (beam wt.) 1.5m

yR = 40N

x = 5.4m 3morigin

O

O

O

2.52

x

y

origin1m 2m

100N

150N

30N/m

1m

x

y

origin

100N

150N

W = 90N (beam wt.) 1.5m

yR = 40N

x = 5.4m 3morigin

O

O

O

2.52

x

y

origin1m 2m

100N

150N

30N/m

1m

x

y

origin

100N

150N

W = 90N (beam wt.) 1.5m

yR = 40N

x = 5.4m 3morigin

O

O

O


2.21

2.53

origin

y

x

A B200#100#

400#

4’ 4’ 4’ 4’

20#/ft

originx

A B200#100#

400#

W = 320#

y

Total beam weight equals (20#/ft)(16’) = 320#at the center of the beam length.

For the beam to remain stationary and horizontal,the moments taken about points A and B shouldbe balanced by the opposing moments due to Band A respectively, resulting in no resultant moment.

€

ΣMA = 0[ ] + 200# 4'( ) +100# 8'( ) − 320# 8'( ) − 400# 12'( ) − B 16'( ) = 0

∴ B = 360#

ΣMB = 0[ ] + 400# 4'( ) + 320# 8'( ) −100# 8'( ) − 200# 12'( ) − A 16'( ) = 0

∴ A = 60#

Ry = ΣFy = +A +100# +100# − 320# − 400# + B = 0

= +60# + 200# +100# − 320# − 400# + 360# = 0

Force Fx_______ Fy_________

AB

€

−1213

AB

€

+5

13AB

AC

€

−45

AC

€

−35

AC

W 0 -5k

€

ΣFx = 0[ ] − 1213

AB−45

AC = 0; 1213

AB = −45

AC = 0

∴ AB = −1315

AC

ΣFy = 0[ ] + 513

AC −35

AC − 5k = 0; + 513

−1315

AC⎛

⎝ ⎜

⎞

⎠ ⎟ −

35

AC = +5k

−AC3

− 3AC5

= 5k; AC = −5.36k compression( )

AB = −1315

−5.36k( ) = +4.64k tension( )

2.54

2.54

125

34

AB

AC W = 5k

ABy

ABx

ACx

ACy


2.22

2.55

2.55

3

4

4

3

60°

y

x

F = 2kN

Fx = 1.2kN

Fy = 1.6kN

AC

ACx

ACy

BC

BCx

BCy

3

4

60°

4

3

F = 2kN

BC = 2.1kN

AC = 0.27kN

Tip-to-tail

2.55

3

4

4

3

60°

y

x

F = 2kN

Fx = 1.2kN

Fy = 1.6kN

AC

ACx

ACy

BC

BCx

BCy

3

4

60°

4

3

F = 2kN

BC = 2.1kN

AC = 0.27kN

Tip-to-tail

Force Fx_______________ Fy__________________

AC

€

−35

AC

€

−45

AC

BC

€

−BC cos 60° = −0.50BC

€

+BC sin 60° = +0.866BC

F=2kN

€

+35

2kN( ) = +1.2kN

€

−45

2kN( ) = −1.6kN

€

ΣFx = 0[ ] − 35

AC − 0.50BC +1.2kN = 0; 35

AC + 0.50BC = 1.2kN

ΣFy = 0[ ] − 45

AC + 0.866BC −1.6kN = 0; − 45

AC + 0.866BC = 1.6kN

Solving simul tan eously;BC = 2.1kN (compression)AC = 0.27kN (tension)

Force Fx______________ Fy_______________

BA

€

−1213

BA

€

−5

13BA

DB

€

+DBsin 30° = +0.50DB

€

+DB cos 30° = +0.866DB

W 0 -800#

€

Rx = ΣFx = 0[ ] − 1213

BA + 0.50DB = 0; DB =2413

BA

Ry = ΣFy = 0[ ] − 513

BA + 0.866DB − 800# = 0

−5

13BA + 0.866 24

13BA

⎛

⎝ ⎜

⎞

⎠ ⎟ = 800#

BA = 658.2# DB =2413

⎛

⎝ ⎜

⎞

⎠ ⎟ 658.2#( ) = 1215.2#

2.56

x

y

BA

DB W = 800#

5

12

30°

2.56


2.23

2.57

2.57

x

y

W

CACB

CAx CBx

CBy

CAy

45° 30°

CBy

2.58

3

45

12

BE

BCAB = 1560#

y

xABx

ABy

BCx

BCy

3

4

45°

CB = 1800#CD

W

CDy

CDx CBx

CBy

y

x

Force Fx_______________ Fy__________________

CA

€

−CA cos 45° = −0.707CA

€

CAsin 45° = +0.707CA

CB

€

+CB cos 30° = +0.866CB

€

+CBsin 30° = +0.50CB

W 0 -W

€

ΣFx = 0[ ] − 0.707CA + 0.866CB = 0; CA =0.866CB

0.707= 1.22CB

This relationship indicates the CA > CB, therefore, CA = 1.8kNThen, CB = (1.8kN)/1.22 = 1.48kN

