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.
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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
Statics and Strength of Materials
For Architecture and Building Construction
Fourth Edition
Barry Onouye
Pearson/Prentice Hall
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
Table of Contents
Chapter 2 Statics
• Graphical addition of vectors
• Resolution of forces: x and y components
• Vector addition by components
• Moment of a force
• Varignon’s theorem
• Moment couples
• Equilibrium of concurrent forces
• Equilibrium of rigid bodies
• Supplementary problems
pg 2.1 - 2.2
pg 2.2 - 2.3
pg 2.3 - 2.6
pg 2.6 - 2.7
pg 2.7 – 2.8
pg 2.9
pg 2.10 – 2.13
pg. 2.13 – 2.16
pg 2.16 – 2.26
Chapter 3 Analysis of Determinate Systems
• Cables with concentrated loads
• Equilibrium of rigid bodies with distributed
loads
• Planar trusses – method of joints
• Truss analysis – method of sections
• Diagonal tension counters
• Zero-force members
• Pinned frames – multi-force members
• Supplementary problems
• Retaining walls
Chapter 4 Load Tracing
• Gravity load trace
• Lateral load trace
pg 3.1 – 3.3
pg 3.4 – 3.5
pg 3.6 – 3.8
pg 3.8 – 3.10
pg 3.10 – 3.12
pg 3.12
pg 3.13 – 3.15
pg 3.16 – 3.28
pg 3.29 – 3.32
pg 4.1 – 4.8
pg 4.8 – 4.11
Chapter 5 Strength of Materials
• Tension, Compression and shear stress
• Deformation and strain
• Elasticity, strength and deformation
• Thermal stress and deformation
• Statically indeterminate, axially loaded
members
pg 5.1 – 5.2
pg 5.3
pg 5.3 – 5.4
pg. 5.4 – 5.5
pg 5.5 – 5.6
Chapter 6 Cross-Sectional Properties
• Centroids
• Moment of inertia
• Moment of inertia for composite sections
pg 6.1 – 6.3
pg 6.3 – 6.7
pg 6.7 – 6.9
Chapter 7 Bending and Shear Diagrams
• Equilibrium method for shear and moment
diagrams
• Semi-graphical method for shear and
moment diagrams
pg 7.1 – 7.4
pg 7.5 – 7.10
Chapter 8 Bending and Shear Stress in Beams
• Bending stress
• Bending and shear stresses
• Deflection in Beams
pg 8.1 – 8.5
pg 8.6 – 8.12
pg 8.13 – 8.15
Chapter 9 Column Analysis and Design
• Euler buckling loads and stresses
• Axially loaded steel columns - analysis
• Design of steel columns
• Axially load wood columns
pg 9.1 – 9.2
pg 9.3 – 9.4
pg 9.5 – 9.6
pg 9.6 – 9.9
Chapter 10 Structural Connections
• Bolted steel connections
• Framed connections
• Welded connections
pg 10.1 – 10.3
pg 10.3
pg 10.4 – 10.5
Prob. 2.3
y
x
F1
17
y
10
=
0
R
2.1#
50˚
#
200
Fa
y
x
=
10
0#
=
R
50˚
20˚
Fb =
0#
0#
72
R = 173#
θ = 50˚ from horiz.
φ = 40˚ from vert.
40˚
#
200
40˚
60
40˚
R = 173#
θ = 50˚ from horiz.
φ = 40˚ from vert.
or
17
20˚
0#
40˚
Fb =
x
50˚
10
0#
=
=
17
3#
=
x
F2
Fa
40˚
y
60
=
0#
=
F2
10
=
R = 930#
=
F1
=
Fa
or
#
200
Fb = 20˚
40˚
y
17
3#
x
50˚
R
40˚
3#
=
3#
Chapter 2 Problem Solutions
2.3
72
0#
Prob. 2.3
R = 930#
y
R
Fa
#
200
Fb = 20˚
40˚
2.4
Prob. 2.4
4
3
x
T3 = 910#
Prob. 2.4
2.2
4
3
3
ob. 2.2
T3 = 910#
50
00
#
4
35˚
F2 =6kN
θ = 5˚
#
00
50
x
10˚
#
910
x
5
=5
R = 10.2kN
30˚ F3 =5kN
12
T2
F1 = 3kN
#
R = 10.2kN
910
30˚ F3 =5kN
F1 = 3kN
=
y
T1
10˚
=5
35˚
3
4
T2
F2 =6kN
R = 10,000#
Prob. 2.2
R = 10,000#
T1
=
y
θ = 5˚
2.1
12
5
Prob. 2.5
Prob. 2.6
2.5
2.6
y
T2 =
240
F=1000 lb.
lb.
3
x
4
20°
10°
x
A
Fx
W = 1,200 lb.
T1
Fy
5
B
F=1000 lb.
3
5
4
θ
By similar triangles:
Fx Fy F
=
=
5
4
3
∴ Fx =
Fy =
€
sinθ =
€
4
F=
5
3
F=
5
3
5
4
5
3
5
(1000#) = 800#
(1000#) = 600#
and
cos θ =
∴ Fx = F cos θ = (1000#)
4
5
( 45 ) = 800#
Fy = F sinθ = (1000#)( 35 ) = 600#
Prob. 2.7
€
y
y
Tx
x
Tx
x
10°
Ty
Ty
10°
T
2.2
T
Prob. 2.7
2.9
y
F1y = +F1 cos 30° = 10k (0.866) = 8.66k
y
F1x = +F1 sin 30° = 10k (0.50) = 5k
30
y
Problem
Tx
x Solutions: 2.3.11-2.3.15
Tx
F2 = −F2x = −12k
1
18k
F3x = +
(F3 ) = +
2
2
1
18k
F3y = −
(F3 ) = −
2
2
x
F2 =12k
O
10°
Prob. 2.3.11
F1 =10k
x
Ty
T
4
250N
= 254N
0.985
y
y
F2 =12k
O
x
k
10°
1
2.8
Prob. 2.3.13
θ = tan−1
Px
F2 = 12k
4
(
4
12.65
.0F30
23 =12k
k
€
x
x
ResultantR
=7
θ = 35.1°
F3 =18k .03k
1
R y 4.07
=
= 0.710
R x 5.73
y
θ = tan−1 (0.710) = 35.4° from
horizontal
R
sinθ = y
R
Ry
Ry
R=
=
sinθ sin 35.4°
4.07k
∴ R=
= 7.03k
(0.579)
Rx = +5.73 k
x
θ = 35.1°
R
=7
.03
k
)
Py = P
y
k
18
.03
k
x
3
Py
4
P = 600 lb.
Px
12
A
30
Qy
x
Q = 600 lb.
9
B
k
18
=7
θ = 35.1°
1
=
R
1
F3
€
F2 = 12k
12
30
(12.65
) = ( 300#)( 0.949) = 285#
Prob. 2.3.14
Purlin Detail
€
T
= ( 300#)( 0.316) = 94.9#
1
=
12
€
F3
θ
P
=P
1x
y
(124 ) = 18.43°
0k
θ
Ty
=1
Rafter
=7
k
Resultant
P=300 lb.
F1
Py
R1
Tx
F3 =18k
x
θ = 35.1°
1
18
1
30
=
1
€
€
F2 = 12k
O
x
tanθ =
y
F1 =10k
Rx = +5.73 k
F3
cos10°
=
y
.03
0k
Ty
Prob. 2.3.12
30
=7
=1
∴ T=
F1 =10k
0k
Ty = T cos10°
€
=1
B
y
R
18k
R x = ΣFx = +5k −12k +
= +5.73k
2
Resultant
18k
R y = ΣFy = +8.66k −
= −4.07k
2
F3 =18k
T
x
TFxx= T sin10°
A
1
4
3
R
=7
θ = 35.1°
.03
k
Graphical solution using
the tip-to-tail method
Qx
16'
16'
x
θ = 35.1°
F1
3
1
10°
F1
F
Ty y
Ry = -4.07 k
F=1000 lb.
Rx = +5.73 k
Ry = -4.07 k
y
Ry = -4.07 k
2.7
2.3
2.10
2.10
60˚
40˚
TACx
TABy
40˚
0N
y
TAC=800N
y
+TABx = +TAB cos 40° = +0.766TAB
R x = ΣFx = −( 0.5)(800N) + θ( 0.766)( 600N) = 59.6N
Rx = 59.6N
θ
R y = ΣFy = −( 0.866)(800N) − ( 0.642)( 600N) = −1078N
θ = 86.8˚
⎛ Ry ⎞
−1 ⎛ 1078 ⎞
−1 φ = 3.2˚
θ = tan−1⎜
⎟ = tan (18.1) = 86.8°
⎟ = tan ⎜
⎝ 59.6 ⎠
⎝ Rx ⎠
⎛R ⎞
⎛ 59.6 ⎞
−1
φ = tan−1⎜⎜ x ⎟⎟ = tan−1 ⎜
⎟ = tan ( 0.055) = 3.2°
⎝
⎠
R
1078
y
⎝
⎠
θ = 86.8˚
y
x
40˚
60˚
TA
B=
60
60˚
0N
Rx = 59.6N
0N
80
60
0N
x
40˚
TA
B=
R = 1079N
60
0N
φ = 3.2˚
R = 1079N
φ = 3.2˚
y
Rx = 59.6N
Scale 1mm = 10N
Scale 1mm = 10N
Scale 1mm = 10N
φ = 3.2˚
θ
θ = 86.8˚
φ = 3.2˚
R = 1079N
Ry = 1078N
2
R = 59.6 2 +R1078
= 1079N R = 1079N
y = 1078N
φ
B=
φ = 3.2˚
x
−TABy = −TAB sin 40° = −0.642T
AB
y
€
R = 1079N
TAB=600N
TACy
−TACy = −TAC sin 60° = −0.866TAC
€
TAB=600N
TACy
y
TACy
−TACx = −TAC cos 60° = −0.5TAC
TA
C=
TABy
TAC=800N
x
x
TA
TAB=600N
x
y
80
TABy
€
TABx
y
Graphical
x Solution:
40˚
60˚
60˚
TAC=800N
y
40˚
C=
60˚
x
80
0N
TABx
C=
TACx
x2.10 cont’d
TA
2.10
TABx
TA
2.10
TACx
φ
R = 1079N
Ry = 1078N
φ
2.4
2.11
2.12
2.12
ax
is o
x
fb
y
oo
m
2.11
Fy
€
fb
oo
m
Wx
#
F1
2.12
30˚
R = resultant reaction
W=200#
#
156
F=
-F2x = -F2cos 25˚
W=200#
ax
is o
25°
93
30˚
R
Fx
−Fx = −F cos 40° = −0.766F
+Fy = +F sin 40° = +0.642F
F2
A
=2
Wy
40˚
F
−Wx = −W cos 30° = −0.866W
−Wy = −W sin 30° = −0.50W
y
y
-F = -F sin 25˚
F2
2y
2
Tip-to-tail
Since the
Scale:
1” resultant
= 100# must be vertical,
R y = ΣFy = 0;
− 0.50( 200#) + 0.642F = 0
-F2x + F1 = 0
100#
= 156#
0.642
R = R x = ΣFx = −0.866( 200# ) − 0.766(156# )
∴ F=
25˚
is o
x
fb
oo
m
R = −173# −120# = −293#
€
A
Then: Rx = ΣFx = 0
ax
y
F2cos25˚ = F1
From this equation,
it is seen that F1 =2.95kN
7kN of F , therefore, R
Is only aFfraction
1=
2
F2 = 7kN.
25°
F1
R = resultant reaction
F2 = 6.34kN
Then, F1 = F2cos25˚ = (7kN)(0.706)
F1 = 6.34kN and F2 = 7kN
R = F2y = (7kN)(sin25˚) = 2.95kN.
Wy
25˚
93
=2
30˚
30˚
W=200#
R
Fx
#
40˚
F1
Wx
W=200#
F=
#
156
Tip-to-tail
Scale: 1” = 100#
kN
R =2.95kN
=7
F2 = 6.34kN
Graphical solution
2.5
2.13
2.14
2.14
2.13
2.13
y
x
30°
A
4’
5’
−T2x = −T2 cos 60° = −0.50T2
−T2y = −T2 sin 60° = −0.866T2
F = 8k
y
T2
−T1x = −T1 cos 30° = −0.866T1
−T1y = −T1 sin 30° = −0.50T1
45°
60°
T1
2.14
W = 25#
A
x
MA = -20#(5’) + 25#(4’) = -100 #-ft + 100 #-ft = 0
4’
The box is just on the verge of tipping over.
60°
T1 −Fy = −F sin 45° = −.707(8k) = −5.65k
F = 8k
But TT1 = T2
5’
2.15
2
For the resultant to be vertical,
W = 25#
y
€
R x = ΣFx = 0
∴ − 0.866T − 0.50T + 5.65k = 0
T = 4.14k
T
x
y
TR
x
€
TR
A
B
T
2.15
2.15
8k
2m
1m
W=700N
C
x
800N
A
8k
2m
AB = BC
so that x-components
cancel
C
B
x
A
R = 11.3k
T
W=700N
A
800N
R = R y = ΣFy = −0.50( 4.14k ) − 0.866( 4.14k − 5.65k) = −11.3k
T
F = 20#
8’
30°= +F cos 45° = +0.707
+F
45° (8k) = +5.65k€
x
A
F = 20#
8’
A
( ΣM A = 0)
+800N(1m) − 700N( x ) = 0
(800N)(1m) = 1.14m
x=
( 700N)
Using the parallelogram law
R = 11.3k
AB = BC
so that x-components cancel
Using the parallelogram law
1m
Since x > 1m, the man is OK.
€
2.6
2.16
2.18
8”
W = 200#
P
A
F
A
F
17”
15”
12”
12
F = 39#
5
5
36”
14”
4”
Fy = 15#
[ ΣM A = 0]
[ ΣM B = 0]
12
Fx = 36#
− 200# (12") = F( 26") = 0;
+ F( 4") − P( 36") = 0;
∴ F = 92.3#
∴ P = 10.3#
2.19
ΣM A = −36# (15") + 15# (8") =
€
−( 540# −in) + (120# −in) = −420# −in.
2.19
5kN
€
10kN
2.17
9kN
18.8”
W = 100#
26.3”
P
8kN
20˚
8kN
28”
A
6”
8kN
[ ΣM A = 0]
4m
4m
4m
24m
4m
4m
4m
20”
A
M A = −(5kN)(24m) − (10kN)(20m) − (9kN)(16m) − (8kN)(12m)
ΣM A = −W(18.8") = −100# (18.8") = −1880# −in. ( clockwise)
−(8kN)(8m) − (8kN)( 4m)
− 100# (18.8") + P( 45.1") = 0
M A = −(120kN − m) − (200kN − m) − (144kN − m)
1880# −in.
P=
= 41.7#
45.1 in.
€
B
−(96kN − m) − (64kN − m) − ( 32kN − m)
M A = −656kN − m
2.7
€
2.22
2.20
2.22
5
Fx
5’
A
B
12’
30˚
Fx = 12
13 (1300# ) = 1200#
D
C
12’
Fx
Fy = 135 (1300# ) = 500#
m
12
F = 1.5kN
Fy
Fy
0m
F = 1300#
dy
20
2.20
€
Fx = F cos 30° = (1.5kN)(0.866) = 1.3kN
M B = −Fx (5') + Fy (0) = −1200# (5') = −6000# −ft
A
M C = −Fx (5') + Fy (12') = −1200# (5') + 500# (12')
Fy = F sin 30° = (1.5kN)(0.50) = 0.75kN
60˚
120mm
M C = −(6000# −ft ) + (6000# −ft ) = 0
d x = dcos60° = (200mm)(0.50) = 100mm
dx
d y = dsin60° = (200mm)(0.866) = 173mm
M A = −Fx (d y ) + Fy (120mm + d x )
€
M A = −1.3kN(173mm) + 0.75kN(220mm€
)
2.21
M A == −60kN − mm = −0.06kN − m
2.23
€
B
25˚
2.23
F = 30#
Fy
25˚
6”
A
Fx
3’
Tx
4”
Fy = F sin25° = ( 30# )(0.423) = 12.7#
dy
6”
€
M A = Fy (6") = (27.2# )( 4") = 108.8 #−in
W
Ty
T = 2000#
Ty = T sin 30° = 2000# ( 0.50) = 1000#
d x1 = 7' cos 60° = 7' ( 0.50) = 3.5'
d x 2 = 10' ( 0.50) = 5'
d y = 7' sin 60° = 7' (.866) = 6.06'
7’
60˚
CCW
M B = −Fy (6") − Fx (2") = −(12.7# )(6") − (27.2# )(2") = −130.6
€
30˚
Fx = F cos25° = ( 30# )(0.906) = 27.2#
A
Tx = T cos 30° = 2000# (.866) = 1732#
B
€
#−in.
dx1
[ ΣM C = 0]
Tx d y − Ty ( d x1) − W( d x 2 ) = 0
( )
(1732#)( 6.06') − (1000#)( 3.5') − W( 5') = 0
5W = 10,500 #−ft − 3500 #−ft
dx2
W=
€
€
7000 #−ft
= 1400#
5'
2.8
2.24
2.26
2.26
2.24
A
8’
7.07 k
10k
B
M = 90kN(0.305m) = 27.45kN-m
90kN
7.07 k
7.07 k
3’
8’
.
11
6’
8’
10k
7.07 k
4k
CL
8’
M A = −(10k )(11.3') − ( 4k )(14') = −113k−ft − 56 k−ft = −169 k−ft
M B = −(10k )(11.3') − ( 4k )(14') = −169 k−ft
CL
2.27
2.27
2.25
2’
2’
RBx = 25#
B
R Ax and R Bx form a couple
Fy = 69.6#
M couple 1 = 25# (12') = +300 #−ft
R Ay and 150# man form a couple
150#
€6’
C
M couple2 = −(150# )(2') = −300 #−ft
A
F = 85#
12"
3"
B
30°
55°
Fx = 48.7#
4"
€
90kN
180mm
125mm
€
Fx = (85 lb.) cos 55° = 48.8 lb.
6’
RAx = 25#
A
RAy = 150#
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.
Since the moment due to a couple is constant,
€
MA = MB = MC = +300#-ft - 300#-ft = 0
€
2.9
2.28
2.29
2.28
2.29
y
B
F = 500N
1000#
AC
Cx
45º
Ay
RC
Cy
Ax = Acos60° = 0.50A
Ay = Asin60° = 0.866A
[ΣFx = 0]
[ΣF
y
Fy____________________
F
500N
-500cos20º = -470N
-500sin20º = -171N
AC
?
+ACsin10º = +0.174AC
+ACcos10º = +0.984AC
+BCsin30º = +0.50BC
-BCcos30º = -0.866BC
?
y
Cy
Ay
Ax
€
= 0] −171N + 0.984AC − 0.866BC = 0 ... Eq (2)
0.866 × [0.50BC + 0.174AC = 470N] ... Eq (1)
0.50 × [−0.866BC − 0.984AC = 171N] ... Eq (2)
x
Therefore,
∴ 1.37A = 1000#
1000#
− 470N + 0.174AC + 0.50BC = 0 ... Eq (1)
Solving equations (1) and (2) simultaneously,
RA
0.707(0.707A) + 0.866A = 1000#B
A = 732#
Fx__________________
[ΣF
= 0] + C y + Ay C
−1000#
=0
x
C = 0.707( 732# ) = 518#
Magnitude
[ΣFx = 0]
y
− C x + Ax = 0
−0.707C + 0.50A = 0
0.50A
= 0.707ARC
C=
0.707
Force
BC
C x = Ccos 45° = 0.707C
C y = Csin 45° = 0.707C
€
BC
Free-body diagram of joint C
RA
€
10˚
30˚
60º
Ax
x
C
20˚
€
+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)
2.10
2.30
2.30
2.30
y
F2 = 150#
2.30
F1 = 50#
3
P
4
25˚
€
F1 = 50#
y
α
x
3
F1 = 50#
α
x
Magnitude
Fx__________________
Fy__________________
F1
50
-50 cos25° = −45.3
+50#sin25° = +21.1
F2
150#
+150#(3/5) = +90# W = 60#
+150#(4/5) = +120#
W
60#
0
-60#
P
?
+Pcosα
+Psinα
#
#
#
[ ΣFx = 0]
F2 = 150#
− 45.3# +90# +P cos α = 0 ..... (1)
[ΣFy = 0]
+ 21.1# +120# −60# +P sinα = 0 .... ( 2)
4
α =61.1˚
P = 925#
−81.1#
P=
....
sinα
−44.7#F1 =−81.1#
= 50#
cos α
sinα 25˚
3
F1 = 50#
4
25˚
α =61.1˚
W = 60#
25˚
x
α =61.1˚
W = 60#
W=6
W = 60#
P = 925#
P = 925#
Final Free Body Diagram
W = 60#
F2 = 150#
F2 = 150#
P = 925#
W = 60#
3
W = 60#
4
3
α =61.1˚
P = 925#
4
sinα
−81.1#
= tan α =
= +1.81Graphical check
cos α
−44.7#
α = tan−1(1.81) = 61.1°
F2 = 150#
4x
4
( 2)
x
y
3
α
3
−44.7#
.... (1)
cos α
Equating :
#
y
P
3
25˚
25˚ F1 = 50#
Force
P=
€
F1 = 50#
P
4
W = 60#
25˚
€
F2 = 150#
−44.7#
−44.7#
y =
= −92.5#
cos 61.1°
0.483
F2 = 150#
Note that the negative sign for P
indicates ythat it 3was initially assumed
in the wrong direction.
4
F2 = 150#
P=
F1 = 50#
P = 925#
25˚
α =61.1˚
F1 = 50#
25˚
α =61.1˚
Graphical check
2.11
€
Graphical check
2.31
2.32
2.31
2.32
W = 2.5kN
y
T
5˚
B
30˚
2.31
Tx = Tsin5° = 0.087T
Ty = Tcos5° = 0.996T
Px = P cos20° = 0.940P
Py = P sin20° = 0.342P
x
20˚
€
A
P
75˚
W = 2000#
FBD of the sphere
W = 2.5kN
[ΣFx = 0]
[ΣFx = 0] Acos 75° − Bsin60° = 0
B(0.866)
A=
= 3.346B
0.259
[ΣF
y
€
= 0] Asin
+ Bcos60° − 2.5kN = 0
B 75°
= 0.67kN
90˚B
3.346B(0.966) + 0.50B = 2.5kN
A = 2.24kN
30˚
3.732B
= 2.5kN
B = 0.67kNA
90˚
75˚
A = 2.24kN
15˚
substituting;
0.996(10.8P) − 0.342P = 2000#
2.33
y
P = 192#
C
x
AB = T = 10.8(192# ) = 2074#
3
CD
€
DC
CA
€
y
4
60°
60˚
FBD of the sphere
Forces exerted bt the sphere
onto the smooth surface.
− 0.087T + 0.94P = 0
0.94P
T=
= 10.8P
0.087
[ΣFy = 0] + 0.996T − 0.342P − 2000# = 0
4
DE
3
BC
FBD(a)
B = 0.67kN
90˚
15°
D
FBD(b)
W = 200 lb.
A = 2.24kN
15˚
90˚
60˚
Forces exerted bt the sphere
onto the smooth surface.
2.12
2.34
2.33
2.34
2.33
3’
y
C
x
y
4
60°
Bx
300#
3
CD
4
DC
CA
FBD(b)
FBD(a)
Force
_________Fx_________
DC
−
DE
+DE cos15° = +0.966DE
0
∑ Fx = − 0.80DC + 0.966DE = 0 ;
Ax
DC = 1.21DE
€
Writing equations of equilibrium for FBD(b);
DE = +203lb.;
CA
−CA cos 60° = −0.50CA
0
∑ Fx = − 0.50CA + 197lb. = 0 ;
€
[ΣF
−CA sin 60° = −0.866CA
y
∑ Fy = − 0.866( 394lb.) + BC − 148lb. = 0 ;
− 40kN(2.5m) − 50kN(5.0m) + By ( 7.5m) = 0
= 0] + A − 40kN − 50kN + 46.7kN = 0
No horizontal reaction is necessary for this load case.
€
2.36
BC = +489lb. (C)
3k
2k
€
€
B
∴ = 43.3KN
+BC
∴ CA = 394lb. ( T)
D
∴ By = 46.7kN
3
− ( 246lb.) = −148lb.
5
€
C
[ΣM A = 0]
FxFy
2.5m
2.5m
2.5m
A
DC = 1.21( 203lb.) = 246lb.
€
CD
50kN
40kN
_____________________
0.985DE = 200lb. ;
4
( 246lb.) = +197lb.
5
€
2.35
-200 lb.
€
+
= 0] + Ay − 300 # = 0
∴ Ay = +300 #
2.35
( DCx )
€
[ΣF
Ay
∑ Fy = + 0.60(1.21DE)+ 0.259DE − 200lb. = 0
€BC
∴ Ax = +150 #
4’
W = 200 lb.
+DE sin15° = +0.259DE
€
+ Ax −100 # − Bx = 0
+Ax −100 # − 50 # = 0
3
+ DC = +0.60DC
5
___________________
€
Force
[ΣFx = 0]
100#
__________Fy__________
4
DC = −0.80DC
5
+ 100 # ( 4') − 300 # ( 3') + Bx (10') = 0
∴ Bx = +50 #
y
Solving FBD(a) first:
€
15°
D
[ΣM A = 0]
6’
DE
3
BC
€W
2’
Ax
20’
20’
20’
15’
4k
20’
2.13
20’
15’
2.36
2.36
2.38
3k
2k
Ax
20’
15’
20’
15’
1kN
By
Bx
5
5
12
m
2.5
Ay
12
20’
20’
20’
1kN
2.38
4k
[ΣFx = 0]
[ΣM A = 0]
By =
Ax
− 2k (20') − 3k ( 40') − 4k (60') + By (80') = 0
[ΣM A = 0] + B(12m) − (0.923kN)(2.5m) + (0.385kN)(2.5m) −
(0.385kN + 0.923kN)(6m) = 0
+ 4k (20') + 3k ( 40') + 2k (60') − Ay (80') = 0
Solving for B;
2.37
By = ( 12
13 )(0.767kN ) = 0.708kN
1500#
3000#
Reverting back to the unresolved forces;
60°
6’
€
8.6
1500#
90°
6’
8.6
30°
[ΣM A = 0]
[ΣFx = 0]
+ Ax + 1kN − (0.295kN) = 0
∴ Ax = +0.705kN
90°
30°
Ay
Dx
[ΣF
y
Dy
30’
( Bx )
= 0] + Ay −1kN + (0.708kN) = 0
∴ Ay = +0.292kN
( By )
−1500 # (17.33') − 3000 # (8.67') + Dy ( 30') = 0
∴Dy = +1733#
y
B = 0.767kN
Bx = ( 135 )(0.767kN) = 0.295kN
2.37
[ΣF
B
Ay
∴ Ay = +4k
€
By
(135 )(1kN) = 0.385kN
(1213 )(1kN) = 0.923kN
€
40 k−ft + 120 k−ft + 240 k−ft
= +5k
80'
[ΣM B = 0]
5
12
6m
No force to balance Ax , ∴ Ax = 0
6m
€
= 0] −1500 # cos 30° − 3000 # cos 30° −1500 # cos 30° + Ay + 1733# = 0
∴ Ay = +3463#
[ΣFx = 0]
2.38
+ 1500sin 30° + 3000sin 30° + 1500sin 30° − Dx = 0
∴ Dx = +3000 #
1kN
2.14
12
€
1kN
5
12
5
Bx
2.39
2.39
2.40
8k
12’
20k
12’
Dy
Cy
36’
24’
36’
48’
Ay
16k
4k
4k
12’
12’
12’
By
Ey
Fy
C y = +480 #
− 20k (24') − 4k (60') + By ( 48') = 0
[ΣF
y
= 0] + Ay − 20k + 15k − 4k = 0
2.40
500#
400#
300#
Cy
By
5’
Cy
By
+ 400 # (12') − C y (10') = 0
Dx
Dy
+ 300 # + Dx = 0
Dx = −300 # (←)
Lower beam:
€
[ΣM B = 0]
Ay = −240
Cy
− 4k + E yB−x 16k + 12k − 4k = 0
∴ E y = +12k
10’
Ay
Dy
= 0] − 400 # + C y + Dy = 0
[ΣFx = 0]
+ 4k(12') − 16k( 24') + F2’y ( 48') − 4k ( 60') = 010’
∴ Fy = +12k
[ΣFy = 0]
10’
Dy = −480 # + 400 # = −80 # (↓)
Right beam:
[ ΣM E = 0]
Cy
Upper beam:
∴ By = +15k
€
10’
Ay
[ΣM D = 0]
∴ Ay = +9k
2’
Bx
Left beam:
y
Dx
36’
Cy +Dy =8k; Cy = Dy = 4k
[ΣF
400#
300#
4k
Upper beam:
[ΣM A = 0]
500#
2.40
5’
[ΣF
y
− C y (5') − Ay (10') = 0
#
(↓)
= 0] − 240 # + By − 480 # = 0
By = +720 # (↑)
€
[ Σ x = 0]
€
Bx = 0
2.15
2.42
2.41
2.42
FDy
FDx
FD x = ( 4 5) FD
Ax
8k
€
[ ΣM A = 0] + FD x ( 4') + FD y ( 20') − 2k(16') − 8k( 32') = 0
Ax
y
FD x = ( 4 5)(18.9k ) = 15.2k
FD
FD y = ( 3 5)(18.9k) = 11.3k
FDx
= 0] + A x − FD x = 0
[ ΣFxDC
θR = 72.5°
#
00
=4
2
F
θ2 = 30°
FDy
x
DFx = 15.2k
DF = 18.9k
DFy = 11.3k
∴ A x = +15.2k (→)
500#
∴ FD = 18.9k
F1 =
( 4 5) FD( 4') + ( 3 5) FD( 20') − 32 k−ft − 256 k−ft = 0
#
2k
Ay
€
y
FD y = ( 3 5) FD
720
2.41
R=
FD
2k
[ΣFyA=y 0] + A yBD− 2k − 8k + 11.3k = 0
8k
θ1 = 75°
x
Parallelogram Method
∴A y = −1.3k (↓)
y
€
500#
⎛ DC ⎞ ⎛ 5BD ⎞
−⎜
⎟+ ⎜
⎟− 11.3k = 0
⎝ 3.16 ⎠ ⎝ 5.1 ⎠ ( DFy )
θ1 = 75°
( DCy )
( BD x )
θR = 72.5°
( BD y )
Solving the two equations simultaneously;
€
720
#
F1 =
⎛ 3DC ⎞ ⎛ BD ⎞
−⎜
⎟+ ⎜
⎟+ 15.2k = 0
⎝ 3.16 ⎠ ⎝ 5.1 ⎠ ( DFx )
( DC x )
[ΣFy = 0]
θ2 = 30°
DF = 18.9k
DFy = 11.3k
BD
[ ΣFx = 0]
=4
F2
R=
DC
#
00
x
DFx = 15.2k
Tip-to-Tail Method
BD = 17.9k
DC = 19.7k
2.16
2.45
2.45
2.43
2.45
y
A
C
x
O
R y = −5.2k − 1.5k = −6.7k
x
30°
€
Rx = -0.4k
θ = 86.6°
3k
6k
R x = −3k + 2.6k = −0.4k
Rx = -0.4k
= 86.6°
R y = ΣFy = −( 6k) sinθ 60°
− ( 3k ) sin 30°
3k
6k
5
R x = ΣFx =R−x (=6k-0.4k
) cos 60° + ( 3k) cos 30°
θ = 86.6°
3k
60°
2kN
12
30°
A
B=
A=
x
y
1.8
30°
x
60°
2.45
kN
4k
N
3.
=
θ = 52°
N
1k
1
y
y
30°
A
60°
=
1
R
2.43
R = R 2x + R 2y =
6k
( 0.40)
2
R = 6.71k
2
+ ( 6.7) = 6.71k
Ry = -6.7k
€
2.44
R = 6.71k
€
S = 20.5k
y
30°
θ = 86.6°
R = 42.5k
(vertical)
60°
x
30°
3k
60°
30°
x
6k
3k
6k
60°
R = 6.71k
P = 16k
45°
3k
y
θ = 86.6°
R = 6.71k
75°
x
y
30°
θ = 86.6°
Q = 22k
Ry = -6.7k
R = 6.71k
⎛ Ry ⎞
−1 ⎛ 6.7 ⎞
−1
θ = tan−1⎜
⎟ = tan (16.75) = 86.6°
⎟ = tan ⎜
⎝ 0.4 ⎠
⎝ Rx ⎠
Tip-to-Tail Method
Ry = -6.7k
Tip-to-tail
O
x
Tip-to-tail
6k
R = 6.71k
2.17
Tip-to-tail
2.47
2.46
2.46
2.47
y
y
x
D
F1 = 800#
F1y
1
4
3
2.46
P = 500#
2.47
F1x
30°
y
30°
x
F2x
30°
F1x
F2x
Fx______________________
Fy­­­­­__________________________
P=500#
0
-500#
P = 500#
F2y
F1
F2 = 1200#
+F1cos30°=(800#)(0.866)=692.8#
+ F1sin30°=(800#)(0.50)=400#
F2
+ F2cos30°=(1200#)(0.866)=1039.2# - F2sin30°=-(1200#)(0.50)=-600#
(1732#)
2
2
Rx = 1732#
+ (−700# ) = 1868#
θ = 22°
Rx = 1732#
R
Ry = 700#
Ry = 700#
R=
186
8#
Ry
€
;
Rx
⎛ 700 ⎞
−1
θ = tan−1⎜
⎟ = tan (0.404 ) = 22°
⎝1732 ⎠
−⎜
⎟( 90kN) = −
⎝ 2 ⎠ 30°
y
1
4
AD = 90kN k
€
€y
868
#
R
sinθ = y ;
R
Ry
700
700
R=
=
=
= 1867#
sinθ sin22° 0.375
x
90kN
= −63.6kN
2
Rx = +4.7kN
BD = 45kN
CD = 110kN
Fy
x
− 3 5 ( 45kN) = −27kN
BD = 45kN € 3
Alternate way to find the resultant R:
=1
θ = 22°
D⎛ 1 ⎞
CD = 110kN
θ = 88.3˚
R x = ∑ Fx = +4.7kN
⎛ 1 ⎞
−⎜
⎟( 90kN) = −63.6kN
⎝ 2⎠
− 4 5 ( 45kN) = −36kN
€
+(110kN) cos 30° = +95.3kN
Ry€= -154.6kN
R x = ΣFx = +692.8# +1039.2# = +1732# (→)
R y = ΣFy = −500# +400# −600# = −700# (↓)
R=
Fx
y
1
Force
CD = 110kN
BD = 45kN
AD = 90kN
F2 =x 1200#
30°
AD = 90kN k
Component
F1 = 800#
F1y
F2y
€
30°
1
€
€
€
2
2
Rx = +4.7kNR =x154.7kNR = R x + R y =
Resultant
θ = 88.3˚ €
Ry = -154.6kN
−(110kN) sin 30° = −55kN
R y = ∑ Fy = −154.6kN
( 4.7)
2
2
+ (154.6) = 154.7kN
⎛ Ry ⎞
−1 ⎛ 154.6 ⎞
θ = tan−1⎜
⎟
⎟ = tan ⎜
⎝ 4.7 ⎠
⎝ Rx ⎠
θ = tan−1( 32.9) = 88.3°
€
tanθ =
€
R = 154.7kN
Resultant
2.18
48
2.48
2.50
4’
100#
250#
T1x
F
20°
d2
d1 =
=
14
’
d2y
10’
5’
T1y
T2x
T2 = 700#
T2y
T1 = 500#
6’
45°
C
d1x
d2x
30’
d 2x = d 2 cos 45° = 14' ( 0.707) = 9.9'
d 2y = d 2 sin 45° = 14' ( 0.707) = 9.9'
A
d1x = d1 cos 20° = 10' ( 0.94) = 9.4'
+ 250 # ( d1x ) − 100 # d 2y − F( d 2x ) = 0
[ ΣM C = 0]
€
2.49
€
F=
#
( )
#
250 ( 9.4') − 100 ( 9.9')
= 137.4 #
9.9'
Force
Fx_________________
Fy_________________
T1 = 500#
(500#)cos15o = 483#
(500#)sin15o =129.5#
T2 = 700@#
(700#)cos10o = 689.5#
(700#)sin10o = 121.8#
MA = +T2x(30’) + T2y(6’) – T1y(35’) – T1y(4’) =0
2.49
Ax
1.33m
MA = 689.5#(30’) + 121.8#(6’) – 483#(35’) – 129.5#(4’) = +3990#-ft
MR A
Ay
3
4
Fy = 3.21kN
F = 5kN
40°
1.1m
1m
Fx = 3.83kN
1m
[ ΣFx = 0]
+ A x − 3.83kN = 0
A x = 3.83kN
[ΣFy = 0]
+ A y − 3.21kN = 0
A y = 3.21kN
( Fx )
( Fy )
M A = −3.21kN( 3.1m) − 3.83kN(1.33m)
€
€
M A = −9.95kN − m − 5.09kN − m = −15.04kN − m
2.19
2.52
2.51
y
2.52
2.51
100N
y
y
18#
7#
4”
O
O
origin
10#
2.51
6#
8”
5”
1m
18#
R = 5#
y
7#
6#
R(x)O= 224 #-in;
origin
4”
O
origin224 #−in
5#
5”
44.8”
8”
x
O
x
origin
x
O
= 44.8"
origin
y
x
2m
30N/m
150N
x
ω = 30N/m
Assume the member weight is
located at the
center of the length.
x
150N
2m
100N
1m
150N
100N
1m
1.5m
100N
1m
1.5m
x
150N
W = 90N (beam wt.)
x
R = ΣFy = -100N – 90N + 150N = -40N
R = 5#
44.8”
y
1m
Weight of wood member:
150N
30N/m
2m
30N/m
O
10#
MO = +(7#)(4”)
+ (6#)(9”) – (18#)(17” = - 224#-in.
O
1m
100N
origin y
R = ΣFyy = 10# + 7# + 6# - 18# = +5#
origin
100N
origin
x
y
€
y
O
origin
x=
150N
2.52
x
x
€
W = 90N (beam wt.)
O
x
R = 40N
MO = - (100N)(1m) – (90N)(1.5m) +y(150N)(3m) = +215 N-m
origin
1.5m
R(x) = MO
W = 90N (beam
wt.)
O
R = 40N
215N − m
∴ x=
= 5.4m
origin y
40N
3m
x = 5.4m
O
R = 40N
origin y
x = 5.4m
x = 5.4m
O
origin
3m
3m
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.
2.20
2.53
2.54
y
200#
A
B
100#
20#/ft
origin
12
4’
4’
ACx
3
4
4’
AC
Total beam weight equals (20#/ft)(16’) = 320#
at the center of the beam length.
y
400#
200#
B
100#
5
ABx
x
4’
A
ABy
AB
400#
Fx_______
AB
−
AC
€
x
origin
Force
W
12
AB
13
4
− AC
5
€
0
ACy
W = 5k
Fy_________
+
5
AB
13
3
− AC
5
-5k
€
€
12
4
12
4
AB = − AC = 0
[ΣFx = 0] − AB − AC = 0;
13
5
13
5
13
∴ AB = − AC
15
⎞ 3
5
3
5 ⎛ 13
[ΣFy = 0] + 13 AC − 5 AC − 5k = 0; + 13 ⎜⎝− 15 AC⎟⎠ − 5 AC = +5k
3AC
AC
−
= 5k;
AC = −5.36k (compression)
−
5
3
13
AB = − (−5.36k ) = +4.64k (tension)
15
W = 320#
For the beam to remain stationary and horizontal,
the moments taken about points A and B should
be balanced by the opposing moments due to B
and A respectively, resulting in no resultant moment.