€

ΣFy = 0[ ] 0.707CA + 0.50CB − W = 0

W = 0.707 1.8kN( ) + 0.50 1.48kN( ) = 2.0kN

2.58

2.57

x

y

W

CACB

CAx CBx

CBy

CAy

45° 30°

CBy

2.58

3

45

12

BE

BCAB = 1560#

y

xABx

ABy

BCx

BCy

3

4

45°

CB = 1800#CD

W

CDy

CDx CBx

CBy

y

x

Force Fx______________ Fy______________

AB=1560#

€

+1213

1560#( ) = 1440#

€

−5

131560#( ) = −600#

BE 0 +BE

BC

€

−45

BC

€

−35

BC

€

ΣFx = 0[ ] +1440# −45

BC = 0; BC = 1800#

ΣFy = 0[ ] − 600# + BE −35

1800#( ) = 0; BE = 1680#

Joint B:


2.24

2.58 cont’d

2.57

x

y

W

CACB

CAx CBx

CBy

CAy

45° 30°

CBy

2.58

3

45

12

BE

BCAB = 1560#

y

xABx

ABy

BCx

BCy

3

4

45°

CB = 1800#CD

W

CDy

CDx CBx

CBy

y

x

Force Fx_____________ Fy______________

CD -0.707CD +0.707CD

CB = 1800#

€

+45

1800#( ) = 1440#

€

+35

1800#( ) = +1080#

W 0 -W

€

ΣFx = 0[ ] − 0.707CD +1440# = 0; CD = 2037#

ΣFy = 0[ ] + 0.707 2037#( ) +1080# − W = 0; W = 2520#

Joint C:

2.59

€

ΣM A = 0[ ] − 500# 10'( ) +5B13

10'( )Bx( )

+12B13

24'( )By( )

= 0

50B13

+288B

13= 5000#−ft ; B = 192.3#

Bx = 74#; By = 177.5#

€

ΣFx = 0[ ] + Ax − 74#Bx( )

= 0; Ax = 74#

ΣFy = 0[ ] + Ay − 500# +177.5#

By( )= 0; Ay = 322.5#

2.59

34

B3

4 10’

500#10’

24’

Ax

Ay

By

Bx

26’


2.25

€

ΣFx = 0[ ] Bx = 0

ΣM B = 0[ ] − Ay 4.5m( ) +1.8kN 2.5m( ) = 0; Ay = 1kN

ΣFy = 0[ ] +1kN −1.8kN + By = 0; By = 0.8kN

€

ΣFx = 0[ ] Cx = 0

ΣFy = 0[ ] − 0.8kN − 2.7kN + Cy = 0; Cy = 3.5kN

ΣM C = 0[ ] − M RC + 2.7kN 3m( ) + 0.8kN 6m( ) = 0 M RC = 8.1kN − m+ 4.8kN − m = 12.9kN − m

2.60

A

1.8kN 2.7kN

hinge

B C

AyCy

Cx

MRC

2.5m2m 3m 3m

1.8kN

A

AyBy

Bx2m 2.5m

Bx

By

2.7kN

MRC

Cx

Cy

3m3m

2.60

2.60

A

1.8kN 2.7kN

hinge

B C

AyCy

Cx

MRC

2.5m2m 3m 3m

1.8kN

A

AyBy

Bx2m 2.5m

Bx

By

2.7kN

MRC

Cx

Cy

3m3m

Beam AB:

Beam BC:

2.61

2.61

5k

4k

3k

2k 2k3k

Ay

Ax

By

20’

20’

20’ 20’ 20’

10’ 10’

2.62

Ay

Ax300#240#

180#

By

4’ 6’ 7’

4’

5’

5’

200#

80#

60#

By

Dy

Dx

Cy

€

ΣFx = 0[ ] − Ax + 4k = 0; Ax = +4k ←( )ΣM A = 0[ ] − 2k 20'( ) − 3k 40'( ) − 2k 60'( ) − 3k 40'( ) + 4k 20'( ) + By 80'( ) = 0

∴ By = +4k ↑( )ΣFy = 0[ ] + Ay − 2k − 3k − 2k − 3k + 4k

By( )= 0

∴ Ay = +6k ↑( )


2.26

2.61

5k

4k

3k

2k 2k3k

Ay

Ax

By

20’

20’

20’ 20’ 20’

10’ 10’

2.62

Ay

Ax300#240#

180#

By

4’ 6’ 7’

4’

5’

5’

200#

80#

60#

By

Dy

Dx

Cy

€

ΣM A = 0[ ] − 300# 5'( ) + By 8'( ) = 0; By = +187.5# ↑( )ΣFx = 0[ ] + Ax −180# = 0; Ax = +180# →( )ΣFy = 0[ ] +187.5# − 240# + Ay = 0; Ay = +52.5# ↑( )

€

ΣM D = 0[ ] + 200# 4'( ) + Cy 6'( ) −187.5# 9'( ) − 80# 13'( ) = 0

∴ Cy = +322# ↑( )ΣFx = 0[ ] + Dx − 60# = 0; Dx = +60# →( )ΣFy = 0[ ] − 200# + Dy + 322# −187.5# − 80# = 0

∴ Dy = +145.5# ↑( )

2.62

Upper Beam:

Lower Beam:

Documents

Fourth Edition - Frat Stock · Chapter 5 Strength of Materials • Tension, Compression and shear stress pg 5.1 – 5.2 • Deformation and strain pg 5.3 • Elasticity, strength