[ΣM A = 0]
+ 200 # ( 4') + 100 # (8') − 320 # (8') − 400 # (12') − B(16') = 0
[ΣM B = 0]
+ 400 # ( 4') + 320 # (8') −100 # (8') − 200 # (12') − A(16') = 0
∴ B = 360 #
∴ A = 60 #
€
R y = ΣFy = +A + 100 + 100 − 320 − 400 + B = 0
#
#
#
#
= +60 # + 200 # + 100 # − 320 # − 400 # + 360 # = 0
€
2.21
2.56
2.55
2.55
2.56
y
y
x
Fy = 1.6kN
F = 2kN
3
60°
4
12
5
3
F
ACy
BC
3
N
AC
2k
N
.1k
4
BCy
=
=2
60°
3
x
BCx
BC
ACx
30°
BA
4
Fx = 1.2kN
4
DB
W = 800#
AC = 0.27kN
Tip-to-tail
Force
Fx_______________
AC
3
− AC
5
BC
Fy__________________
−
+BC sin 60° = +0.866BC
−BC cos 60° = −0.50BC
€
€
60°
€
F=2kN
−
4
N
2k
N
AC = 0.27kN
+DBsin 30° = +0.50DB
€
−
5
BA
13
+DB cos 30° = +0.866DB
0
€
-800#
€
−
(
AC = 0.27kN (tension)
€
DB
12
BA
13
12
24
BA + 0.50DB = 0;
DB =
BA
13
13
5
[R y = ΣFy = 0] − 13 BA + 0.866DB − 800 # = 0
⎛ 24
⎞
5
− BA + 0.866⎜ BA ⎟ = 800 #
⎝ 13
⎠
13
⎛ 24 ⎞
BA = 658.2 #
DB = ⎜ ⎟ 658.2 # = 1215.2 #
⎝ 13 ⎠
.1k
Solving simul tan eously;
Tip-to-tail
BC = 2.1kN (compression)
−
[ R x = ΣFx = 0]
=
€
BA
Fy_______________
€
3
3
3
AC − 0.50BC + 1.2kN = 0;
AC + 0.50BC = 1.2kN
5
5 4
4
4
− AC + 0.866BC − 1.6kN = 0; − AC + 0.866BC = 1.6kN
5
5
−
Fx______________
W
4
( 2kN) = −1.6kN
5
F
[ΣFy = 0]
3
( 2kN) = +1.2kN
€3
5
=2
[ ΣFx = 0]
+
BC
€
4
AC
5
Force
)
€
2.22
W
2.57
2.57
2.58
y
CA
2.58
y
CAy
CB
CBy
45°
CAx
Joint B:
30°
BCx
CBx
x
ABx
12
4
x
5
3
ABy
BCy
BC
AB = 1560#
BE
W
2.58
Force
Fx_______________
Fy__________________
Force
CA
−CA cos 45° = −0.707CA
CA sin 45° = +0.707CA
AB=1560#
+
BE
0
CB
W
BCx
€
+CB cos 30° = +0.866CB
€
y
5
− 0.707CA + 0.866CB = 0;
x
CA =
ABy
BE
[ΣFy = 0]
CDx
€
-W
[ ΣFx = €0]
0.866CB
= 1.22CB
0.707
[ΣFy = 0]
AB = 1560#
This relationship indicates the CA > CB, therefore, CA = 1.8kN
BC
Then, CB = (1.8kN)/1.22 = 1.48kN
BCy
€
€
ABx
12
4
[ ΣFx 3= 0]
+CBsin 30° = +0.50CB
Fy______________
y
CD
BC
0
€
Fx______________
CDy
12
5
#y
CB
1560
= 1440 # CB = 1800#
−
1560 # = −600 #
13
13
4
(
)
(
+BE
3
45°
CBx
4
− BC
5
€
)
x
3
− BC
5
4 W
#
BC = 0;
€BC = 1800
5
3
− 600 # + BE − 1800 # = 0;
BE = 1680 #
5
+ 1440 # −
(
)
€
0.707CA + 0.50CB − W = 0
W = 0.707(1.8kN) + 0.50(1.48kN) = 2.0kN
y
€
CD
CDy
CB = 1800#
CBy
4
2.23
3
CDx
45°
CBx
x
3
ABy
BCy
BC
AB = 1560#
2.59
BE
2.58 cont’d
2.59
Joint C:
10’
y
CD
CDy
500#
3
CB = 1800#
CBy
3
CBx
45°
Ax
x
4
’
26
4
CDx
Bx
Ay
By
3
4
B
10’
24’
W
Force
Fx_____________
Fy______________
CD
-0.707CD
+0.707CD
CB = 1800#
+
W
0
€
[ ΣFx = 0]
[ΣFy = 0]
€
4
1800 # = 1440 #
5
(
)
(
3
1800 # = +1080 #
5
(
)
#
) + 1080
#
€
CD = 2037 #
− W = 0;
W = 2520
− 500 # (10') +
5B
12B
(10')+
( 24') = 0
13
13
50B 288B
+
= 5000 #−ft ;
13
13
B x = 74 # ;
B y = 177.5#
-W
− 0.707CD + 1440 # = 0;
+ 0.707 2037
+
[ ΣM A = 0]
+ A x − 74 # = 0;
[ΣFy = 0]
+ A y − 500 # + 177.5# = 0;
( Bx )
(By )
B = 192.3#
[ ΣFx = 0]
€
#
(Bx )
A x = 74 #
( By )
A y = 322.5#
€
2.24
2.7kN
1.8kN
2.61
.60
2.60
B
A
hinge
Ay
2.5m
hinge
2.5m
10’
MRC
Cx
2m
3k
A
Bx
1.8kN
=0
[ΣFy = 0] + A y − 2k − 3k − 2k − 3k + (4k
By )
2.62
∴ A y = +6k (↑)
2.7kN
2.5m By
2m
Ay
MRC
3m
3m
By
MRC
3m
[ ΣFx = 0] Bx = 0
[ ΣM B = 0] − BAx y ( 4.5m) + 1.8kN( 2.5m) = 0;
[ΣFy = 0]
+ 1kN − 1.8kN + B y = 0;
3m
[ ΣFx = 0]
[ΣFy = 0]
[ ΣM C = 0]
5’
Cy
B y = 0.8kN
5’
− 0.8kN − 2.7kN + C y = 0;
180#
By
Cx
By
Dx
80#
4’
6’
4’
Dy
Cx = 0
300#
240#
200#
A y = 1kN
Beam BC:
€
Ax
Cy
2.7kN
Beam AB:
€
Cx
By
Bx
Ay
Bx
5k
[ ΣFx = 0] − A x + 4k = 0; A x = +4k (← )
[ ΣM A = 0] − 2k( 20') − 3k( 40') − 2k( 60') − 3k( 40') + 4k( 20') + By (80') = 0
∴ B y = +4k (↑)
By
Ay
By
4k
3m
3m
2.5m
20’
10’
20’
Ay
1.8kN
A
20’
20’
20’
2k
Cy
Ay
2m
Ax
C
3k
2k
3m
3m
B
A
2.61
Cx
Cy
2.7kN
1.8kN
2m
MRC
C
7’
60#
Cy
C y = 3.5kN
− M RC + 2.7kN( 3m) + 0.8kN( 6m) = 0
M RC = 8.1kN − m + 4.8kN − m = 12.9kN − m
2.25
€
2.62
2.62
Ay
Ax
5’
300#
240#
5’
180#
By
200#
By
Dx
80#
4’
6’
4’
Dy
7’
60#
Cy
Upper Beam:
[ ΣM A = 0] − 300 # ( 5') + By (8') = 0; By = +187.5# (↑)
[ ΣFx = 0] + A x − 180 # = 0; A x = +180 # (→)
[ΣFy = 0] + 187.5# − 240 # + A y = 0; A y = +52.5# (↑)
Lower Beam:
€
[ ΣM D = 0]
+ 200 # ( 4') + C y ( 6') − 187.5# ( 9') − 80 # (13') = 0
∴ C y = +322 # (↑)
[ ΣFx = 0]
[ΣFy = 0]
+ D x − 60 # = 0;
D x = +60 # (→)
− 200 # + D y + 322 # − 187.5# − 80 # = 0
∴ D y = +145.5# (↑)
€
2.26
3.1
3.2
Chapter 3 Problem Solutions
3.1
A
1
10’
2
4’
5
10’
300#
10’
10’
2
€
10’4’
2E
10’ Ex
10’
E y = hc
29
D
C
 2E 
#
[ ΣM A = 0] + 
) − 300 # ( 20') − 300 # (10') = 0
( 40') − 300 ( 30'
300#
 29 
300#
300#
4’
CBy hc
CB
E = +225 29 = +1212 #
y
D
C
∴ E x = 1125# ;
E y = 450 # CBx
5E 4’
Ex =
;
5
29
B
10’
)
(
10’
10’
300#
€
y
4’
CBy hc
CB
[ΣFy = 0]
Ay
)
Ey=450#
€
Ey=450#
CBx
CBy hc
10’
C
#
1125 #
€
€
CBy hc
CBx
4.05’
10’
D
300#
C
Ey=450#
4.05’
D
E
slope is
y
10
Ey=450#
10’
CB
y
Ex=1112#
y
slope is
10
slope is
10’
Ex=1112# 10’
4.05’
CB x = 1112 #
D
C #
#
#
x
[ΣFCB
y = 0] + CB y − 300 − 300 + 450 = 0
300#
#
∴ CBy = +150
300#
CB = 1112 # < 1200 # ∴OK
CB
[ ΣFx = 0]
CBy hc
y
E=1200#
10
E=1200#
10
10’
300#
(150 )(10') = 1.33'
h = 5.33'
CB
10’
y
10
y = 4.05'
y’
y’
Ey=450#
Ex
Ex=1112#
E x = 1112 #
10’
y’
E
300#
300#
y
450
=
;
10' 1112
Ex=1112#
300#
300#
300#
D
Ex
C
E
+ CBy − 300 # − 300 # + 450 # = 0
h c = 4'+y;
Ey
10’
∴ CBy = +150 #
€
300#
C
Ex
E
#
D # )(20') + ( 300
− ( 300 # )(10') + ( 300
( 40') = 0
)(30') + E yE=1200#
Ey=450#
E x = E 2 − E 2y ;
€
CB x = E x = 1125
y=
[ΣM
B A = 0]
300#
E
Ex=1112#
C
hD
hC
hD
10’
D
E
Ey
hB
B
Ey
10’
10’
Assume
300#Emax = 1200#
300#
Cable
hD
hB force DE h=C E
E y = +450 #
#
y
10'
=
;
CBy CBx
5
5
5
hC
10’
€
300#
Ey=450#
2
2
hB
10’
B
10’
10’
300#
300#
[ ΣFx = 0]
(
10’
D
C
CBx
)
2
Ax
E
Ax
Ay
Ax
300#
10’
300#
Ey
(
3.2
A
Ay300#
Ax
A
D
C
A10’
y
10’
E
Ex
4’
hc
B
A
10’
Ey
10’
10’
3.2
Ay300#
Ax
A
10’
10’
3.2
E
Ex=1112#
Ex=1112#
300#
300#
Ey=450#
E
Ex=1112#
y' CBy
=
; y'= 1.35'
10' CBx
h c = y'+4.05'= 5.4'
3.1
y
10
E
CB
3.3
20’
12’
A
12’
kN)
N)
kN)
20’
10’
Ax
(5.49m)
EyhC=12’
B
B
C
3k
(13.35kN)
3k
hC=12’
(13.35kN)
D
[ΣFx = 0]
E
[ΣF
y
Ex
2k
E y = 5.73k (25.5kN)
10’
A y = 6.27k (27.9kN
18’
CB
D
C
CBx
2k
D
C
7k
Ex
2k
(8.90kN)
B
3k
(13.35kN)
20’
(6.1m)
CDy
C
CD
CDx
7k
(31.14kN)
Ey=5.73k (25.5kN)
E
[ ΣFx = 0]
[ΣFy = 0]
CD x = 11.7k ( 52.1kN)
CD y = 3.73k (16.6kN)
∴ CD = 12.28k ( 54.5kN)
Ex
€
(31.14kN)
FBD (a)
(8.90kN)
7k
18’
Ey=5.73k (25.5kN)
CBy hc=12’
E
CBx
CBy hc=12’
CB
Ay
(3.67m)
A=13.28k (59.1kN)
Ax=11.7k (52.1kN)
Ay=6.27k (27.9kN)
− 3k(12') − 7k( 32') − 2k ( 42') + E y ( 60') = 0
[ΣFy = 0] + A y − 3k − 7k − 2k + 5.73k = 0
CB = 12.15k (54.1kN)
Ax
Original FBD
Original FBD
10’
A
(31.14kN)
(31.14kN)
€
= 0] + CBy − 7k − 2k + 5.73k = 0
12’
(8.90kN)
2k
(8.90kN)
FBD (a)
CBx = E x = 11.7k (52.1kN)
CBy = 3.27k (14.4kN)
€
2k
Ex
(31.14kN)
D
7k
(8.90kN)
7k
[ ΣM A = 0]
C
Ey
Ex
E
7k
FBD(a)
18’
(5.49m)
18’
(3.05m)
(6.1m)
Ay
Ay
Ay
10’
(3.05m)
(6.1m)
(3.67m)
D
C
CBx
3.3
(3.67m)
CBy hc=12’
(31.14kN)
FBD (a)
[ ΣM C = 0]
(3.67m)
E x = 11.7k ( 52.1kN
= 13.03k( 58.0kN)
); E = ED(6.1m)
A
Ay
12’
(3.67m)
€
€
− 2k(10'12’
− E x (12') = 0
) + 5.73k( 28')20’
20’
From
the original FBD:
(6.1m)
Ax
[ ΣFx = 0] A x = 11.7k ( 52.1kN)
A=13.28k (59.1kN)
B
∴
A (52.1kN)
= AB = 13.27k ( 59.1kN)
Ax=11.7k
Ay=6.27k (27.9kN)
3k
CD
CD(13.35kN)
y
B
3k
(13.35kN)
C
7k
(31.14kN)
CDx
CDy
C
CD
CDx
7k
(31.14kN)
3.2
CBy CDy
CB
C
CBx
3.4
[ ΣFx = 0]
From the solution for Prob. 3.3, it appears that
the
3.4force in cable AB is maximum. Therefore,
assume the following:
reaction @ A = 20k and AB = 20k.
Axx=19k
Ay=6.27k
A=20k
C
3k
Ax=19k
C
A x =3kA 2 − A 2y = 19k
12’
12’
Axx=19k
1
1
Ay=6.27k
=20k
12’
Ax=19k
1
3.1
9
3.03
20’
33..11
99
hB
B
= 0]
[ ΣFx 3k
CBx=19k
20’
1
C
CDx
20’
20’
B
3.03
3.03
5.8
€
7k
Ayy=6.27k
A=20k
B
hc-hb
h C − h B CBy 3.27k
=
=
20'
CBx
19k
 3.27k 
( h C − h B ) = 20
 = 3.44'
19k 
∴ h C = h B + 3.44' = 3.96' +3.44' = 7.40'
CDx
7k
CD
CDy
If A = B20k and Ay = 6.27k, then
€
€
CD
CDyy
B
7k
CB x = 19k
[ΣFy = 0] + CBy + 3.73k − 7k = 0
Ayy=6.27k
A=20k
CDx
CBy = +3.27k
Using the original FBD of Prob. 3.3, Ay = 6.27k
4
CD
hB
B
C
hC
C
3k
CDyy=3.73k
CD
CDx=19k
CDy=3.73k
7k
CD
hC
C
− 19k+ CD x = 0;
(Ax )
CDCDx=19k
x = 19k
7k + CD y = 0
[ΣFy = 0] + 6.27k − 3k − 7k
CD y = 3.73k
€
h B 12'
=
;
1
3.03
€
CB
h B = 3.96'
CBy CDy
C
CBx
CByy CDyy
CB
C
CBx
CD
CDx
7k
CD
CDx
7k
B
hcc-hbb
B
5.8
5.8
CBxx=19k
20’
1
1
C
3.3
3.5
3.5
3.7
3.5
ω = 300 #/ft
2k
6’
12’
ω = 300 #/ft
A
B
[ ΣM A = 0]
A
− 300 # ft (18')( 9') − 1200 # (18') + B(12') = 0
B
[ ΣM A = 0]
1200#
B = 3.7k
3.6
ω = 145 N/m
800N
A
ω = 145 N/m
1.52m
[ΣFy = 0] + A − 2k − 2k − (0.4 k ft)(4') + 3.7k = 0
3.8
3.8
B
2’
1000#
900#
2.13m
B
1.52m
2.13m
Ay
− 800N(1.52m) − (145 N m )(1.52m + 2.13m)(1.825m) + B( 3.65m) = 0
2k
2k
€
3.7
5’
4’
ω = 0.4 k/ft
6’
6’
6’
B y = +1550 #
3.9
€
By
[ ΣFx = 0] A x = 0
[ ΣM A = 0] − 1000 # ( 6') − 900 # (14') + By (12') = 0
A = 731N
3.7
A
Ax
[ΣFy = 0] + A − 800N − (145 N m)(3.65m) + 598N = 0
3’
− 2k( 3') − 2k(8') − 0.4 k ft ( 4')(14') − B(12') = 0
€
800N
B = 598N
B
A = 1.9k
€
[ ΣM A = 0]
ω = 0.4 k/ft
4’
A
6’
A = 750 # (↑)
A
5’
4’
[ΣFy = 0] + A − 300 # ft (18') − 1200 # + 5850 # = 0
3.6
2k
3’
1200#
12’
B = 5850 # (↑)
3.6
3.7
R=60kN
[ΣFy = 0] + A y − 1000 # − 900 # + 1550 # = 0
ω = 15kN/m
AAy x= +350 #
By
Ay
2m
1m
1m
2m
3.4
Ay
1000#
6’
2’
6’
900#
A
Ax
3.9
6’
3.11
R=60kN
6’
6’
ω = 15kN/m
Ax
Ay
1m
1m
2m
2m
8.2’
13.375’
ω = 15kN/m
[ΣM A = 0]
By ( 3m) − 60kN(B4m
By = 80kN
y ) = 0;
Ay
2’ − 60kN
= 0; Ay = 20kN
P2=900#
[ΣFy = 0] − Ay + 80kN
1m
2m
€
3.10
2m
P2=900#
4’
P1 = (100 # ft )(6') = 600#
P2 =
1
2
2
E = 1900
#
2
2
− 1200 # ( 3') − 1400 # (8.2') − 1200 # (13.375') + E(16.375') = 0
A − 1200 # − 1400 # − 1200 # + 1900 # = 0
A = 1900 #
(300 # ft )(6') = 900#
100#/ft
[ ΣM A = 0]
[ΣFy = 0]
By
3’
3’ P1=600#
300#/ft
3
3
€
2’
16.375’
3
100#/ft
Ax
Ey
(128 )(1')(150 # ft )(6') + (100 # ft )(1')(6') = 1200 #
R 2 = ( 12
1' 150 # ft )( 5.6') + (100 # ft )(1')( 5.6') = 1400 #
12 )( )(
R 3 = ( 128 )(1')(150 # ft )( 6') + (100 # ft )(1')( 6') = 1200 #
R1 =
P1=600#
300#/ft
Ay
3.10
1m
R3
C
3’
R=60kN
A[xΣF = 0] A = 0
x
x
3.10
R1
R2
By
Ay
3.9
3.11
By
Ay
3.9
4’
3’
3’
6’
€
€
Ax
By
Ay
3.11
R1
4’
3’
3’
R2
R3
[ ΣFx = 0] A x = 0
#
#
A
y
[ ΣM A = 0] − 900 ( 2') − 600C ( 3') + By (10') = 0
3.11
B y = 360 #
3’
8.2’
16.375’
A y = 1140 #
€
Ey
#
600
[ ΣM B = 0] + R
) + 900 # (8') − A y (10') = 0
1 ( 7'
13.375’
Ay
R2
R3
C
3’
8.2’
13.375’
Ey
16.375’
3.5
EF
E
500#
EC
DC
#
AB BA
= −(−0.707 ×BD
2030
= 1433#
) = +1433# (tension)
DE = 500#
F
E
30°
BD
EF=1000#
F
B
500#
EF
Dy
EC
BD = +1433# ( T)
BA
Ax
BC
CF
1000#
€
Ax
A
E
30°
AB
Joint D:
Dy
#
15
14
866#
866#
1433#
866#
#
1433#
30
500#
EC
BC
1000#
F
500#
1000#
1000#
C
A
Force Summation Diagram
BA
Joint A:
20
866#
#
500#
0
B
[ΣFy = 0] +BDBD − 1433# = 0
BC
D
Dx
C
BC B:
=0
[ ΣFx = 0]Joint
15
866#
15
14
1000#
E
5#
500#
DE
30°
DC
DE = 500#
BA
Joint E:
1000#
DE
1000#
14
1
E
AC = −2030 # (comp.) D D
BC DE = 500#
DC
Dx = 0
[ΣFy = 0] + AB +DxAC
y
14
0#
20
3
#
14
€
1000#
Joint B:
Joint F: €
CF = −1415# (C)
DiagramCF
1000#
1000#
+ 1433 + 0.707AC
Dy Dy = 0
BD = 1433#
866#
#
15
14
866#
866#
#
30
866#
20
Force Summation
D
30°
[ΣFy = 0] −BD1000 # − 0.707CF = 0
866#
30°
866#
1433#
1433#
866#
30
EC
Joint E:
Joint B:
F
866#
Joint D:
#
Dx
C
Force Summation Diagr
Dy
EF = +1000 # ( T)
1000#
#
ABD = 1433#
1000#
BA
EF=1000#
A
− 500 # − 500 # + EF = 0
[ ΣFx = 0]
EF
Joint F:
EF
Ax
Joint E:
1000#
1000#
500#
30
€
[ΣFxBD= 0]
A
500#
EF
EC
€
Joint E:
AC x = 0.707AC;
3.12
AC
Joint
B:
y = 0.707AC
DE
#
20
15
#
866#
1000#
C
0
B
Joint D:
EC = −866 # (C)
EC
A
BA
500#
AC
E
30°
20
Dx
− 1000 = 0 E
BC
Ax
1433#
ACD
#
AB E
F
DE = 500#
2#
000#
Ax
Dy
BD
500#
500#
[ΣFy = 0] − 866 # − EC = 0
1000#
61
F
CF
AB
[ΣFy = 0] + D y − 866
#
BD = 1433#
Joint D:
F
EF=1000#
DE
Joint A: 30°
1000#
1000#
30°
1000#
E
D
Dx
Joint1000#
E:
Diagram
Joint E:
1433#
AC
D x = +933#
D
1000#
C
DE = 500#BC
D
+1866 #
Ay =
DC
Force Summation Diagram
Ax
Step 3: Isolate a joint and solve two
unknown member forces.
€
Joint F:
Joint A:
866#
EF=1000# Dy F
A
1000##
Joint 0E:
Joint
JointF:D:
#B
+ 1433B:
=0
[ ΣFx = 0] − D x − 500 # Joint
DA
x
x
CF
Ax
1000#
866#
A
500#
500#
+ A x ( 20') − 866 (10') − 1000 ( 20') = 0
0
866#
A
Ax
[ ΣM D = 0]
D
E
BC = 0
A
BD
BD = 1433#
1433#
AC
#
DCSummation
Force
BD = 1433#
Ax
1000#
DC
Force Summation
Diagram
DAx
#
C
Ax CF
#
B A1000#
x = +1433
1433#
B
support
reactions.
AB
Dx
30°
F
Force Summation Diagram
DE = 500#
A
Joint B:
DE
D
2#
Joint
StepA:2: Solve for the external
1000#
DFy
A
BD = +1433# ( T)
€
C
61
C
A
2#
Step 1: FBD 1000#
of the entire truss.
E
500#
F
30
D
D
61
DAx
A x Dx
500#
866#
Ax
E
500#
Ax
500#
Dy
#
=0
[ΣFy =C0] + BD − 143330°
DE
D
Dx
1000#
F
[ ΣFx = 0]
Dy B
1000#
B
Joint F: Joint D:
1000#
Dy
30°
EF=1000#
F
20
500#
30°
Dy
AC
1433#
E
DFy
866#
866#
AB
1000#
1000#
0
Joint A:
30°
2#
C
61
B
3.121000#
F
500#
A
1433#
A
E
500#
1433#
Ax
D
DxAx
30°
866#
F
500#
AC
1433#
866#
E
AB
Dy
Ax
E
C
Joint D:
1000#
30°
1000#
1000#
0
B
Joint A:
D
500#
2#
D
30°
Dx
E
Dy
2#
Dy
C
500#
61
B
1000#
D
A
1000#
1000#
D
Dx
Ax
3.12
Dx
Dy
F
500#
61
Dx
3.12
E
D
3.6
EF
2kN
B
0.58kN
C
3.13
3.13
3.15
N
8k
kN
0.5
0.5
0.5
60°
60°
0.29kN
1600
1.5kN
60°
0.87kN
E
A
0.5kN
800
1600
400
800
400
400
400
1600
800
400
400
600
0
0
600 600
1200
800
400
600
800
1600
1600
800
N
3k
N
8k
8k
N
D
1.7
0.87kN
E
0.5
60°
400#
400
600
1200
1200
8k
N
0.5
8
N
N
400#
C
60°
60°
A
3k
8k
0.29kN
0.5kN
0.58kN
60°
400#
1.7
0.5
B
60°
A
2kN
800
800
400
0
0
400
800
800
800
800
400
800
B
800
400#
800#
D
400
0
1.5kN
3.16
3.14
3.16
3.14
A
3kN
5.7
5
B
2.57
0
3kN
C
5k
3kN
07
k
A
k
D
8.
8.
58
Ey = 4.43
8kN
12kN
C
9kN
3kN
12kN
k
2k
9kN
5k
D
k
E
5k
5k
k
Ex = 5.15
21
0.
58
k
2.42k
B
E
07
5
4kN
7.
5.7
12kN
B
7.
21
0.
5.15
4kN
4kN
2.42k
3.145.15
D
4kN
C
2.57
2k
12kN
A
Ay = 12kN
Fy = 12kN
F
Fx = 12kN
3.7
Ex = 5.15
Ay = 12kN
Ay
Fy = 12kN
3.18
3.17
a
A
Ax
C
3.18
C
2000
B
3464
40
00
D
0
2000
G
0
40
00
3464
Ay
E
ABxx
By
F
Bx
2000#
2000
a
A
B
0
G
D
C 4k
a
F
2k
4kE
G
B
By
E
D
4k
a
F
2k
4k
A
4000
0
AC
C
E
BC
D
BD
AC
B
Ax = 0 A
Ay=2000#
G
A
H
C 4k
Hy=4000#
BC
[ΣM B = 0]
+
B
+
4kE
2AC
AC
( 3') + (6') − 4k (6') − 4k(12') − 2k (18') = 0
5 BD 5
D
F
( AC x )
4k
4k
2BC
BC
(3') + (6') + 4k (6') − 2k(6') = 0
5
5
( BC x )
2k
G
( ACy )
AC = +9 5 = +20.1k ( T)
[ΣM F = 0]
F
2k
( BCy )
BC = − 5 = −2.24k (C)
[ΣM C = 0] − BD(3') − 4k(6') − 2k (12') = 0
BD = −16k (C)
€
3.8
3.19
3.20
3.19
.19
B
Ax
K
E
D
C
3.20
1.6kN
1.2kN
500#
2000#
a
y
]
A y = 5.24kN
[ ΣFx = 0]
− A x + 1.2 = 0
A x = +1.2kN
G
BC
c
B
Hx
€
B
BC
H
H
C
43kN
4
F
b
800#
a
D
D
c
CH
.4
6
5
J
5
Hy
C
Jy
E
C
BC
1.6kN
[ ΣM C = 0]
− FH( 2m) + 1.2( 2m) − 1.6( 2.5m) = 0
c
500#
BC
A
HG
€
Hx
B
E
BC
B
CE Hy
F
BE
EA
E
600#
A
800#
E
[ΣM A = 0]
BE = −500
+ 500 # (5') + BE (5') = 0
#
(C)
800#
500#
BE
J
A
600#
Jy
800#
EF
[ΣM A = 0]
E
600#
A
B
BC
€
FJ
500#
Hx
C
800#
600#
F
JG
Section
HG b-b:
800#
A
EA
E
G
B
A
B
BC
EF
F
G
600#
BC = +7.75kN ( T)
FH = −0.8kN (C)
J
BE
Jy
Hy
EF
F
+ BC( 2m) − 3( 2.5m) − 1.6( 5m) = 0
B
E
( CH y )
€
500#
CE
4
[ΣFy = 0] − 3kN − 1.6kN − 6.4 CH = 0
[ ΣM H = 0]
E
C# = 0 BC
= 0] − H x + 600
E
H x = 600 #
a
CE
F a-a:b
Section
Hx
1.6kN
1.2kN
F
FH
c
1.2kN
F
4
FH
J y = 4600 #
500#
a
#
=0
[ΣFyC= 0b] − H y − 2000# B− 500# − 800#A+ 4600
600#
#
500#
H y = 1300
y
G
D
500#
[ΣM H = 0] J y (5') − 500 # (10') − 800 # (15') − 600 # (10') = 0
[ΣF
6.
CH
CH = −7.35kN (C)
E
3kN
b
Jy
Hy
2000#
a
J
600#
A
800#
c
Hx
B
3.20
y
F
c
E
a
b
K[ ΣM = 0] H ( 2.5') − ( 3kN)( 5m) + (1.2kN)( 2m)F− (1.6kN)( 7.5m
) =E0
A
y
C
D
H y = 9.84kN
Hy
1.6kN
ΣF = 0 − A − 3− 1.6 + 9.84 = 0 a
[
1.2kN
F
Hy
G
600#
A
B
3kN
B
Ax
500#
a
b
C
D
D
C
a
Ay
2000#
3kN
a
Ay
3.20
EA
800#
+ 500 (5') − CE y (5') − CE x (2.5') = 0
#
2
1
CE; CE y =
CE
5
5
2
1
CE (5') +
CE (2.5') = 500 # (5')
∴
5
5
But; CE x =
CE = 250 5 #
F
JG
€
FJ
G
3.9
6
800
3.22
EF
F
2k
3.22
Section c-c:
G
3.21
F
JG
HG
Hx
− 600 # (5') + FJ(5') + 4600 # (5') = 0
a
2k
a
€Jy
Section a-a:
C
Ay = 6k
H
a
B
G 3.22
C
Ay = 6k
E
EH
F
GH
Fx = 2k
Fy = 4.73k
F
Ey = 18k
b
2k
G
A
c
J
Ay = 2.27k A
Ay = 2.27k
2k
D
ED
GD
BC
C
D
C
y
EH y = +2.73k;
H
GH
€
BI
I
BC
E2k
H
G
a
Ay = 6k
A
Ay = 6k
B
C
ABx = 4 5 AB
HG
G
B
ABy = 3 5 AB
H
a
Fy = 4.73k
I
b
3k
JI
Fx = 2k
F
Fy = 4.73k
G
ED
D
DC
D
2k
DI
F
Ay = 2.27k
Fx = 2k
Fy = 4.73k
F
BC
EH
BI
GD
JC
GH
H
I
A
Ay = 2.27k
JI
C
[ΣFy = 0] + 4.73k − 2k − 3k + HC y = 0
DC
D
E
A
C
J
2k
2k
JFx = 2k
A
JC
I
H
C
HC y = +0.27k;
HC DI
G
G
H
HI
3k
H
HC = 5 4 ( 0.27k) = +0.34k ( T)
I
€HC
Section c-c:
Fy = 4.73k
2k
E
2k
c
2k
Ay = 2.27k
E
G
€
G
Section b-b:
Ey = 18k
F
− 35 ( AB)(12'BH
[ ΣM H = 0] AB
) − 6k(12') = 0
−6(12)( 5)
AB A
=
= −10k (C)
3(12)
H
HG
BH = 0
[ ΣM A = 0] BH(12') = 0;
Ay B==6k
0] + HG ( 9') − 6k(12') = 0
[ ΣM
HG = +8k ( T)
€
D
c
Fx = 2k
BH
H
a
E
24k
B
AB
F
b
B
JI 5 4 ( 2.73k
EH =
J
) = 3.41k ( T)
Fy = 4.73k
A
2k
B
− 2k − EH
0
[ΣFy = 0] + 4.73k
JCy =BI
(F )
GD EH
G
I
3k
ED
2k
Fx = 2k
F
D
B
J
c
2k
H
a
D
24k
2k
c
I
3k
G
Fy = 4.73k
2k
E
a
B
C
b
a
E
3.21
A
b
D
H
G
F
Fx = 2k
Fy = 4.73k
F
Ay = 6k
c
C
Fx = 2k
FJ = 600 − 4600 # = −4000 # (C)
#
B
b
D
E
J
Hy
3.21
[ΣM G = 0]
FJ
a
E
HI
I
3k
B
2k
− BI y = 0
[ΣFy = 0] + 2.27k
(A )
y
J
A
BI y = +2.27k;
BI = 5 4 ( 2.27k) = +2.84k ( T)
Ay = 2.27k €
3.10
3
2
a
3.23
3.23
a
3.23
C
a
D
C
E
B
A
F
A
Ay = 8kN
Ay = 8kN
1
1
Fy = 3.5k
Ay = 3.5k
b
b
b
Fx = 6kN
F Fx = 6kN
Ax = 4k
O
B
2kN
2kN
D
B
1
3
OD = 28’
− 2kN + CE x = 0
E
€
A
C
D
3
Section b-b:
B
E
B
FB
1
3
F
A
A
3
4 Fx =
6kN
Fy = 8kN €
1k
2
E
[ΣFx = 0]
EA
1
A
Ay = 8kN
Ay = 8kN
[ΣM O = 0]
2
F
OD = 16’
€
Fx = 6kN
F Fx = 6kN
1C
Fy = 8kN
Fy = 8kN
8’
3
CE
13
3k
D
1
BF
1
3
4
3
EA
1
DBy =
2DB
;
13
DBx = +1k;
DB x =
O
3DB
13
+ 1k (16') − DBx (16') = 0
DB =
1
( ) = +1.2k (T)
1k 13
2
3
C
€
CE
8’
3k
13
B
OD =
8’
2
E
12’
1k
Only DB can resist the rotational tendency
(counter-clockwise) of the 1-k applied load.
[ ΣM O = 0]
E
3
B
OD = 28’
2
FB = +6 2kN ( T)
F
1k
DB
C
( Fx )
D
3
B
+ 6kN− FBx = 0
FBx = +6kN
O
1
8’
E
12’
BF
O
CEAE
= +2 2kNFB
( T)
FB
AE
CE x = +2kN;
B
B
E
[ ΣFx = 0]
DB
DB
E
Ay = 8kN
Fy = 3.5k
Ay = 3.5k
Section a-a:
C
AE
F
A
Fy = 8kN
Fy = 8kN
D
CE
CE
3k
E
3
Section a-a:
C
2
a
B
F
A
1k
a
b
E 4kN
4kN
3
2
b
a
D
C
3
Ax = 4k
b
b
3.24
1
2kN
2kN
a
a
B
b
D
3k3.24
E
B
DB
2
3
E
3.11
1
D
1
A
Fy = 3.5k
Ay = 3.5k
Section b-b:
Zero Force Members:
P
I
3.25
3.25
O
3.25
P
I
0
G
3.25
P
E
0
G
OE
OD = 28’
0
0
0
E
1
H
0 0
C
3
I
00
G
1
H
0
H
F
0
0
0
0
OD = 16’
C
F
D B
00
D
D
3
B
BF
1
C
[ ΣM O = 0]
8’
3k
+ 1k( 28') + 3k( 36') − EA x ( 36') = 0
EA x = +3.8k
4EA
5EA x
EA x =
;
EA =
5
4
EA = +4.7k ( T)
CE
13
B
3.263
4
3
3.26
0
C
L
C
K
B P
B
0
L
P
0
K K
0
D
F
0
0
E
E
I
0
G
H
F
0
F
0 0
0
0
J J
0
G
0
G
H
H
II
P P
P
3.27
P
3.27
0
0
P
L
D
J
0
A
F
3.27
E
B
E0
A
A
B
C
3.26
2
1
D
DB
EA
1
1k
A
D
8’
3.26
€
3
A
2
E
12’
1k
B
0
A
C
D
0
CA
2
F
E
D
3.27
C
B
P
M
L
C
C
B
A
A
B
0
0
M0
0
O
N
0
0
A
F
0
P
E
E
D
0
D
P
N 0
N
0
L
G
0
0
K
O
O
0
0
F
J
F0
0
0
K
I
G
0 0
0
H
0
0
0
G
0
0
J
H
0
H
0
3.12
I
I
3.29
28
3.28
3.28
6’
By
3.29
6’
Bx
B
Bx
ω = 400#/ft
By
9’
Ax
150#
Ay
150#
B
C
200#
Ay
4’
12’
4
200#
[ΣM B = 0]
Bx
+ Ax (9') −150 # (6') − 200 # (16') = 0
By
for member AC :
150#
[ΣFx = 0]
C x = 455 #
[ΣFx = 0]
Bx = 455 # (←)
BC x = 3 5 BC;
For member BC :
150#
B
Cy
Cx
BC = 4167
Cy
Cx
Ax Ay A
[ ΣFx = 0]
Ay
200#
A y = 666 # (↑)
€
For equilibrium in member BC:
− 200 # (16') + C y (12') = 0
∴ A y = C y − 200 # ;
[ΣFy = 0] + A y + 3334 # − 400 # / ft (10') = 0
200#
Member AC:
C y = 267 #
− A x + 2500 # = 0
BC y = 4 5 ( 4167) = 3334 #
A x = 2500 # (← )
Cy
Cx
( BCy )
#
BC x = 3 5 ( 4167) = 2500 # ;
Cx
A
BC y = 4 5 BC
− 400 # / ft (10')( 5') + 4 5 BC ( 6') = 0
[ ΣM A = 0]
€Cy
[ ΣM A = 0]
B
Ax = +455 # (→)
By
BCx
4’
By
B
Bx
BC
3
12’
Bx
4’
BCy
BC
A
Ax Ay A
€
BCx
BCy
9’
Ax
Ax
6’
B y − 150 # − C y = 0
B y = 412 # (↑)
A y = −67 # (↓)
€
3.13
3.30
3.31
Bx
B
3.30
4’
3.30
3.31
C
8m
8m
6.67m
8m
8m
[ΣFy = 0] + 153.4kN − 135kN − 180kN + By = 0
B y = 161.6kN
Ax
Bx
Ax
y
€
Ay
= 0] + Ay − 500 # − 200 # = 0
Ay = +700 # (↑)
E
6’
D
By
200#
Bx =586#
B
Cx
C
(a)
180kN
Cx
Ay
B
Cy
Cy
By
+ 135kN( 6.67m) − 153.4kN(13.33m) + A x ( 6.67m) = 0
D
(a)
Bx
Cy
Ax = 586#
Dy
D
Dy
By
A x = 171.6kN
Ay = 700#
+ 171.6kN − C x = 0
#
CAxx =
= 820
586#;
C x = 171.6kN
D x = 234 # ;
[ΣFy = 0] + 153.4kN − 135kN − C y = 0
Cy
1
(b)
1
Cx
2-force
member
1
(b)
500#
Dx
Dx
(c)
Dy
500#
1
Ex
Ey
Ey
Ex
200#
E
Ex
Ey
E
Ey
C y = 820 # Dx
Ex
D y = 820 #
#
E x = 820 # ;Ay =E700#
y = 820 (c)
C y = 18.4kN
2-force
member
Dx
Dy
Bx
Cx
C
180kN
Cy
Bx =586#
Cx
C
Cx
200#
+ 171.6kN − B x = 0
B x = 171.6kN
€
4’
+ Ax − 586 # = 0
= 586 # (→)
[ΣF
200#
500#
6’
Cx
Cy
[ ΣFx = 0]
E Ax
6’
C
Cy
Ay
€
10’
Ay
Cy
[ ΣFx = 0]
[ΣFx = 0]
4’
Cx
Ax
[ ΣM C = 0]
500#
6’
+ Bx (14') − 500 # (10') − 200 # (16') = 0
Bx = 586 # (←)
D
Bx
By
[ΣM A = 0]
135kN
135kN
€
Ax
6.67m
6.67m
Ay
Ax
[ ΣM B = 0] + 180kN(8m) + 135kN( 22.67m) − A y ( 29.34m) = 0
A y = 153.4kN
Ay
6.67m
180kN
Ax
6.67m
4’
10’
C
C
135kN
Bx
B
180kN
135kN
6.67m
C
3.31
€
3.14
3.33
3.33
3.32
10kN
10kN
1.33m
1.33m
FBD(a):
Ax
[ΣM A = 0]
By = 8kN
C
C
3.33
Ax
2m 2m
−10kN(2m) + By (2.5m) = 0
Cx
[ΣFy = 0] −10kN + Ay + 8kN = 0
2.5m2.5m
Ay = 2kN
Cx
4’
R = 3k
4’
R = 3k
Ay
B
FBD of the entire frame:
B
[ ΣM C = 0]
Ay
E
D
Cy
E1k
D
Cy
2m
2m
3.323.32
€
A
A
Ax Ax
Ay Ay
Bx Bx
(a) (a)
B
B
FBD(c):
By By
[ΣM C = 0]
Bx = 10kN
[ΣF
y
[ΣFx = 0]
Cy Cy
Cx
Cx
Cy Cy
Cx Cx
C
C
2.5m2.5m
A
Ax Ax
Ay Ay
B
Cy
[ ΣM E = 0]
BD = +6k
BD
By
BD B
BD
(b)
(d) DB
(d)
+ BD( 6') − 3k(12') = 0
[ ΣM B = 0] + 3k( 2') − A y ( 6') = 0
A y = 1k (↑)
Bx
[ΣFy = 0] + 1k + C y − 3k − 1k = 0
Bx Bx
E y = +4k
[ ΣFx = 0]
By By
E x = +4k
Bx
Bx
By €
(c)
Bx
2-force
members
(c)
Ex
Ey
2-force
members
Ex
Ey
D
E
D
E
Ey
1k
Ex
Ey
Ex
FBD(c):
1k
[ΣFy = 0] + By − 4k = 0
B y = +4k
[ ΣFx = 0]
B x = +4k
+ 4k − E x = 0
€
€
C y = 3k (↑)
By
[ΣFy = 0] + 3k − 6k + E y − 1k = 0
5
B
By
DB
FBD(d):Cy
(c) (c)
4
BD
DB
Cx
2m 2m
5
B
DB
Cx
4
FBD(a):
(b)
2-force
2-force
member
member
(b) (b)
A
(a)
or, since member CB is a two-force
member, use the
slope relationship for Cx and Cy.
€
C
(a)
Ay
Ay
+ C x −10kN = 0
4’
R = 3k
C x = 10kN
2m 2m
C
Ax
= 0] − C y + 8kN = 0
C y = 8kN
10kN
10kN
Ax
− Bx (2m) + 8kN(2.5m) = 0
R = 3k
A x = +4k (← )
[ ΣM A = 0] − 3k( 4') − 1k(12') + C x ( 6') = 0
C x = +4k (→)
1k
€
4’
− 3k( 4') − 1k(12') + A x ( 6') = 0
− B x + 4k = 0
3.15
3.36
3.36
3.34
3.36
4
5k
1m R1 = 4.5kN
R2 = 9kN
1kN/m
1kN/m
Ay
3.5m
R1 = 4.5kN
1m
5k
By
Bx
4k
3
2k/ft
4k
3k
4
Cx
3k
Cy
Dy
Cy
Dy
3
R2 = 9kN
5’
5’
Bx
Ay
3.5m
By
4k
Equivalent FBD
[ ΣFx = 0] Bx = 0
[ ΣM B = 0] + 9kN( 3.5m) + 4.5kN( 5m) − A y ( 6m) = 0
A y = 9kN
[ΣFy = 0] + 9kN − 4.5kN − 9kN + By = 0
By
R = 20k
R = 20k
Bx
Bx
By
Ayy
A
By
Ay
Cx
3k
Cx
3k
Cy
Dy
Cy
Dy
[ ΣFx = 0] Bx = 0
[ ΣM B = 0] − A y ( 20') + R(15') = 0
3.35
A y = 15k
[ ΣM A = 0]
Floor reaction
Bx
)
) × 1' = 262.5
p = γ soil × h = 35 # ft 3 × 7.5' = 262.5 # ft 2
ω = p × 1' =
8’-0”
5’-0”
(
B y = 5k
€
(
262.5 #
ft 2
#
ft
ωL ( 262.5 # ft )( 7.5')
F=
=
= 984 #
2
2
€
€
B x = 308
Ax
[ ΣFx = 0]
Slab
#
A x = 676
p=γxh
D y = 8.5k
#
+ B x (8') − 984 ( 2.5') = 0
#
− 4k( 5') − 15k(10') + D y ( 20') = 0
[ΣFy = 0] + C y − 4k − 15k + 8.5k = 0
C y = 10.5k
#
− A x + 984 − 308 = 0
#
€
€
+ C x + 3k = 0
C x = −3k (← )
[ ΣM C = 0]
[ ΣM A = 0]
− 20k( 5') + B y ( 20') = 0
Lower Beam:
[ ΣFx = 0]
F
2’-6”
€
4k
Bx
By
Ay
Upper Beam:
B y = 4.5kN
Bx
2k/ft
Cx
3.16
3.37
Overhang beam
C
hinges
E
ω = 320 #/ft
F
D
E
F
ω = 320
E #/ft
Fy
20’
Dy = 3200#
3200#
25’ Fy
20’
Dy = 3200# Fy
ω = 320 #/ft
25’
Fy
3200#
Cy = 3200#
Cy = 3200#
Ey
5’ +9600 #
By =
25’
9.6k
ED
Ay
9.6k 9.6k
Ay
#
By
(320 )( 30') + 9600
[ΣFEy= 0] − 3200 −4.8k
#
y
4.8k
#
ft
E
3k
9.6k ........
+ Ay = 0
9.6k
9.6k 14.4k
9.6k
24’
9.6k ........ 28.8k
24’
[ ΣFx = 0]
B x = 2.91k (→);
ED
E
28.8k
By
A
Joint E:
D
2k
Ay
ED
ED=7.8k
E
DC
DF
EF
EF
B
Ax
ED x = 12
ED;
13
EF=7.2k
B
y
ED
=
y
Bx
ED = +7.8k ( T)
[ ΣFx = 0]
CD=13k
CF=7.2k
DF=12k
CF
5
ED
13F
[ΣFy = 0] +D135 ED − 3k =2k0
ED=7.8k
DF=12k
EF=7.2k
FA
+ 12
7.8kDC
) + EF = 0
13 (
€
D
ED=7.8k E
DF
3k
D
k
7.8 2k
DC
7.2k
2k
DC x = 0.707DC;
[ ΣF12.2k
x = 0]
7.2k
DC = +13k ( T)
F
CF
k
7.8
FC
FA
2k
12.2k13k = 0
( )
[ΣFy = 0] − 135 (7.8k) − DF − 0.707
E
CF=7.2k
7.2k
F
7.2k
CA
C
3k
12.2k
2k
Ax=0.91k
k
A
7.6
D
C
k
7.6
12.2k
€
CD=13k
D =0
− 12
7.8k) − 2k + 0.707DC
13 (
DF = −12.2k
3k
DF=12k
DCy = 0.707DC
EF=7.2k
C
FA
DF
EF = −7.2k (C)
Joint D:
C
CF CA
F
k
Girder with Columns @ 24’ o.c.
Ay
B y = 7k (↓)
Bx
13
28.8k
B
Ax
− 2k − A x + 2.91k =12’
0
24’
28.8k
24’
By A
k
13
24’
Bx
B 10') = 0
[ ΣM A = 0] + 3k(12') + 2k(17'F) − 12
13 (
3k
Girder with Columns @ 24’ o.c.
14.4k
€ EF
#
24’
B
C
5’
A y = +10k (↑)
ω = 320 #/ft
5’
Ay
2k
[ΣFy = 0] − 3k + A y − 7k = 0
25’
By
Ax
A x = +0.91k (← )
ω = 320 #/ft
5’
A
5’
[ΣM A = 0] + (3200 # )( 30') + (320 # ft )( 30')(15') − By (25') = 0
Ay = 3200
€
3200#
5’
12’
12’
B = 7.6k;
C
5’
F
12’
A
3k
ω = 320 #/ft
12’
3k
A
3200#
5’
E
3.38
2k
C
5’
F
3k
E
B
ω = 320 #/ft
E
B
C
12’
D
C
D
DL+SL = 40psf
DL+SL = 40psf
Beams @ 8’-0” o.c. ω = 320 #/ft
Column
D
5’
E
DL = 15psf
SL = 25psf
B
2k
k
D
B
Overhang beam
Beams @ 8’-0” o.c.
Column
simple beam
C
hinges
DL = 15psf
SL = 25psf
2.4
D
E
D
3.38
k
simple beam
2.4
3.37
E
3.38
3.38
B
Bx=2.91k
3.17
CB
3k
12’
3.38 - cont’d
A
Joint F:
DF=12k
EF=7.2k
2k
3.39
3.39
Bx
Ay = 10k
[ ΣFx = 0B]y + 7.2k + CF = 0
Ay
CD=13k
CF = −7.2k (C)
[
CF
F
C
B
Ax
]
FA = −12.2k (C)
FA
Ax = 10k
ΣFy =CF=7.2k
0 − 12.2k − FA C= 0
10
A
10
10
10
CB
CA
B
3.33
10
13.3
10
Joint C:
D
EF
k
F
13
k
7.2k
CCA
CF
[ΣFy = 0] +F0.707(13k) − 1213 CA − 1213 CB = 0
C
DC
DF
+ 7.2k − 0.707(13k)
− 135 CA + 135 CB =
CB = +7.6k ( T)
FA
0
CF=7.2k
Ex = 10k
C
CA
CA = +2.4k ( T)
CB
CD=13k
Force Summation Diagram
k
7.2k
A = 10 2k;
C
[ ΣFx = 0]
3.40
By=7k
A x = 10k;
+ E x − 10k = 0
Ax=0.91k
Ay=10k
Force Summation Diagram
(Ay )
A y = 10k
[ΣFy = 0] − E y + 10k − 10k = 0
18kN
Ey = 0
B
9kN
A
A
A
(10') + ( 4') = 0
2
2
E x = +10k (→)
k
7.6
12.2k
− 10k(14') + +
(Ax )
k
3k
Bx=2.91k
F
2.4
2.4
Ay=10k
7.2k
12.2k
[ ΣM E = 0]
k
Ax=0.91k
E
B
2k
13
k
7.8
A
D
Solving for the support reactions;
D
k
Bx=2.91k
7.6
12.2k
13.3
Ey = 0
CB
10k
10
10
E
€
rce Summation Diagram
=7k
ΣFx = 0]
[EF=7.2k
2k
CF=7.2k
ED=7.8k
12.2k
0
DF=12k
D
CD=13k
2k
13.3
€
ED
C
10
3.33
B
€
7
Bx=2.91k
7
7
By=7k
Ax = 9kN
4
7
16
Ay = 3kN
16
4
3
4
F
C
20
15
4
3
3
20
20
4
16
15
E
20
D
Dy = 15kN
Force Summation Diagram
3.18
3.41
3.41
3.40
3.40
3.41
18kN
B
9kN
7
7
7
4
Ax = 9kN
7
16
16
4
Ay = 3kN
3
4
C
F
15
3
20
0
D D
D
0
0
20
15
E
0
150#
150#
20
4
16
200#
200#
150#
150#
4
3
B B
200#
200#
Ax =Ax200#
= 200#
20
150#
150#
Ay =Ay150#
= 150#
150#
150#
0# # 200#
200#
0
25 250
200#
200#
0
150#
150#
E E
150#
150#
Dy = 15kN
C C
200#
0# # 200#
25 250
200#
200#
Force Summation Diagram
150#
150#
Fx =Fx200#
= 200#
Solving for the support reactions;
Solving for the support reactions;
[ ΣM A = 0]
[ ΣM A = 0]
+ D y (12m) − 18kN(8m) − 9kN( 4m) = 0
Fx = +200 # (→)
D y = +15kN (↑)
+ 200 # − A x = 0
[ΣFy = 0] + A y − 18kN + 15kN = 0
[ ΣFx = 0]
[ ΣFx = 0]
[ΣFy = 0] − 150 # + A y = 0
A x = +200 # (← )
A y = +3kN (↑)
− A x + 9kN = 0
A y = +150 # (↑)
A x = +9kN (← )
€
− 150 # (8') + Fx ( 6') = 0
3.42
3.42
€
3.42
8k 8k
8k 8k
8
16 16
8 8
12 12
4k 4k
4 4
8 8
24 24
24 24
0 0
4
12 12
C C
A A
24k24k
16k16k
D D
8
16 16
8
4
F F 24k24k
8
8
8k 8k
16 16
E E
8
12 12
8k 8k 4
4k 4k
4
24
24 24
24
8 8
8 8 8
0 0 12 12
4 4
16 16
G G
8
24k24k
H H 24k24k
B B
16k16k
3.19
3.44
3.43
10k
5k
a
a
F
a
G
F
D
Ay = 10k
Ax = 0
E
D
E
B
a
[ ΣFxA= 0] C x = 0
10k( 24') + C y ( 48') = 0 B
[ ΣM A =A0y]=−10k
a
C y = +5k (↑)
5k
C
Cy = 5k
[ΣM A = 0]
F
[ΣF
y
Cx =0
D
Ay = 10k
B
DG = −8.33k (C);
DG x = −6.66k;
DG y = −5k
[ ΣFx = 0]
AB
B
B
60°
Ay = 1.25k
A 60°
[ΣM H = 0]
FG = +1.87k ( T)
60°
Ay = 1.25k
Fy = 4.75k
a
F
Fy = 4.75k
HDy
3
1
HDx
CDH
C
I
D
HDy
3
1
HD
G
HG
HD
HDx
I
H
G
HG
( AB)
[ΣM C = 0]
(
)
−1.25k (2.5') + GH 5 3k = 0
( Ay )
(
)
− CD 5 3 −1.25 k (20') = 0
CD = −2.89k (C)
+ FG + (−6.66k)+ ( 4.8k) = 0
( DG x )
G
F
D
( Fy )
GH = +3.61k ( T)
AB = +4.8k ( T)
Ay = 10k
Ax = 0
[ΣM D = 0]
[ ΣM G = 0] + 5k( 24') − 10k( 24') + AB( 25') = 0
A
E
Using the left side of the section cut;
DG x = 4 5 DGF
( DG y )
AB
− 2k (25') − 4k ( 35') + Fy ( 40') = 0
A 60°
Using the left half of the section cut;
DG y = 3 5 DG;
G
H
2
[ΣFy = 0] + 3 5 DG − 5k + 10k = 0
DGx
A
Ay = 1.25k
I
= 0] + Ay −B 2k − 4k + 4.75
k = 0 CD
C
Ax = 0
DG
60°
2
DGx
DGy
D
H
a
Ay = +1.25k (↑)
C
Cy = 5k
G
C
I
4k
2k
Fy = +4.75k (↑)
DG
D
FG
Ay = 1.25k
[ΣFAxx == 00 ] Ax = 0
G
DGy
€
60°
B
60°
Cx =0
€
FG
E
Support Areactions for the truss:
[ΣFy =5k0] − 5k − 10k + 5k + A y = 0
F
a
A 60°
H
Support reactions for the entire FBD:
A y = +10k (↑)
D
H
10k
5k
4k
2k
C
B
G
A
3.44
3.44
( Ay )
(
)
−1.25k (15') + 3.61k 5 3 +
HD = −1.44k (C)
( GH)
1
2
(HD)(5
( HD x )
)
3 +
3
2
(HD)(5') = 0
( HD y )
€
€
3.20
.45
1k
1k
1k
B
J
G
1k
3.45
Ox = 0
a
A
O
Oy = 3k
Ox = 0
1k
1k
H
1k
a
G
1k
a
4kN
C
B
K
E
3.46
3.46
C
a
Oy = 3k
3
9.4
4’
Solve for BG:
A
4
HE
G
[ ΣM B = 0]
= 2kN( 4m) + 4kN( 2m) − DE( 3m) = 0
[ ΣM Z = 0]
+ 2kN( 2m) − AB( 3m) = 0
DE
E
AB = +1.33kN ( T)
1k
E
GB
[ΣFy = 0] − 2kN − 4kN + 3 5 BC = 0
B
HB
10
3
Oy = 3k
1k
H
1k
8
5
D
DE = +5.33kN (C)
HB
10
O
10’
Z
B
GB
9.4
N
4
D
K
G
1k
C
2kN
Dy = 1k
1k
1k
A
E
B
BC
Dy = 1k
H
AB
D
J
A
O
A
8
BC = +10kN ( T)
€
4’
5
BG y = 2 5.4 BG
O
H
HE E
#
[ ΣM H = 0] − 300010’
( 20') − ( 5 5.4 BG )(12') + 1000 # ( 20') + 1000 # (10') = 0
Oy = 3k
BG = −2700 # (C)
N
BG x = 5 5.4 BG;
€
Solve for HE:
[ ΣM B = 0]
− 3000 # ( 30') + HE(16') + 1000 # (10') + 1000 # ( 20') + 1000 # ( 30') = 0
HE = +1875# ( T)
€
Solve for HB:
HBy = 8 9.43 ( HB)
[ ΣM N = 0]
+ 3000 # (10') +
HB = +1179 # ( T)
8
9.43
( HB)( 30') − 1000 # (10') − 1000 # ( 20') − 1000 # ( 30') = 0
€
3.21
3.48
3.47
Ay = 20k
a
F
D
3.47
B
B
a
Ay = 24k
C
E
A
C
Ay = 24k
E
F
H
G
a
J
K
I
48k
G
a
I
K
1
O
312’
1
O
12
A
30’
24’
B
24’
C
In this example, a moment equation will be used Cto
determine the312’
effective tension counter.Ay = 24k
−
4DG
5DG
( 30') −
( 360') + A y ( 312') = 0
41
41
( DG x )
DG = +25k ( T)
Ly = 24k
24’
30’
24’
E
DF = −54.2k (C)
[ ΣM D = 0]
€
a
b
C
F
G
5k
By = 0
1
b
Bx = 20k
Bx = 20k
F
[ΣM B = 0]
G
G
G
b
5k
Ay = 20k; D A = 20HD2k
( Ay )
By = 0
[ΣFx = 0]
10k
b
a
+ Ax (20') − 5k (10') −10k (20') − 5k ( 30') = 0
10k
HD
+ Bx − 20k = 0
ED
( Ax )
F
D
E
EF
ED y = 5k;
€
a
5k
H
HD
ED
ED = 5 2k ( T)
F
H
HF
E
EF
F
CD
E
HF
ED y − 5k = 0
[ΣFy = 0] +5k
C
5k
HF
D
Bx = 20k
F
H
[ΣFy = 0] + 20k− 5k − 5k −10k + By = 0
G
E
F
By = 0
Ax = +20k;
H
5k
B
5k
By = 0
a
1
a
1
10k
B
D
H
E
F
1
b
D
C
Bx = 20k
ED
5kEF
E
5k
D
H
DC
( DFy )
− 24k( 48') + EG ( 30') = 0
EG = +38.4k ( T)
1
1
C
FC
To solve for EG:
€
H
1
1
EF
EG
A
€
12DF
DF
[ ΣM G = 0] −
( 30') −
( 24') − 24k( 72') = 0
145
145
( DFx )
1 D
Ax = 20k
1
Ly = 24k
DF
EF
DG
EG
E
Ay =a20k
bA
( DG y )
To solve for DF:
€
Ay = 20k
1
Ax = 20k
C
From (ΣΜΟ=0), Ay causes a counter-clockwise rotation
about point O. The only tension counter capable of
resisting in the clockwise direction about O is member DG.
Therefore:
[ ΣM O = 0]
3.48
DG
D
Ay = 24k
12
3.48
L
DF
D
12
A
1
B
B
1
A
L
48k
1
12
Ax = 3.48
20k
J
D
A
H
G
FG
D
CD
C
DC
E
F
CD
FC y = 15k;
€
G
FG
FC = 15 2k
FC
E
F
G
H
DC
FC5k
10k
H
[ΣFy = 0] + FC yD− 10k − 5k = 0
FG
F
5k
E
3.22
800#
C
3.49
3’
3.49
400#
693# 30°
3’
D
3.50
3’
800#
B
C
400#
D
3.50
Ay
Ay
800#
693# 30°
3’
3.50
Ax
Ax
A
A
FBD of the entire frame:
3’
[ ΣM G = 0]
3’
B
Ax
5’
800#
C
[ΣFy = 0] + A y + 400
MRA
A y = +400
Ax
[ ΣFx = 0]
A
#
#
− 800 = 0
Gx
Gx
(↑)
[ ΣM A = 0]
#
#
Ax = 320
Cy = 400#
C
Cy = 400#
Cy = 400# By = 800#
B
C
Bx = 800#
Cx = 107#
= 800#
By =Bx800#
B
Bx = 800#
A
D
B
400#
800#
By = 800#
MRA = 3144#-ft
C
Fy = 720
Dx = 800#
C
Fy = 720
Gx = 320
E
€E
360#
By = 1080
Dy = 720
Cy = 400
360#
Cy = 400
Cx = 320
Cx = 320
€
F
Gx = 320
By = 800#
By = 1080
Dy = 720
800#
Cx = 320
Cy = 400
B Bx = 0
D
Cx = 320
Cy = 400
B Bx = 0
D
D
D
C
G
+ 1080 # (2') − Ay (6') + 320 # (6') = 0
( By )
( Ax )
Ay = +680 # (↑)
Bx = 0
C
Dy = 800#
B
[ΣM C = 0]
By = 1080
Bx = 0
B
693#Dx30°
= 800#
Dx = 800#
Using the result for By = 1080#;
proceed to FBD of the inclined
member ABC:
€
400#
Dy = 800#
800#
C
MRA = 3144#-ft
Bx = 800#
B
Ay = 400#
A
A
D
Dx = 800#
Cx = 107#
[ ΣFx = 0] Bx = 0
[ ΣM B = 0] + D( 2') − 360 # ( 4') = 0
B y = 1080 #
Ay = 680
A
693# 30°
C
FBD of the horizontal beam DBE:
#
− 360 # = 0
[ΣFy = 0] + By − 720
( D)
By = 1080
800#
(Ax )
D = F = 720 #
Ay = 680
Ax = 320
Cy = 400#
F
#
− M R A + 693 (8') + 400 ( 6') − 800 ( 6') = 0
€
€
Gy
M R A = +3144 #−ft. clockwise
Ay
360#
360#
F
G
G x − 320 # = 0
G x = 320 # (→)
G Gy
+ A x − 693# = 0
A x = 693# (→)
MRA
#
A x = +320 # (← )
[ ΣFx = 0]
C
Ay
Ax = 693#
E
B
D
A
5’
Ax = 693#
E
B
D
+ A x ( 9') − 360 # (8') = 0
F
= 0] (+680 # )−1080 # + C y = 0
[ΣF
y
(Ay )
C y = +400 #
Return to the FBD of the entire
frame and solve for Gy;
[ΣF
y
= 0] + (680 # )− G y − 360 # = 0
(Ay )
G y = +320 # (↓)
Use the FBD of the inclined
member CFG;
[ΣFx = 0]
[ΣF
y
G Gy = 320
( By )
+ ( 320 # )− C x = 0
(Gx )
= 0] −320 # + 720 # − C y = 0
( Gy )
C x = 320 #
C y = 400 #
Gy = 320
€
3.23
Ax
3.51
D
A
2-force members
C
3.51
F
3.52
R = 400#
2’
hinge
2’
MRC
4’
B
10’
Ay
Ay
3.52
200#
A
Ax
Cx
Bx
FBD (a)
E
2-force members
FBD (a)
C
Cy
8kN
D
B
F
2m
2m
Ay = 8kN
B
Bx
2’
Bx
10’
Ay
FBD (b)
Bx
By
By
2’
MRC
4’
FBD (c)
Cx
[ΣFx = 0]
Cy
A
Cy
y
Cx = 16kN
DE
Cy
€
[ ΣM C = 0]
E
DF
E
+ 80 # ( 6') + 200 # ( 4') − M R C = 0
M R C = +1280
DF = EF
8kN
D
DF
#−ft.
FBD Joint F
F
EF
€
Bx = 12kN
Joint F:
[ΣFx = 0]
EF
Cx = 0
FBD (c)
C
ED = 8kN
Cy = 12kN
FE = 5.66kN
D
FBD (b)
D
B
DF = 5.66kN
F
[ΣFy = 0] − 80 # − 200 # + C y = 0
Bx = 12kN
Ay =C8kN
(↑) C
x
FBD(c):
C y = 280 #
E
= 0] + Ay − 8kN = 0
C
ED = 8kN
Cy = 12kN
FE = 5.66kN
( Bx )
A y = +320 #
[ ΣFx = 0]
DE
FBD=(b)
− Ax + 12kN
0
Ax = 12kN A x = 12kN (←)
[ΣFy = 0] + A y − 100 # ft (4') + 80 # = 0
€
Cx = 16kN
C
BA=y =12kN
8kN (→)
FBD(b):
B y = +80 #
Cx
2m
2m
DF = 5.
[ΣM A = 0] − 8kN(6m) + B(4m) = 0
[ΣF
[ ΣFx = 0] Bx = 0
[ ΣM A = 0] − 100 # ft ( 4')( 2') + By (10') = 0
FBD (a)
Support reactions:
200#
D
E
A
Ax = 12kN
R = 400#
8kN
B
FBD (c)
− 0.707DF + 0.707EF = 0
[ΣFx = 0] + 0.707DF + 0.707EF − 8kN = 0
2(0.707DF) = 8kN
DF = 5.66kN ( T); EF = 5.66kN (C)
€
8kN
E
FBD Joint F
3.24
500#
F
8kN
Bx
3.52 - cont’d
B
Dy
3.53
E
FBD (a)
2m
BAy
BAy
B
2m
BAx
BAx
B
D
A
C
CCx
DF = 5.66kN
Cx
Cx = 16kN
C
DE
Cy
E
Bx = 12kN
D
(
Dy
FBD (c)
B
y
C y = 12kN
)
[ΣF
y
Figure (c):
€
( DE)
F
EF
E
FBD Joint F
Dy
3.53
BAy = 55.5
ABy = +55.5B#
B
BAx = 44.4
[ΣM B = 0]
500#
Ay
Figure (b):
[ΣFx = 0]
y
Ay = 333.3# (↑)
BAy
B
€
BAx
)
BAy = 55.5
Cx = 44.4
Cy = 222.2
FBD(a)
Dy
BAy
A
FBD(c)
A
ABx = 44.4
4
ABy = 55.5
ABx
ABy
C x = +44.4 #
C
= 0] + 166.7 # − 500 # + Ay = 0
2-force
member
5
2-force
member
4
ABy
BAy = 55.5
BAx = 44.4
B
BAx = 44.4
ABy = 55.5
[ΣFx = 0] Ax = 0
[ΣM A = 0] − Dy (12') + 500 # ( 4') = 0
Dy = 166.7 # (↑)
B
− C x + ( 44.4 # ) = 0
Support reactions:
[ΣF
Dy
5
ABy = 55.5
(
]
B
D
B
C
ABx
A
Cx
Cx
[ΣFx = 0C] + BAx − ABx = 0
FBD(c)
C
A
Cx = 44.4
BACx y= +44.4 #
ABx = 44.4
Cy
Ay
Cy = 222.2ΣF = C
y=
222.2+ 55.5 # = 0
0
−
BA
y
y
500#
FBD(a)
Ay = 333.3
500#
FBD(a)
BAy = +55.5 #
FBD(b)
(Note: resultant
Dy
FBD(b)
vertical = 227.8#)
€
Ax
+ 55.5 ( 4') − ABx (5') = 0
#
C
B
A
BAy
55.5
BAy =
BA
x
BAx = 44.4
BAx
ABx = +44.4 #
€
C
ABx
ABy
Ay
Ay
500#
FBD(b)
#
y
= 0] + 222.2 #BA
− 500
+ 333.3# − ABy = 0
[
8kN
3.53
A
FBD(c)
ABx
Ax
A
C y = 222.2 #
DF
= 0] + 8kN
− C y + 8kN− (0.707)(5.66kN) = 0
€
Dy
C x −12kN − (0.707)(5.66kN) = 0
( Ay )
500#
4
Figure (b): [ΣM A = 0] − C y (9') + 500 # ( 4') = 0
C x = 16kN
[ΣF
Cy
FBD(a)
ABy
A
C
Cx
Cy
+ 12kN(2m) − 8kN(2m) + DE (2kN) − 5.66kN 2 2m = 0
DE = 8kN ( T)
[ΣFx = 0D]
C
ED = 8kN
Cy = 12kN
FE = 5.66kN
FBD (b)
[ΣM C = 0]
2-force
member
5
Ay = 8kN
Ax = 12kN
B
C
A
5
4
FBD(c)
ABx = 44.4
Cy = 222.2
500#
FBD(b)
Ay = 333.3
2-force
member
A
ABx = 44.4
ABy = 55.5
(Note: resultant
vertical = 227.8#)
3.25
3.54
3.54
FBD (b):
[ ΣM D = 0]
C
B y = 3k
− B y (18') + 7.2k( 4') − 3.6k( 3') = 0
[ΣFy = 0] + 3k + D y − 3.6k − 7.2k = 0
D y = 7.8k
3.54
C
9’
€
9’
R1 = 7.2k
Cy
B
D
A y = 3.24k
[ ΣFx = 0]
A xAy= 0
Cx
C
FBD (b):
[ΣFx = 0]
− 0.333k + Dx = 0
Dx = 0.333k
Ey
FBD(a)
C
€
FBD(c)
Cy
€
€
Cy
Cx
+ Bx − 0.333k = 0
Bx = 0.333k
R2 = 3.6k
3’
+ C x (16.2k ) − 3.24k (1') + (−0.24k )(9') = 0
C x = 0.333k
€
Ey
9’
[ΣFy A=x0] + A y − 7.2k − C3.6k + 7.56k = 0
= 0] + 3.24k − 3k − C y = 0
[ΣFx = 0]
= 7.2k
E y ( 20') − 7.2k( 4') − 3.6kR(122'
)=0
[ ΣM A = 0]
E y = 7.56k
9’
y
[ΣM B = 0]
D
Ay
[ΣF
C y = 0.24k
R1 = 400 # ft (18') = 7200 #
R 2 = 600 # ft ( 6') 3600 #
B
Ax
R2 = 3.6k
3’
FBD (a):
Cy
Cx
Cx
R1 = 7.2k
C
R2 = 3.6k
By
Dy
Bx
B
Ax = 0
B
Bx
Ay = 3.24k
By
FBD(a)
Ax = 0
Bx
Ay = 3.24k
Bx
B
By
FBD(c)
Dy
FBD(b)
Summary:
By
B
D
Dx
Ey = 7.56k
R1 = 7.2k
Ax = 0, Ay = 3.24k
Ey = 7.56k
Bx = 0.33k, By= 3k
Cx = 0.33k, Cy= 0.24k
Dx = 0.33k, Dy = 7.8k
FBD(b)
Dx
R2 = 3.6k
Dy
Dx
D
Dy
Dx
Ey = 7.56k
Summary:
Ax = 0, Ay = 3.24k
Ey = 7.56k
Bx = 0.33k, By= 3k
Cx = 0.33k, Cy= 0.24k
Dx = 0.33k, Dy = 7.8k
3.26
3.55
3.55
3.55
From FBD (b):
B
B
600#
600#
A x = 2714 # (← )
R2 = 1560#
R2 = 1560#
3000#
3000#
[ΣFy = 0] − 3000 # + 2714 # − A y = 0
1440#
1440#
A y = 286 # (↑)
(
ft
R1 (
=
2
2
R1
#
=
= 3000
2
100 #
R1x =
Cy
Cy
FBD(a)
FBD(a)
Ay
Ay
)
€
)( 30 2')
= 3000
R1 = (100 # ft ) 30 2' = 4242.6 #
Cx
Cx
Ax
Ax
R1y
#
Bx
Bx
By
By
3000#
3000#
R1 = 4243#
R1 = 4243#
Bx
Bx
3000#
3000#
€
By
By
C x = 1726
+ 1440 # + 286 # − C x = 0
#
(←)
3.56
3.56
€
600#
600#
R2 = 1560#
R2 = 1560#
2kN
C
1440#
1440#
FBD(c)
FBD(c)
B
Cx
Cx
FBD(b)
FBD(b)
Ax
Ax
[ ΣFx = 0]
C y = 2114 # (↑)
R 2 x = 1213 ( R 2 ) = 1440 #
R 2 y = 513 ( R 2 ) = 600 #
From FBD (c):
[ΣFy = 0] − 2714 # + 600 # + C y = 0
R 2 = ( 60 # ft )(1312 × 24') = 1560 #
B
B
+ 3000 # − 286 # − A x = 0
[ ΣFx = 0]
3000#
3000#
R1 = 4243#
R1 = 4243#
1kN
Cy
Cy
Ay
Ay
From FBD (b):
[ ΣM A = 0]
From FBD (c):
#
#
[ ΣM C = 0Summary:
( 5') A−y1440
(12') − Bx ( 24') + By (10') = 0
2714#,
= 286#
A] x −= 600
Ax = 2714#, Ay = 286#
€
− 3000 # (15') − 3000 # (15') + B x ( 30') + B y ( 30') = 0
Summary:
Ax
Bx = 286#, By = 2714#
Bx = 286#, By = 2714#
x = 1726#, Cy = 2114#
1726#,
Cy := 2114#
Cx =for
equationsC
(FBDb)
Solving the two
simultaneously,
€
1kN
A
Ay
B x = 286 # (← )
FBD(a)
D
Dx
Dy
FBD (a):
B y = 2714 # (↑)
[ ΣM A = 0]
The same forces are equal and opposite for FBD(c).
+ 2kN( 4m) + 1kN( 2m) − 1kN(1.33m) − D y ( 2.66m) = 0
D y = 3.26kN (↓)
€
[ΣFy = 0] + A y − ikN − 3.26kN = 0
A y = 4.26kN (↑)
2kN
€
3.27
C
By = 4.26kN
3.57
3.57
3.56 - cont’d
3.57
D
C
By = 4.26kN
FBD(c)
9.5’
FBD(b)
Ax = 0.33kN
1kN
A
Ay = 4.26kN
Ax
9.5’
D
A
Dy = 3.26kN
3#
[ ΣFx = 0]
− 2kN − 1kN + 3.33kN − A x = 0
03
03
9.5’
€
Dy = 752#
FBD(b)
Ax = 1524#
F
Gx = 376#
G
Gy = 752#
[ ΣM D = 0]
G x = 376
F
752#
Gy =G
x = 376#
Ax = 1524#
Ay = 752#
A
From FBD (c):
E
G
Ay = 752#
€
#
FBD(c)
A
A x = 0.33kN
E
700
FBD(c)
FBD(b)
9.5’
700 A y = 752 # (↓)
#
#
Go back to FBD (b) and solve for Ax.
Dy = 752#
#
44
€
B
H = 1900#
(↑)
14
D x = 3.33kN
B
H = 1900#
G y = 752
#
− 3.33kN + D x = 0
#
3
209
− 1900 # ( 9.5') + G y ( 24') = 0
44
[ ΣFx = 0]
D
Dx = 376#
Dy = 752#
Dy = 752#
D
Gx
[ ΣM
A = 0]
[ΣFy = 0] − A y + 752 # = 0
D
14
C
Dx = 376#
D
209
C
G
Gy
Ay
B y = 4.26kN
− 4.26kN( 2.66m) + 1kN(1.33m) + B x ( 3m) = 0
Gy
Ax
Ay
From the FBD of the
entire framework;
Gx
G
FBD(a)
A
Dx = 3.33kN
[ΣFy = 0] + By − 1kN − 3.26kN = 0
B x = 3.33kN
€
FBD(a)
Using FBD (c):
[ ΣM D = 0]
F
B
H = 1900#
H = 100 # ft (19') = 1900 #
F
B
H = 1900#
Horizontal force on the
frame is equal to:
E
#
1kN
E
C
B
#
Bx = 3.33kN
22
B
D
C
By = 4.26kN
22
2kN
#
+ 752 # (12') − G x ( 24') = 0
(Gy )
(←)
D x = 376 # (→)
[ ΣFx = 0]
#
− Dy = 0
[ΣFy = 0] + 752
(G )
y
D y = 752
#
(↓)
From FBD (b): [ ΣFx = 0] + 1900 # − A x − 376 # = 0
€
€
A x = 1524 # (← )
3.28
3.58
3.58
Bearing pressure check:
1’
0.5’
x=
W
W = (1')( 3')(1')(150 # ft 3 ) = 450#
3’
P
3.58
A
pmax
(toe)
€
1’
P = ( 1 2)p max × h ×1'= ( 1 2)(105 # ft 2 )( 3')(1') = 157.5#
p max =
1’
M RMA = W
3’ × 0.5'= ( 450# )(0.5') = 225# −ft.
S.F. =
M RMA
M OTM
A A
=
W2 = ( 4')(1.5')(1')(150 # ft 3 ) = 900#
2’
W3 = (1')(5.5')(1')(150 # ft 3 ) = 825#
W2
pmax
does not
meet the
.75’
x=0.15’
2
2
R = pmax
(W ) + (P) =
€
W1 = 1 2 (1.5')(1.5')(1')(150 # ft 3 ) = 168.75#
W3
overturning requirement.
(toe)
W1
P
225# −ft.
= 1.43 <h/3
1.5= 1’
157.5# −ft.
P=157.5#
(toe) wall
Retaining
A
€
2W 2( 450# )
=
= 2000psf < 3000psf (allowable)
3x 3(0.15')
3.59
M OTMA =W=405#
P ×1'= (157.5# ) ×1'= 157.5# −ft.
R=477#
(225# −ft.) − (157.5# −ft.) = 0.15'
( 450# )
€
0.5’ about the toe @ A:
Moments
W
€
=
Maximum bearing pressure at the toe is:
∴ OK
3.59
p max = ω' × h = ( 35 # ft 3 )( 3') = 105 # ft 2
W
x = 0.15'< a = 1' = 0.33'
3
3
€
h/3
€= 1’
M RMA − M OTMA
(450)
2
P
2
+ (157.5) = 477#
€
h/3=1.83’
W=405#
p max = ω' × h = ( 30 # ft 3 )(5.5') = 165 # ft 2
R=477#
P = 1 2 p max × h ×1'=
P=157.5#
A
(toe)
pmax
WT = 168.75# +900# +825# = 1894#
x=0.15’
2
(165 # ft )(5.5')(1') = 454#
2
M OTM = P × h = ( 454# )(1.83') = 831# −ft.
3
M A = (W1 )(1') + (W2 )(0.75') + (W3 )(2')
€
M A = (168.8# )(1') + (900# )(0.75') + (825# )(2')
€
b=1.32’
€
1
=1894#−ft.
MA W
= T2494#
3.29
3.60
3.60
h/3=1.83’
3.59 cont’d
3.60
p max = ω' ×h
But, MA = WT x b
b=1.32’
WT=1894#
(2494# −ft.) = 1.32'
b=
(1894#)
0.5’
0.5’
P=454#
R=1948#
S.F. =
M RM (2494# −ft.)
=
= 3.0 > 1.5
M OTM (831# −ft.)
A
A
∴ OK, wall is stable
P=
W3
W3
P
P
2.5’
2.5’
W2
W2
R=
€
€
x=
2
+ (P) =
(1894 )
2
WT=4170#
WT=4170#
€
2
+ ( 454 ) = 1948#
W
1894#
= 2T ( 4a − 6x) =
2 ( 4 × 2.5'−6 × 0.88')
a
(2.5')
p max = 1430psf < 3000psf
∴ OK
#
2
ft 2
#
ft 3
#
2
ft 3
#
3
ft 3
WT = 1200# +450# +2520# = 4170#
€
( 4170)
2
2
+ (1280) = 4362#
M RM = M A = W1( 0.5') + W2 ( 2.5') + W3 ( 2.5')
M A = (1200#)( 0.5') + ( 450#)( 2.5') + ( 2520#) = 8025# −ft.
M A = WT × b = 8025# −ft.
M RM − M OTM (2494# −ft.) − (831# −ft.)
=
= 0.88'
WT
1894#
a = 2.5' = 0.83'< x = 0.88'< 2a = 1.76'
3
3
3
∴ within the middle third
p max
€
(WT )
2
1
€
R=
b=1.92’
b=1.92’
x=0.88’
× h × 1'
1
pmax
pmax
€
2 p max
(
)
P = ( 320
)(8')(1') = 1280#
W = (1')(8')(1')(150
) = 1200#
W = ( 3')(1')(1')(150
) = 450#
W = ( 3')( 7')(1')(150
) = 2520#
2.67’
2.67’
h/3=1.83’
1
p max = 40 # ft 3 (8') = 320 # ft 2
€
€
W1
W1
M RM = M A = 2494# −ft.
€
2.5’
2.5’
P=1280#
P=1280#
R=4362#
R=4362#
x=1.1’
x=1.1’
A
A
€
b=
8025# −ft.
= 1.92'
4170#
2.67’ M
h
OTM = P × 3 = (1280# )( 2.67') = 3418# −ft.
2.67’
S.F. =
€
€
€
€
M RM
8025# −ft.
=
= 2.34 > 1.5
M OTM 3418# −ft.
a = 4' = 1.33'
3
3
∴ OK
2a = 2.67'
3
M RM − M OTM (8025 − 3418)
=
= 1.1'
WT
4170
x < a3
x=
p max =
2W 2( 4170)
=
= 2527 # ft 2 < 3000psf
3x
3(1.1')
∴ OK
€
3.30
3.62
3.61
3.61
3.62
(
)
p max = ω'×h = (5.5 kN m 3 )(5.5m) = 30.25 kN m 2
W1 = (1')( 3')(1') 150 # ft 3 = 450#
3.5’
(
)
W2
WT = 450# +1200# = 1650#
(
)
320
(
)(8')(1') = 1280#
p max = 40 # ft 3 (8') = 320 # ft 2
€
P
1.5’
2.67’
W1
€
A
pmax
€
P=
A
2
#
(1650)
0.75m
2
M RM = M A = W1(1.5') + W2 ( 3.5')
M A = ( 450#)(1.5') + (1200#)( 3.5') = 4875# −ft.
1.5m
W3
P=1280# €
€
€
1.83m
A
pmax
4875# −ft.
= 2.95'
1650#
M RM
4875# −ft.
=
= 1.43 < 1.5
M OTM 3418# −ft.
Not stable for overturning
b = 1.67m
€
2
2
+ (83.3) = 286kN
2W 2(1650)
=
= 1250psf < 3000psf
3x
3( 0.88)
WT = 274kN
P = 83.3kN
M RM − M OTM ( 4875) − ( 3418)
=
= 0.88'
WT
1650
x < a 3 = 1.33'
a = 4' = 1.33'
3
3
(274 )
M€A = (58.8kN)(0.75m) + ( 35.3kN)(1.5m) + (180kN)(2m)
M A = 457kN − m
€
x=
∴ OK for bearing pressure
x=0.88’
R = WT2 + P 2 =
M RM = M A = W1 (0.75m) + W2 (1.5m) + W3 (2m)
M OTM = P × h 3 = (1280#)( 2.67') = 3418# −ft.
p max =
2
WT = 58.8kN + 35.3kN + 180kN = 274kN
P
W2
S.F. =
WT=4170#
(30.25 kN m )(5.5m)(1') = 83.2kN
W3 = (2m)(5m)(1m)(18 kN m 3 ) = 180kN
2
+ (1280) = 2088#
2
W2 = (0.5m)( 3m)(1m)(23.5 kN m 3 ) = 35.3kN
€
W1
ft 2
1
W1 = (0.5m)(5m)(1m)(23.5 kN m 3 ) = 58.8kN
2m
M A = WT × b = 4875# −ft.
€
R=2088#
1
R=
b=
b=2.95’
P = 1 2 p max (h)(1') =
W2 = (1')(8') 150 # ft 3 = 1200#
R = 286kN
€
A
x = 1.11m
€
2a = 2.67'
3
€
€
M A = WT × b = 457kN − m
457kN − m
= 1.67m
b=
274kN
M OTM = P × h 3 = (83.2kN)(1.83m) = 152kN − m
S.F. =
M RM 457kN − m
=
= 3.0 > 1.5
M OTM 152kN − m
∴ OK
a = 3m = 1m
2a = 2m
3
3
3
M RM − M OTM (457 −152)kN − m
x=
=
= 1.11m
WT
274kN
a < x = 1.11m < 2a
3
3
∴ p max =
WT
(274kN) 4 × 3m − 6 ×1.11m
(4a − 6x) =
)
2 (
a2
(3m)
p max = 163 kN m 2 ( 3400psf ) > 143.6 kN m 2 ( 3000psf )
Overstressed in bearing
€
3.31
3.63
W1 = (1.33')(14.67')(1')(150 # ft 3 ) = 2927#
W2 = (1.33')(8')(1')(150 # ft 3 ) = 1596#
5.33’
2’
W3 = (5.33')(14.67')(1')(150 # ft 3 ) = 8992#
W1
4’
WT = 2927# +1596# +8992# = 13,515#
W3
P
€
W2
5.33’
A
pmax
p max = ( 40 # ft 3 )(16') = 640 # ft 2
P = 1 2 (640 # ft 2 )(16')(1') = 5120#
R = WT2 + P 2 =
b = 4.45’
WT = 13,520#
€
P = 5120#
R = 14,450#
A
x = 2.43’
€
€
€
€
(13,520)
2
2
+ (5120) = 14,450#
M RM = M A = W1 (2') + W2 ( 4') + W3 (5.33')
M A = (2927# )(2') + (1596# )( 4') + (8992# )(5.33')
M A = 60,200# −ft.
M A = WT × b = 60,200# −ft.
60,200# −ft.
= 4.45'
b=
13,520#
M OTM = P × h 3 = (5120# )(5.33') = 27,300# −ft.
M RM 60,200# −ft.
=
= 2.2 > 1.5
M OTM 27,300# −ft.
∴ OK for overturning
S.F. =
a = 8' = 2.67'
2a = 5.33'
3
3
3
M RM − M OTM 60,200 − 27,300
=
= 2.43'< a 3
x=
13,520
WT
p max =
2W 2(13,520# )
=
= 3709psf > 3000psf
3(2.43')
3x
∴ exceed the allowable bearing pressure
3.32
Chapter 4 Problem Solutions
4.2
4.1Problem Solution 5.1.1
ωsnow= (25 lb./ft.2)x(2') = 50 lb./ft.
12
sf
7
ω = (50 psf)x(5') = 250 lb./ft.
0p
€
L=10 ft.
Beams B-1, B-2, B-3
Wall A
777 lb.
R
R
ωsnow= 11'
(25 lb./ft.2)x(2') = 11'
50 lb./ft.
(1,250 lb.)
(1,250 lb.)
ω DL = 10 lb. ft.2 × ( 2') = 20 lb. ft.
(
)
⎛ trib. ⎞
⎜
⎟
⎝ width ⎠
ω Snow = 25 lb. ft.2 × ( 2') = 50 lb. ft.
Wall B
777 lb.
(
)
⎛ trib. ⎞
⎜
⎟
⎝ width ⎠
€
'
ω Total
Snow
DL =2(=50
ω ==ω(10
lb.+
/ft.ω2)x(2')
20lb.
lb./ft.
ft. ) + ( 20.6 lb.12ft.) = 70.6 lb. ft.
FBD of beams B-1, B-2, B-3
Tributary width for beams B-1, B-2, B-3
Rafters
T
T
12.37
12
3
12.3
=5
LL
+
DL
ωDL= (10 lb./ft.2)x(2') = 20 lb./ft.
5'
Tributary
width
Slope conversion :
DL
ω = (15 lb./ft. )x(2') = 30 lb./ft.
3
2.37Ceiling
€projection of the roof while the dead loads
Snow loads are assumed to be on the horizontal
1
Joists
12'
€
Gi
rde
Col. A
am
Be
rG
1
B-
10'
am
Be
330 lb. horizontal
Wall
B and added to the snow load. Determination of
Wall A into an equivalent
converted
load
777 lb.
777 lb.
the dead load 11'
as an equivalent
horizontal
load requires a slope conversion.
11'
Beam B-1
(1,250 lb.)
Beam B-1
(1,250 lb.)
-1
10'
10'
are applied 10'
along the length of the rafter.
T
T Interior Wall
Wall A
Wall B Rafters
Both150
load
conditions
are
combined to simplify
lb.
180 lb. the computations. Generally, the dead load is
150+180=
463 lb./ft.
478 lb./ft.
165 lb./ft.
1
B-
Girder G-1 (partial framing)
Col. D
(Col. A-1)
(Col. D-1)
(1,250 lb.)
(1,250 lb.)
ω = (15 lb./ft.2)x(2') = 30 lb./ft.
FBD of girder G-1
8'
Wall A
150 lb.
1250 lb.
ω = 250 lb./ft.
Wall A
543 lb./ft.
463 lb./ft.
ω = 250 lb./ft.
Wall
12' 80 lb./ft.
Wall
lb./ft.
8010'
Interior Wall
150+180=
330 lb.
Int. Wall
245 lb./ft.
165 lb./ft.
Walls
Ceiling
Joists
Wall B
180 lb.
€
Wall B
558 lb./ft.
478 lb./ft.
10'
Col D-3
8'
FBD of girder G-3
FBD of girder G-2
B
B3
1250
3750
1250
2500
Open
B1
B1
1250
1250
2500
1250
1250
B2
1250
3750
G2
Walls
Wall
Wall
80 lb./ft.
80 lb./ft.
ω = (60 lb./ft.2)x(1') = 60 lb./ft.
Wall A 10'
543 lb./ft.
Wall A
300 lb.
D
G1
1250
G3
C
Int. Wall 12'
245 lb./ft.
Inter. Wall
660 lb.
€
3rd floor walls:
Wall A:
Roof = 777 lb. 2' = 388 lb. ft.
Ceiling = 150 lb. 2' = 75 lb. ft.
463 lb. ft.
3rd Flr
Wall B
360 lb.
Wall B:
Roof = 777 lb. 2' = 388 lb. ft.
Ceiling = 180 lb. 2' = 90 lb. ft.
Wall B
558 lb./ft.
Interior Walls:
478 lb. ft.
€
Ceiling = 330 lb. 2' = 165lb. ft.
G3
A
1250
2500 lb.
1250 lb.
Col. D-3
3750 lb.
3
Col. D-1
1250
Col. A-3
3750 lb.
2
ω = 15 lb. ft.× 2ft. = 30 lb. ft.
L = 20'
L=30 ft.
1
Ceiling Joist: DL + LL = 15 lb./ft.2
3rd Floor: ω = (60 lb./ft.2)x(1') = 60 lb./ft.
€
3rd Flr
Wall B = 10'
ω × 6' = ( 60 lb. ft.) × 12'
( 6') = 360 lb. ft.
Interior
× ( 5' +6'
ft.)B× (11') = 660 lb. ft.
Wall) = ( 60 lb.Wall
Wall AWall = ( ω) Inter.
Summary of Column Loads
300 lb.
€
€
660 lb.
360 lb.
4.1
ω=
3
84
4.2 cont’d.
/ft.
lb.
ft.
b./
l
05
=9
ω
ω=
Inter.
Wall
Wall A
8
91
/ft.
lb.
ω=
.
ω
2nd Floor Walls:
ω
€
ω=
( wall
above)
( 3rd
.
floor )
Wall B
ft.
b./
360 lb. 998 l
Interior
ω = Wall: 245 lb. ft.+ =660 lb. ft. = 905lb. ft.
ω=
( wall
€
above)
ω
ft.
Inter../Wall
b
5 l lb.
660
8
9
( 3rd
€
ω=
( wall
€
floor )
( beam
weight)
Beam - 28’ span:
End reaction R Int.
2nd Flr
12'
ω =61 lb./ft.
Wall B
Wall A ωbm=43 lb./ft. Inter. Wall bm
above)
( 2nd
floor )
( beam
weight)
ω × L (1706 lb. ft.) × ( 28')
= total
=
= 23,880lb.
2
2
€
360 lb.
660 lb.
.
( 2nd
1
10'
RA=12,660 lb.
2nd floor:
above)
( wall
./ft
ω=
300 lb.
RInt.=23,880 lb.
ω total = ω total = ( 985 lb. ft.)+ ( 660 lb. ft.)+ ( 61 lb. ft.) = 1706 lb. ft.
lb
b
3 l = (60 lb./ft.2)x(1')6=
4560 lb./ft.
2ω
12
ωbm=61 lb./ft.
ω × L (1266 lb. ft.) × ( 20')
End reaction R A = total
=
= 12,600lb.
2
2
.
.
1
ω total = ω total = ( 923lb. ft.)+ ( 300 lb. ft.)+ ( 43lb. ft.) = 1266 lb. ft.
floor )
./ft
/ft.
lb.
Beam - 20’ span:
2nd Flr
12'
10'
l
9/8ft.
=8 l9b.
9ω1
Wall B
558 lb. ft.+ 360 lb. ft. = 918 lb. ft.
WalllbA./ft
23 lb.
9300
ω=
RA=12,660 lb.
ft.
b./
.
./ft
Inter.
Wall A
Wall A: 543lb.ωft.=+ (60
300lb.
lb./ft.Wall
ft.
= 843lb.
2)x(1')
= 60ft.lb./ft.
( wall above)
( 3rd floor )
ωwall = 80 lb./ft.
Wall B:
1
5
64
ωbm=43 lb./ft.
b
l
8/f5t.
=lb9.
5
ω
0
=9
/ft.
lb.
Wall B
ωwall = 80 lb./ft.
/ft
lb.
2./3ft.
9
b
ω4=3 l
=8
3
22
Wall B
360 lb.
Inter. Wall
660 lb.
Wall A
300 lb.
4.2 cont’d.
2nd Flr
12'
10'
RInt.=23,880 lb.
t.
/ft= 60 lb. ft.2 × ( 5'l)b.=/f 300 lb. ft.
Wall A = ω l×b.5'
45
23
12
(
)
6
=1
=
Wall ωB = ω × 6' = ( 60 lb. ωft.) × ( 6') = 360 lb. ft.
€
€
€
Interior Wall = ( ω) × ( 5' +6') = ( 60
lb.=61
ft.)lb./ft.
× (11') = 660 lb. ft.
ωbm
ωbm=43 lb./ft.
RA=12,660 lb.
RInt.=23,880 lb.
4.2
2460 lb.
82
ω
=3
4.3
ω = 66 lb./ft.
4'
8'
14'
ω=
Roof
Rafters
ft.
ω
=
2
38
12'
/ft.
424 lb. 396 lb.
Roof beam
lb.
.
ω
=4
ω
/ft
lb.
46 ft.
4
/
ω =82 lb.
=3
ft.
b./
l
10
ωwall = 64 lb./ft.
8'
396 lb. per 2'
ft.
Back wall
b./
ω
Front wall
m
ea
fb
o
Ro
'
12
(
=
8
19
ω=
l
6
=
36
€
+3
44
ω
ft.
b./
2l .
6
2 ./ft
b
ω =92460
8 l lb.
1
(Col.)
=
.
./ft
8=
g
tin
ω = 624 lb./ft.
ωOccupancy
= 96 lb./ft. (LL) = 40 psf
l
6
33
6+
ω
=7
62
4
=4
t
ron
( joist
spcg )
ng
oti ω = 96 lb./ft.
88
62
fo
2
ω=
14'
12'
F
€
672 lb. per 2'
(front footing)
1248 lb. per 2'
(floor beam)
2460 lb.= 782
(Col.)336
ω=
6+
44
/ft.
lb.
ω g= 624 lb./ft.
n
Fro
6'
6'
ω
6 + nt fo
ng
Fro
oti
4
=4
t fo
k
n
Fro
c
Ba
€
2
26
ω = ( 446 lb. ft.)+ ( 336 lb. ft.) = 782 lb. ft.
( Wall
load )
( floor
joists)
Back footing:
ω = ( 262 lb. ft.)+ ( 288 lb. ft.) = 550 lb. ft.
( Wall
load )
( floor
joists)
k
c
Ba
.
./ft
b
0l
55
8
+2
g
4332 lb.
tin
foo
2460 lb.
(Col.)
2460 lb.
(Col.)
ω = 624 lb./ft.
ω = 624 lb./ft.
6'
6'
6'
6'
3744 lb.
4920 lb.
(Col.)
4920 lb.
(Col.)
€
8664 lb.
(Int.
8664post)
lb.
(Int. post)
6'
8664 lb.
6'
3744 lb.
Ext.
post
4332
lb.
Int.
post
3744
lb.
Int. post
8664 lb.
3744 lb.
Int. post
Critical footing
Ext. post
Int. post
Int. post
Int. post
Critical footing
8664 lb.
(Int. post)
4920 lb.
(Col.)
6'
8664 lb.
3744 lb.
Int. post
Int. post
Critical footing
+2
g
tin
oo
f
ck
Ba
ft.
b./
0l
5
=5
Floor
Joists
576 lb. per 2'
(back footing)
ft.
b./
oti
fo
nt
lb
82
=7 g
6
n
33
oti
Front footing:
/f
lb.
2
4
/2ft.
ω DL+LL
lb. × ( 2ft.) = 96
ω =lb. ft.
ω = = 48 lb.82 ft.
)
ω
.
./ft
(back tfooting)
.
(floor beam)
f
b./
(
ω=
Back wall
Floor 6'
6'
Floor = 5 psf
Joists
Joists
=
3
psf
12'
14'
8' ωwall = 64 lb./ft.
8' ωwall = 64 lb./ft.
3744 lb.
Total DL = 8 psf 4332 lb.
Ext. post
Int. post
lb. per
2' psf
672 lb.
per 2'Load = LL
1248
2' psf +576
Design
Floor
+ lb.
DLper
= 40
8 psf
= 48
(front footing)
t.
6
33
6+
4
=4
Joists
8' 4920
ωwall
lb. = 64 lb./ft.
Front wall
46
rafter spcg )
(back (footing)
b
2l
78
o
t fo
)
(floor beam)
n
Back
Fro
' wall
12
2460 lb.
Floor Loads:
2
ω
= 33 lb. ft.576
×lb. per
( 2ft.2') = 66 lb. ft.
SL+DL
1248
lb. per 2'
Roof
672 lb. per
2'
(front footing)
Rafters
2460 lb.
764 lb. per 2'
Front wall
ft
b./
2l
8
=7
12'
576 lb. per 2'
(back footing)
576 lb. per 2'
(back footing)
/ft.
lb.
50
5
/ft.
8=
lb.
28
50
+
5
2
26
8 = ng
28 oti
ω=
2 + ck fo
6
g
2
Ba footin
ω=
1248 lb. per 2'
(floor beam)
1248 lb. per 2'
. (floor beam)
Joists
12'
14'
4920 lb.
'
14'
672 lb. per 2'
(front footing)
Floor
Joists
Floor
Design Load = SL + DL = 20 psf + 13 psf
Floor
= 33 psf
'
12
m
a
be
12
4'
Roofing = 5 psf
Sheathing
= 3 psf
ft.
b./
2l
Rafters
= 3 psf
6
2
ω = Ceiling = 2 psf
Total DL = 13 psf
/
lb.
14'
12'
ω = 96 lb./ft.
41
f
lb. per 2'
4.3 672
cont’d
(front footing)
= 20
psf
ωSnow
lb./ft.
wall = 64
6
b.
0l
o
Ro
14'
ωwall = 64 lb./ft. Roof Loads:8'
/ft.
=
ω = 66 lb.ω/ft.
ω = 96 lb./ft.
19
Back wall
396=lb.
44per 2'
ω wall
Back
424 lb. 396 lb.
Roof beam
b.
8l
Front wall
12'
764 lb. per 2'
Front wall
ω = 96 lb./ft.
/ft.
.
/ft
lb.
4920 lb. 550 l
(Col.)288 =
2+
g
26
tin
ω=
foo
k
c
Ba
6'
4332 lb.
3744 lb.
8664 lb.
3744 lb.
Ext. post
Int. post
Int. post
Int. post
8664 lb.
(Int. post)
4.3
h = 30'
4.4
Column
4.4 cont’d
Roof Loads:
Snow = 25 psf
Tributary width = 8'
ft.
b./
5l
3
=3
Beam reactions are treated as
concentrated
spaced
at 6’-0”
o.c.
ω = loads
255 lb./ft.
(trib. width
= 6')
Dead Load = 15 psf
ω
'
24
L=
Girder G-2:
ω total = (SL + DL) ×
4020 lb.
(Wall or
beam)
(8')
( Trib.width)
Truss joists are spaced close together
+ ω beam
(2’ or less),therefore, the reactions may
L = 16'
be treated as an equivalent distributed
Beam B-2
load on girder G-2.
G-2
Wall
ω total = 40 lb. ft.2 × (8') + 15 lb. ft. = 335lb. ft.
(
)
Beam B-1
(2040 lb.)
Beam Reaction: (24’ span)
4020 lb.
(Wall or
beam)
€
R beam =
or wall
2x4020 lb. = 8040 lb.
(every 8')
0'
L
'
0€
=4
Wall
(17k)
ωgirder = 50 lb./ft.
Girder G-1 supports concentrated beam
reactions from both sides plus its own self
2040 lb.
2040 lb.
ωtotal = 566 lb./ft.
L = 30'
12,570 lb.
weight. Since the beam reactions occur at
8’ o.c., they must be represented as
concentrated loads and not reduced to an
P = 42.2k
Girder G-2
12,570 lb.
The tributary width of load from the
truss joists onto the girder equals half
of the span or 12.
equivalent uniform load.
Column:
The column load includes the girder
reactions from both sides plus the tributary
weight of the girder.
Girders
Column
(42.2k)
Wall
(17k)
2040 lb.
2040 lb.
Girder G-1:
Spacing of beam loads
should be treated
as concentrated
loads every 8'
4
L=
ω total × L ( 335 lb. ft.) × 24'
=
= 4020lb.
2
2
(2040 lb.)
ω snow = 25 lb. ft.2 × 12' = 300 lb. ft.
(
)
ω DL = 18 lb. ft. × 12' + ( 50 lb. ft.) = 266 lb. ft.
(
2
)
( girder
weight)
ω total = ω snow + ω DL = 566 lb. ft.
h = 30'
€
Column
Beam B-2: (Tributary width = 6’)
ω = 255 lb./ft. (trib. width = 6')
ω snow = 25 lb. ft.2 × ( 6') = 150 lb. ft.
(
L = 16'
G-2
(
Wall
(2040 lb.)
2040 lb.
(2040 lb.)
2040 lb.
2040 lb.
)
2
)
ω DL = 15lb. ft. × ( 6') + (15lb. ft.) = 105 lb. ft.
Beam B-2
2040 lb.
€
( beam
wt.)
ω DL+SL = 150 lb. ft. + 105 lb. ft. = 255lb. ft.
ωtotal = 566 lb./ft.
4.4
4.5
4.6
ωsnow = 20 psf x (16/12 ft.) = 26.7 plf
ω DL
ωSL = 60 lb./ft.
Critical roof joist: (16” o.c. spacing)
Loads:
lf
0p
=2
1). Rafters
Ridge
beam
3
4
DL = 12 lb. ft.2 × (16 12)' = 16 lb. ft.
(
)
6
ω DL
Joist Wt. = 4 lb./ft.
Wall
€
17'
65
DL =
36 lb
.
Ridge
beam
4
/ft.
Ridge
beam
13
5
12
ωSL = ( 20 lb. ft.) × (16 12)' = 26.7 lb. ft.
( slope
€
ω
lb. / ft.
12.
12
Bearing
wall
ω'DL = ( 5 4) × ( 20 lb. ft.) = 25lb. ft.
FBD of the critical
inclined roof joist
=3
ωSL = 60 lb./ft.
Rafter span = 14'
adj)
FBD of an inclined rafter-left
ω total = ωSL + ω'DL = 26.7 lb. ft. + 25 lb. ft. = 51.7 lb. ft.
Bearing
wall
Rafter span = 16'
FBD of an inclined rafter-right
€
Slope adjusted deald load:
Slope adjusted deald load:
ω'DL = (36 lb./ft.) x (12.65/12)=38 lb./ft.
ω'DL= (5/4)x(20 plf) = 25 plf
€
ωtotal = ωSL + ω'DL
ω'DL = (36 lb./ft.) x (13/12)=39 lb./ft.
ωSL = 60 lb./ft.
ωSL = 60 lb./ft.
ωTotal = 38 lb./ft. + 60 lb./ft. = 98 lb./ft.
ωTotal = 39 lb./ft. + 60 lb./ft. =99 lb./ft.
ωtotal = 26.7 plf + 25 plf = 51.7 plf
17'
439 lb.
439 lb.
L = 14'
FBD of the equivalent,
horizontally projected roof joist
ω = 2x(439 lb./16")x(12/16 ft.)
ω = 659 plf
Ay
22.67'
ωbeam = 40 plf
Col. A
686 lb./24"
Bearing wall
792 lb./24"
Bearing wall
By
Col. B
2) Short wall/roof beam
R = 1 2 × ( 34') × ( 659 lb. ft.) = 11, 200 lb.
∴ B y = 9, 420 lb.
∑ Fy = − (11, 200lb.) − ( 40 lb. ft.)( 34') + ( 9, 420lb.) + A y = 0
792 lb./24"
Ridge beam
FBD of an equivalent
horizontal rafter - right
9
11.33'
ter
FBD of the Ridge beam
96
2
ω
9
=7
4"
lb. / 2
b. / ft.
l
=3
f
(ra
s)
d
loa
ω=
6
(41
1 ft. tributary
width
(wall = 10 psf)
.)
lb. / ft
+
3
(34
2'
16
lb. / ft
.
=4
Loads on the short bearing wall
5
=7
lb. / ft
'
L
0'
1
L=
3795 lb.
.)
lb. / ft
7590 lb.
ω
€
686 lb./24"
Ridge beam
FBD of an equivalent
horizontal rafter - left
Ridge Beam:
The equivalent concentrated load
from the triangular
load distribution is equal to:
17' + B 22.67') = 0
∑ M A = − (11, 200lb.)(17') − ( 40 lb. ft.)( 34'
€ )( ) y (
∴ A y = 3,140 lb.
L = 16'
0
=1
7590 lb.
.
(cont.)
Ridge
beam
Ext.
col.
Int.
col.
Ext.
col.
Ridge beam and column loads 4.5
4.6 cont’d
4.6 cont’d
(3) Walls:
lb. / ft.
3
34
ω=
b. / 2
l
92
ω=
10 psf
(7
)=
(8) Foundation Walls:
lb. / ft.
6
39
ft.
12 / 1
4
.+
10 psf
') =
8
sfx
3
42
8'
lb. / ft.
') =
0p
+(1
3
34
ω=
8
sfx
6
47
23
6
39
ω=
=4
p
(10
2
=8
7
=4
lb. / ft.
lb. / ft
3
77
lb. / ft.
4
53
)=
5'
6
')
fx5
ps
ω
lb. / ft
2 / 16
(1
0
+(1
=
6
87
10 psf
+
ω
76
2
=9
lb. / ft.
lb. / ft.
=8
(8) Continuous Footings
ωwall = (8/12)'x(3')x(150 lb./ft.3) = 300 lb./ft.
ω = 66.7
ω=
lb./
ft.
3
82
L = 14'
L = 16'
534 lb./16"
(Beam)
ω wa
8"
534 lb./16"
(Wall)
FBD of the floor joist -right
15"
(1.25')
/ft.
lb.
0
ll
FBD of the floor joist - left
5')
sfx
lb. / ft.
lb./
ft.
467 lb./16"
(Beam)
ω
3
(5) Floor Joists
467 lb./16"
(Wall)
6
10 psf
5'
lb. / ft.
(4) Interior Columns: See (2) above.
ω =(50 psf )x(16/12)' = 66.7
.+
lb. / ft
0p
+(1
lb. / ft.
lb. / ft
6
(
76
ω
8'
)=
.
.
3
77
0
=3
ft.
b./
6l
lb. / ft.
ω
2
=9
00
lb. / ft.
=3
l
l
ω wa
Footing A
Base = (8/12)'x(150 lb./ft.3)
= 100 lb./ft.2
Footing B
q = 2000psf;
qnet = q - base wt.
qnet = 2000psf - 100psf = 1900psf
Footing A
P/A = (823 lb./ft. + 300 lb./ft.)/(1.25' x 1') = 898 psf < 1900 psf (OK)
(6) and (7) Floor Beam and Post
ω = (467 + 534) x (12/16) = 750 lb./ft.
7590 lb.
(cont.)
'
7590 lb.
L
3745 lb.
0
=1
Footing B
P/A = (926 lb./ft. + 300 lb./ft.)/(1.25' x 1') = 981 psf < 1900 psf (OK)
(9) Critical Pier Footing
P = 15,090 lb.
(Int. post)
0'
1
L=
L=
'
10
Int. post
15,090 lb.
Ext. post
7545 lb.
Int. post
15,090 lb.
Base = (8/12)'x(150 lb./ft.3)
= 100 lb./ft.2
8"
x
x
qnet = 2000psf - 100psf = 1900psf
P/A = (15,090 lb.)/(x2) = 1900 lb./ft.2
x2 = 7.94 ft.2;
x = 2.82' = 2'-10" square
4.6
667 lb.
14.14'
4.7
4.7 cont’d
Jack Rafter (Typical span):
ωsnow = 25 psf x (24/12 ft.) = 50 plf
ω DL
Roof DL:
(
ω DL = 12 lb. ft
ω'DL
lf
4p
( horiz proj)
=2
€
Ridge
3
4
€
2
)
)
ω LL = 20 lb. ft.2 × ( 2') = 40 lb. ft.
= (15 12)( 24 lb. ft.) = 30 lb. ft.
ωSL = 25lb. ft 2 × ( 2') = 50 lb. ft.
(
ω = 54 plf
) × (2') = 24 lb. ft.
Snow:
(
ω DL = 7 lb. ft.2 × ( 2') = 14 lb. ft.
Ceiling Joists: (2' o.c. spacing)
(
12'
8'
324 lb.
Wall
540 lb.
Beam B-1
216 lb.
Wall
Wall
)
ω total = 14 lb. ft. + 40 lb. ft. = 54 lb. ft.
€
10'
FBD of a typical jack rafter
€
Beam B-1:
Beam B-1:
ω
= 270 lb./ft.
ω D+L =
540 lb.
= 270 lb. ft.
2 ft.
L = 8'
ωtotal = 50 plf + 30 plf = 80 plf
ω total = ωSL + ω'DL = 50 lb. ft. + 30 lb. ft. = 80 lb. ft.
1080 lb.
wall
1080 lb.
Col.
€
( horiz proj)
10'
400 lb.
400 lb.
Beam B-2:
FBD of the equivalent, horizontally
€
projected jack rafter
ω = 108 lb./ft. + 27 lb./ft. = 135 lb./ft.
This rafter represents the maximum load
condition onto the hip rafter since other rafters
Hip Rafter
ω = 284 plf
∴ ω=
7.5'
14.14'
1333 lb.
L = 12'
diminish in length.
The spacing of the jack rafter along the length of
the hip rafter is: 2' × 2 = 2.82'
2
× ( 400lb.) = 284 lb. ft.
2.82€
(
)
( 4')
( trib.
width)
= 108 lb. ft.
€
Beam B-3:
Beam B-3:
The load condition on beam B-3 is
identical to beam B-2.
ω = 135 lb./ft.
L = 8'
540 lb.
14.14'
ω = 27 lb. ft.2 ×
810 lb.
Col.
810 lb.
wall
€
667 lb.
Beam B-2:
Joist load: (span = 8’)
ω D+L = 135lb. ft.
540 lb.
€
Ceiling Joists: (2' o.c. spacing)
ω = 54 plf
8'
Interior Column:
Ceiling
joist
12'
B-2
B-1
4.7
L = 8'
540 lb.
4.8
540 lb.
4.7 cont’d
4.8
Interior Column:
Ceiling
joist
B-2
B-1
Loads to the column –
Ceiling joist:
108 lb.
Beam B-1:
1080 lb.
Beam B-2:
810 lb.
Beam B-3:
540 lb.
P = 2538 lb.
H
600#
G
a
a
12’
12’
A
9’
B-3
In addition to the vertically applied loads on
both the jack and hip rafters, truss action
develops due to the ceiling tie condition. An
examination of the truss action for each rafter
case will be performed. The three dimensional
truss solution for the hip rafter was not
covered in the text but can be done relatively
easily using readily available structural
software.
E
F
B
H
600#
C
G
HB
G
600
0
400
Ax = 300
Ay = 400
E
300
400
400
400
300
300
A
D
300
400
400
+ 600# −GAx − EC x = 0
Assuming GA x = EC x
Then GAx = EC x = 300#
€
F
300
[ΣFx = 0]
EC
C
300
0
D
E
FD
B
H
9’
F
GA
A
600#
9’
B
By = 400
C
Cx = 300
Cy = 400
D
Dy = 400
4.8
4.9
4.9
4.9
4.9
a
b
G
1200#
F
1200#E
G
a
H
a
G
1200#
E
G
FH x = 2400#;
[ΣF
y
By
B
A
Dy
D
C
Dy
Dx
D
FH
FHy = 2400#CE
D
C Dy
Dx
I
4k
CE
= 0] −HG y + 2400# −1370# = 0
J
C
FJ
D
L
K
G x = 3000#
GK
[ΣM D = 0]
1030
F
E
3000
2400
1800
2400
G D
G 2400
2400
2400
2400
2400
E
2400
2400
E
B
Bx = 600#
B
Bx = 600#
I
4k
J
[ΣFx = 0]
+ 4k + 6k + CE x = 0
GK y = 4k
L
K
By = 1370#
H
6k
G
= 0] + Dy − 2400# +1370#
=0
By = 1370#
Dy = 1030#
€
F
y
1030
F
2400
2400
1030
[ΣF
1030
= 2400#
BCE
y = y1370#
− Dx + 2400# −600# = 0
Dx = 1800#
Ax = 600#
Ay = 1370#
G 2400
[ΣFx = 0]
Bx = 600#
CE x ( 4') − 600# (16') = 0
CE x = 2400#;
Ay = 1370#
Ay = 1370#
A
B
Ax = 600#
+ 4k − GK x = 0
€
C
A
[ΣFx = 0]
GK x = 4k;
CE= 0
[ΣFx = 0] + 1200# −G x + 2400# −600#
FH
A €Ax = 600#
3000
E
F
Bx
G
Gxy = 1030#
H
3000
G
Bx
By
ΣM G = 0] FH x ( 4'G)x− 600# (16') = 0
[1200#
G
a
H
6k
Dx
Gy
H
L
K
Then; Ax = Bx = 600#
B
Gy
FH
J
a
Assumeb Ax = Bx
C
B
Ay
Gx
I
4k
€
Ax
Gy
1200#
D
By =
Ay By
A
G
a
4.10
D
1200# (16')
= 1370# (↑)
14'
[ΣFy = 0] Ay =C1370# (↓)
[ΣM A = 0]
Bx
A B Ax
Ax
Ay
1200#
E
b
a
4.10
b
F
D
C
b
H
b
F
H
A
a
BH
1030
DF
AG
Assume AG X = CE x
∴ AG X = CE x = 5k
E
F
CE
D 1800
€
1030
D 1800
I
4k
J
4k
0
4k
0
6k
H
0
6k
4k
4k
4k
G
1k
0
4k
F
5k
5k
L
K
4k
5k
E
5k
1k
5k
4.9
H
6k
G
BH
E
F
DF
AG
Prob. 4.11
CE
4.10 cont’d
4.11
I
4k
J
4k
6k
0
H
6k
G
A
5k
5k
Ax = 5k
Ay = 5k
5k
2nd
10’
E
(20psfx10’=200#/ft)
F2 = 200#/ft(40’)=8000#/ft
5k
1k
1k
B
C
By = 1k
8000#
F
5k
5k
(20psfx10’=200#/ft)
FR = 200#/ft(40’)=8000#/ft
0
4k
4k
5k
0
4k
4k
Roof
10’
0
4k
0
8000#
L
K
4k
5k
Cx = 5k
Cy = 1k
5k
D
Dy = 5k
WDL = 10psfx10’x10’ = 1000#
Ro
Dia of
ph
rag
m
ω=
(8000#)
20
VR
0#
VR
/ft
0#
00
=4
W=10psfx10’x10’=1000#
10’
10’
T
#
V2
00
[=ΣM
40 A = 0]
00
=4
€ W
#
00
20
C
T
C
v=200#/ft
− 2000# (10') + 1000# (5') + T(10') = 0
2000#
o
hra or
gm
0#
W=1000#
00
=4
V2
€
0#
00
8
2=
+
VR
20
W=1000#
5’
C
20,000 − 5,000
= 1500#
T=
2n
dF
D 10'
l
ω=
#
00
T
iap
(8000#)
VR 4000#
=
= 200 # ft
20'
20'
20
10’
10’
W
vR =
0#
2000#
W=1000#
W=1000#
0#
/ft
C
W=1000#
C
T
T
V = 4000#
4000#
T
4.10
B
C
#/f
W=1000#
t
(8000#)
10’
W
W=10psfx10’x10’=1000#
4.11 cont’d
10’
C
20
T
C
Snow load = 25 psf
2000#
2n
Dia d Flo
ph or
rag
m
V
ω=
20
0#
#
0
00
=8
v R +2 =
8000#
= 400 # ft
20'
/ft
C
W=1000#
W=1000#
Dead
Snow load = 25 psf
W=1000#
or
12
3
6’
T
C
€
4000#
[ΣM B = 0]
T=
or
4000#
T
T = 5,000#
adj)
ω = 64.4#/ft
t
#/f
37
8
ω=
L=
6’
40’
9207#
1674#
’
22
ω=
9207#
12’
C
− 20,000 #−ft − 4000# (10') + 1000# (5') + 1000# (5') + T(10') = 0
t
#/f
37
x
x
’
22
L = q = 2000psf
€
8
ω=
€
− 2000# (20') − 2000# (10') + 2000# (5') + T(10') = 0
( M)
6’
40’
M = 2,000# x10'= 20,000# −ft.
10”
1674#
= 837 # ft
2'
9207#
x
10”
x
9207#
10”
40,000 + 20,000 −10,000
= 5000#
10'
[ΣM B = 0]
€
B
C
T
W=1000#
sf
1674#
1674# per 2’
4000#
or
=7p
⎛ 12.37 ⎞ #
ω total = 50 # ft + ⎜
⎟(14 ft ) = 50 # ft + 14.4 # ft = 64.4 # ft
⎝ 12 ⎠
2000#
W=1000#
load
Trusses @ 2’ o.c.
Glu-Lam Header
6’
2000#
Dead
ω = 64.4#/ft
4000#
1674# per 2’
M = 20,000#-ft
sf
M = 20,000#-ft
C
T
W=1000#
DL = 7psfx2’ = 14#/ft.
=7p
Trusses @ 2’ o.c.
Glu-Lam Header
( slope
W=1000#
load
W=1000#
B
V = 4000#
(v = 400#/ft)
T
SL = 25psfx2’ = 50#/ft.
4000#
4000#
T
C
C
€
12
2000#
W=1000#
0#
3
W=1000#
Prob. 4.12
00
4
2=
VR
W=1000#
#
4.12
#
0#
00
Prob. 4.12
v=200#/ft
+2
20
C
00
0
=4
(8000#)
#
00
T
10’
T
V2
5’
x
x
q = 2000psf
12’
x
x
⎛ 10 ⎞
q net = 2000psf − ⎜ ft ⎟(150 # ft 3 ) = 1875 # ft 2
⎝ 12 ⎠
P
9207#
A = x2 =
=
= 4.91ft.2
q net 1875 # ft 2
x = 2.21' ≈ 2'−3" square
10”
€
4.11
5.1
Chapter 5 Problem Solutions
5.2
5.1
D
5.2
B
C
TABy
TAB
TABx
1
A
2-force
member
BA
BAx =
2
Ey
€
C
Cx
BAx
BA
Cy
P = 500#
[ΣM C = 0]
Total are of marquee = 20'×10'= 200ft.2
BA
BAy =
2
+ 500# ( 3') − BAx (2') = 0
Total load = 200ft 2 ×100 # ft 2 = 20,000#
€
500# ( 3')
= 750#
2'
BA = 750 2 #
B
BAy
[ΣM C = 0]
P = 750 2 #
€
A
P 750 2 #
=
= 750 2 # in 2 = 1061 # in 2
1 in.2
A
+ 10k (5') − 0.5TAB (10') = 0
( T ABy )
TAB = 10k
€
ft =
Since the framing is symmetrical,
each rod carries an equal amount
of the load.
TABy = TAB sin 30° = 0.5TAB
BAy =
A = 1 2 "×2"= 1 in.2
B
Cy
10k
Ex
D
Cx
30°
1
P = 500#
€
€
ft =
P
;
A
Areq' d =
To the nearest
For
P
10k
=
= 0.46 in.2
ft 22 k in 2
1
16
";
13
16
"φ
"φ rod, A = 0.5185 in.2
10k
ft =
= 19.2ksi < 22ksi
2
( allowable)
( actual) 0.5185 in.
∴ stress is OK, within the stress range
13
16
€
5.1
5.3
5.5
a) P = 20,000#; A = 64 in.2
P 20,000#
fc = =
= 312.5 # in.2
A 64 in.2
P
120k
a) fc = ; Areq' d =
= 8.9 in.2
A
13.5 k in.2
From the steel tables in the appendix;
Use : W8 × 31 (A = 9.12 in.2
b) fp =
€
b) P = 40,000#; A ≠
P
120k
; Ab =
= 266.7 in.2
A
0.45 k in.2
A = 0.302 in.2
P
4000#
∴ ft = =
= 13,250psi
A 0.302 in.2
For a square base plate :
Use : 16.3"×16.3" or
16 1 2 "×16 1 2 " plate.
c) fp =
€
P
;
A
Ab =
120k
= 40ft 2
3 k ft.2
c) P = 40,000#; A = 4"×4"−
Use : 6.32' square or 6'−4" square footing
fbrg =
5.4
e) P = 16,000#; A = 8"×L; Fv = 120psi
P 16,000#
= 133.3 in.2
Areq' d = =
Fv 120 # in.2
a) Taking a 1' (12") strip;
P
f = ; P = A × f = ( 48 in.2 ) ×150 # in 2 = 7200#
A
But : P = γ br × A × h
7200#
P
=
= 180'
h=
48
γ br A (120 # ft 3 )( 144
ft 2 )
€
4000#
= 259.7 # in.2
15.4in.2
2
π( 7 8 ")
= 15.4 in.2
4
d) P = 15,000#; A = 8"×12'= 96 in.2
15,000#
f=
= 156.3psi
96 in.2
€
b) Taking a 1' (12") strip;
P
f = ; P = A × f = 72 in.2 ×150 # in 2 = 10,800#
A
But : P = γ × A × h
10,800#
P
=
= 180'
h=
72
γ × h (120 # ft 3 ) × ( 144
ft 2 )
πD2
since rod is threaded
4
133.3 in.2 = 8"×L;
€
5.6
a) fc =
b)
[ΣF
y
L = 16.7"
5.6
P
10,000#
=
= 4,273psi

A
π × 2 2   π ×12 

 −

 4   4 
= 0] @ joint D :
DB = 10,000#
AD
ADy
CDy
2
ADx
−10,000# +ADy + CDy = 0
5
But, ADy = CDy = 5000#
CD
CDx
1
D
∴CD = 5,000 5 # = 11,180#
€
fc =
P 11,180#
=
= 11,180psi
A 1 in.2
c) fv =
€
P
11,180#
=
2A 2 π × 0.75 2
(
4)
= 12,660psi
5.2
5.7
ε=
5.11
5.11
Consider a 1’ length of wall;
δ 0.0024"
=
= 0.0012 in. in.
L
2.0"
Roof = 1’x10’x100psf=1,000#
Snow = 1’x10’x30psf = 300#
r
’T
10
ry
ta
ibu
th
wid
Dead Load + Snow = 1300#/ft.
€
Brick = 1’x(4/12)’x120#/ft2 = 480#
5.8
ε=
Total load at the base of the wall = 1780#
1’
Bearing area = 4”x12” = 48 in.2
δ
0.125"
=
= 0.0009 in. in.
L 12'×( 12"
1' )
P 1780#
fp = =
= 37.1 psi < 125 psi
A 48 in.2
∴ OK, within stress allowable
€
12’
f
5.9
9
€
P
8”
D = 4”
5.12
δ = 0.0033in./in.
a) wire weight(total) = 0.042 # ft. × 300'= 12.6#
2
πD2 π (0.125)
=
= 0.0123 in.2
4
4
P
12.6#
ft = =
= 1024.4psi
A 0.0123 in.2
A=
ε=
δ
;
L
δ = εL = (0.0033 in in ) × (8") = 0.0264 in.
b) Fallow =
FUlt. 65ksi
=
= 21.67ksi
3
3
(S.F.)
Pallow = Total load = Fallow × A
€
Pallow = (21.67 k in 2 )(0.0123in.2 ) = 266.5#
Wire wt. = 12.6#
∴ Maximum W = P −12.6# = 254#
5.10
ε=
δ
;
L
δ = εL = 0.005 in in (500'×12 in. ft.) = 30 in.
€
€
5.3
5.13
5.15
a) δ =
PL
(29,000#)(25'×12 ft ) = 0.17"
=
AE  π ×1.5 2 
6

(29 ×10 # in 2 )
 4 
L = 90’-10” = 1090”;
P = 60k
Ft = 20 ksi; Upset rods, E = 29x103 ksi
in
a) Areq' d =
PL
PL
; Areq' d =
AE
δE
29,000# )(25'×12 in ft )
(
=
= 3 in.2
(0.10)(29 ×10 6 # in 2 )
b) δ =
Areq' d
πD2
; D=
4
Use : 2"φ rod
A=
4A
=
π
P
60k
=
= 3 in.2
5.15
Ft 20 k in 2
πd 2
= 3 in.2
4
12
Use : d =
= 1.95"
π
A=
4 ( 3)
= 1.95"
3.14
b) δ =
turnbuckle
PL Ft L (20 k in 2 )(1090")
=
=
= 0.75"
29 ×10 3 k in 2
E
AE
Each turn = 1 4 " movement per rod (one thread)
∴ one turn on the turnbuckle = 1 2 " movement
€
5.14
A = 0.006 in.2
P = 16#
Number of turns =
E = 30x106 psi
a) δ =
16# (100' x12 in ft )
PL
=
= 0.1067"
AE (0.006in.2 )( 30 ×10 6 # in 2 )
b) f =
P
16#
=
= 2667psi
A 0.006in.2
0.75"
= 1.5 turns
0.5"
€
5.16
5.16
Both wall
sections move
€
δ = αL∆T = (6 ×10−6 /°F)(2 × 20'×12 in ft )(60°F)
δ = 0.173"
€
5.4
5.17
5.19
δ Al = α AlL∆T = (12.8 ×10−6 /°F)(L )(55°F) = 704 ×10−6 (L )
[ΣF
Assuming unrestrained movement;
5.19
From the equilibrium condition;
y
P = 100k
= 0] fsAs + fc Ac = 100k
δ s = δ c = 0.01"
δ conc = α concL∆T = (6.0 ×10−6 /°F)(L )(15°F) = 90 ×10−6 (L )
δ = 0.01"
fL
δ=
E
Restrained deformation in the aluminum panel :
3
δE s (0.01")(20 ×10 k in 2 )
=
= 2.42 ksi
Ls
120"
δ restrained = δ Al − δ conc = 614 ×10−6 (L )
fs =
Stress required to restrain the aluminum by 614 ×10−6 (L ) :
fc =
δ=
π(15")
− As = 176.6 in.2 − As
4
Substituting int o the equilibrium equation;
f=
δE
PL fL
= ;
f=
L
AE E
614 ×10−6 (L )(10 ×10 6 # in 2 )
L
δE c (0.01")( 3 ×10
=
Lc
120"
Ac =
= 6140 psi
3 k
in 2
) = 0.25
ksi
fs
fc
2
2.42( As ) + 0.25(176.6 − As ) = 100k
2.42( As ) + 44.15k − 0.25As = 100k
€
As = 25.8 in.2
Ac = 150.8 in.2
5.18
a) δ = αL∆T
Set δ = 0.25"
δ
0.25"
∴ ∆T =
=
= 53.4°F
−6
αL (6.5 ×10 /°F)(60'×12 in ft )
€
∆T = Tfinal − Torig. = 53.4°F
∴ Tfinal = 53.4°F + 70°F = 123.4°F (no stress condition)
b) @T = 150°F
€
δ = αL∆T = (6.5 ×10−6 /°F)( 720")(150° −123.4°) = 0.124"
(restrained deformation)
δ=
fL
;
E
f=
6
δE (0.124")(29 ×10 # in 2 )
=
= 4994 psi
720"
L
€
5.5
5.20
5.20
P = 180k
π(12.75"−2 × 0.375")
πD
Acon =
=
= 113 in.2
4
4
2
πD2O .D.
π (12.75)
Ast =
− Acon =
−113 in.2 = 14.7 in.2
4
4
[ΣF = 0] f A + f A
f (113 in. ) + f (14.7
y
c
c
2
c
s
s
= 180k
in. ) = 180k ........ (Eq. 1)
2
s
but; Ps = fsAs and Po = fo Ao
∴ fsAs + fo Ao = 50k
L = 30”
fs
fc
As = 4 in.2
δs = δo
(since
the load is symmetrically applied)
fs
fo
fs
Es
30 ×10 6
× fo =
× fo = 15fo
Eo
2 ×10 6
Substituting int o Eq. 2 :
fs =
3
15fo ( 4 in.2 ) + fo ( 32 in.2 ) = 50k
Substituting Eq. 2 int o Eq. 1;
fo = 0.543 ksi;
113fc + 14.7(9.67fc ) = 180k
113fc + 142fc = 180k
fs = 8.15 ksi
b) δ = δ s = δ o;
f = 9.67(0.71) = 6.87 ksi
fL
E
fsL 6.87 k in 2 ( 30")
δs =
=
= 0.0071"
29 ×10 3 k in 2
Es
Ao = 32 in.2
δ
PL
and ε =
and L s = L o
δ=
L
AE
∴εs = εo
f
f
f
ε=
so s = o
Es Eo
E
Es
29 ×10
× fc =
× fc = 9.67fc .......... (Eq. 2)
Ec
3 ×10 3
fc = 0.71 ksi
δ
..... (Eq. 1)
From elastic deformation;
fs
From the deformation and strain relationship;
fL
δ=
δ = δ s = δ c;
E
fcL fsL
=
;
Ec Es
fs =
P = 50k
[ΣFy = 0] Ps + Po = 50k
From equilibrium;
€
a) From equilibrium;
2
2
5.20
5.21
δs =
fsL 8.15 k in 2 (8")
=
= 0.002"
E
30 ×10 3 k in 2
δ = δs = δc =
€
€
5.6
6.2
6.1
Chapter 6 Problem Solutions
6.1
y
6.2
Y
Y
6.2
4”
y
Y
1”
I
II
6”
III
CG
X
4”
X
X
2”
X
CG
1”
6”
y = 5.67”
X
Ref.
x = 5.33”
2”
Ref.
10”
∆A
20 in.2
2”
x
x∆A
y
5”
100 in.3
9”
1”
16 in.3
4”
64 in.3
4”
3
6”
∆A24 in.2 x
24 in.2
1”
3”
24 in.
8”
Σ∆A = 60 in.2
8.5”
3
204 in.
Σx∆A = 320 in.3
1”
96 in.
2”
6”
Σy∆A = 340 in.3
6”
3”
2”
-2 in.2
9 in.2
9 in.2
x∆A
x∆A
2”
+48 in.3
2.5”
-2 in.2
2”
2
x
∆A
4”
4”
16 in.2
Y
Component
Component6”
x
x
7”
y∆A
180 in.3
y = 2.7”
y = 2.7”
Y
Ref.
2”
8”
7”
x = 2.8”
Y
x = 2.8”
Component
X
CG
2.5”
-5 in.3
5”
+45 in.3
5”
+48 in.y3
3”
-5 in.3
3”
+45 in.3
2”
y
y∆A
y∆A
3”
+72 in.3
+72 in.3
3”
-6 in.3
-6 in.3
2”
18 in.3
18 in.3
3”
= 84 in.3
Σx∆A = 88 in.3Σy∆A = 84Σy∆A
Σ∆A = 31 in.2
in.3
Σx∆A = 88 in.3
Σx∆A 320in.3
x=
=
= 5.33"
A
60in.2
y=
Σ∆A = 31 in.2
3
Σy∆A 340in.
=
= 5.67"
A
60in.2
€
€
x=
Σx∆A 88in.3
=
= 2.8"
A
31in.2
y=
Σy∆A 84in.3
=
= 2.7"
A
31in.2
6.1
6.3
6.4
6.4
6.3
X
7’
Ref.
d+x
W14x90
4’
Y
∆A
x
x∆A
y
Component
y∆A
x
16’
10’
7’
4’
y = 9.4”
Ref.
x
Y
Component
3
5’
800 ft.
3
3.5’
-98 ft.3
160 ft.2
8’
1280 ft.
-28 ft.2
10’
-280 ft.
Σ∆A = 132 ft.2
€
X
y = 5.3’
8’
x = 7.6’
C15x40
(centerted)
X
4’
10’
X
Y
Y
y
Σx∆A = 1000 ft.3
3
x
x∆A
11.8 in.2
0
0
14.02 + .78
=14.8”
174.6 in.3
26.5 in.2
0
0
14.02
2
= 7.01”
185.8 in.3
Σ∆A = 38.3 in.2
Σy∆A = 702 ft.3
x=
Σx∆A 1000ft.3
=
= 7.6'
A
132ft.2
x=0
y=
Σy∆A 702ft.3
=
= 5.3'
A
132ft.2
y=
y∆A
∆A
Σx∆A = 0
y
Σy∆A =360.4 in.3
360.4in.3
= 9.4"
38.3in.2
€
6.2
6.6
6.5
6.6
6.5
6.6
yc1
Y
Y
Y
yc1
dx1 = 0.99”-0.25 = 0.74”
3”
dx1 = 0.99”-0.25 = 0.74”
3”
X
X
xc1
xc1
dy1 = 3.25”-2” = 1.25”
X
Ref.
X
Y
dy1 = 3.25”-2” = 1.25”
X
CG
x =CG
0.99”
x = 0.99”
dx2 = -2”-0.99” = 1.01” dy2 = 2”-0.25 = 1.75”
yc2
y = 2”
Component
x=.7”
tw=.28”
∆A
x
x∆A
y
y∆A
6.09 in.2
0
0
.22 +10”+.28”-.7”
2
= 9.7”
59 in.3
2x5 =
10 in.2
4.71in.2
0
0
0
.22 + 10 = 5.11”
2
2
0
Ref.
51.1in.3
x
∆A2
Component
∆A
2.75 (in.2 )0.25
”
2.0
0
514”
/2”
Σy∆A =110.1 in.3
2.75
”
(in. )
1/2
”
x∆A3.25
3
y
8.94
(in.)
y∆A
3
0.69
3.25
8.94
0.25
Σx∆A = 4.69 in.3
2.0
bh
(in. )
0.50
Σy∆A = 9.44 in.3
0.25
4.0
2
dΣx∆A
y
y
in.3 Iyc
=Ad4.69
Ixcin.2
∆A= 4.75
Component Σ∆A
€
y∆A
3
(in. )
4.0
2.0
4”
x
(in.)0.69
0.25
2.0
Σ∆A = 4.75 in.2
1/2
y
(in.)
(in. )
”
51/2”
1/2”
110.1 in.3
y=
= 5.3"
20.8in.2
x∆A3
4”
(in.)
(in. )
Component
1/2
Σx∆A = 0
xc2
4”
tw=.22”
Σ∆A =20.8 in.2
xc2
dx2 = -2”-0.99” = 1.01”
Ref.
1/2
0
dy2 = 2”-0.25 = 1.75”
yc2
y = 2”
X
3
bh
0.50
2
dx = 9.44
Adx in.3
Σy∆A
3
12
12
Σx∆A 4.69 in.3
9.44in.3
2.75
= 6.93 y =1.25
4.3
x=
=
= 0.99";
= 2.0" = 0.057
2
2
4.75in.
Σ∆A 5 / ” 4.75 in.
I x = ΣI xc + ΣAd 2y = 7.0 + 10.4 = 17.4in.4
dy
Ady 23
∆A 3 Ixc
Component
bh
bh
I y = ΣI yc + ΣAd 2x = 2.7 in.4 +12
3.5 in.4 = 6.2in.4
12
0.74
1.5
12
1/2
€
”
1/2
”
4”
51/2”
2.0
= 0.04
ΣIxc2.75
= 7.0
bh
12
31.75
= 6.93
6.1
= 2.67
1.25ΣIyc = 2.7
4.3
ΣAdy 2 = 10.4
Iyc
1.01 3
bh
12
dx
Ad x 2
0.74
1.5
2.0
= 0.057
ΣAdx 2 = 3.5
6.3
6.7
6.7
6.6 cont’d
Comp.
1/2
∆A
Ixc (in.4)
6.93
51/2”
”
Ady2
(in.4)
”
2.75
1/2
dy (in.)
2
€
0.04
4”
y− y =
1.25
4.3
Iyc (in.4)
0.06
dy (in.)
x−x=
0.74
Adx2
y − y2 =
2.7
1.75
Σ Ixc =
x2 − x =
2.0
Σ Ady =
10.4 in.4
Σx∆A 4.69 in.3
9.44in.3
=
= 0.99"; y =
= 2.0"
2
4.75in.2
Σ∆A 4.75 in.
I x = ΣI xc + ΣAd 2y = 7.0 + 10.4 = 17.4in.4
Σ Iyc =
€2.8 in.4
4
xc1
dy2 = 1”
dy2xc=2 1”
xc2
X
dy3 = 3”
2”
Σ Adx =
2
12”
3.5 in.4
Ref.
12”
4
Ref.
Y
Y
Component
∆A
y
y∆A
3
Component
∆A
(in.2 )
12
y
(in.)
9
y∆A
(in.3 )
108
12
9
108
6
12
5
60
bh
4
12
3
bh
12
36
6
12
5
60
36
24
1
24
24
1
2
(in.2 )
6
2
€
6
2
2
2
12
2
12
Σ∆A = 48 in.2
Σ∆A = 48 in.2
Ixc
dy
Ad y 2
Ixc
dy
Ad y 2
4
24
Σy∆A = 192 in.3
300
1
12
1
12
8
3
216
8
3
216
Component
2
∆A
Iyc
dx
Ad x 2
12
35
0
0
12
4
0
0
2
12
4
0
0
2
24
288
0
24
288
0
6
12
35
0
Σy∆A 12
192 in.3
= 4" ΣIyc = 328 in.4
+
48 in.2
A
ΣAdx
4
y=
0
Σy∆A 192 in.3
+
= 4"
48 in.2
A
I x = ΣI xc + ΣAd 2y = 48 in.4 = 528 in.4 = 576
I y = ΣI yc + ΣAd 2x = 328 in.4 + 0 = 328 in.4
Σy∆A 192 in.3
= 4"
y=
+
0
48 in.2
A
0
2
€
= I0x = ΣI xc + ΣAd 2y = 48 in.4 = 528 in.4 = 576 in.4
ΣAdx 2 = 0
ΣIyc = 328 in.4
I x = ΣI xc + ΣAd = 48 in. = 528 in. = 576 in.4
2
y
ΣAdy 2 = 528 in.4
in.4
Ad x 2
12
ΣAdy 2 = 528 in.4
ΣIin.
xc 3= 48 in.4
Σy∆A = 192
ΣIxc = 48
300
5
dx
6
5
3
Iyc
6
2
(in. )
∆A
6
y=
(in.)
Component
2
2
X
dy3 =X3”
2”
I y = ΣI yc + ΣAd = 2.7 in. + 3.5 in. = 6.2in.
4
xc1
6”
x=
2
x
Y
6”
dy1 = 5”
X
Y
yc
6”
dy1 = 5”
1.01
2
€7.0 in.4
2”
1.5
yc
6”
2”
€
6.1
6.7
4
I y = ΣI yc + ΣAd 2x = 328 in.4 + 0 = 328 in.4
6.4
I y = ΣI yc + ΣAd 2x = 328 in.4 + 0 = 328 in.4
€
2
2
2
8
6.8
20
10
6.8 cont’d
Y
yc
6.67
ΣIxc = 346.7 in.4
-6
720
ΣAdy 2 = 1440 in.4
yc
Component
xc
2
10
dy = 6”
10
X
X
dy = 6”
Iyc
dx
20
166.7
0
0
40
13.3
4
640
20
166.7
0
0
2
2
2
Ad x 2
∆A
10
ΣIyc = 346.7 in.4
ΣAdx 2 = 640 in.4
xc
dx = 4”
I y = ΣI yc + ΣA2x = 347 + 640 = 987 in.4
dx = 4”
Y
Component
2
10
10
€
∆A
Ixc
dy
Ad y 2
20
6.67
+6
720
40
333.33
0
0
20
6.67
-6
720
2
2
2
10
ΣIxc = 346.7 in.4
ΣAdy 2 = 1440 in.4
I x = ΣI xc + ΣAd 2y = 346.7 + 1440 = 1787 in.4
Iyc
dx
∆A
Component
By formula
:
2
20
10
3
166.7
0
0
13.3
4
640
3
bh 3 − b1h13 (10)(14 ) − (6)(10)
Ix =
=
= 1787 in.4
12
12
10
40
2
6.5
2
2
€
Ad x 2
10
20
166.7
0
0
6.10
6.9
6.9
6.10
Y
X
X
xc
groove
tongue
dy1
dy2
CG
X
xc
dy3
y = 5.74”
X
X
X
dy1 = y1 - y
xc
dy2 = y - y2
dy3 = y - y3
Ref.
Component
Y
9”
Component
y
∆A
2
y∆A
Ixc
dy
Ady
5.25
10.05
55.13
0.98
4.76
119.0
2x16.88
= 33.76
5.63
190.1
356.0
0.11
0.4
5.25
1.75
9.19
0.98
3.99
83.6
Σ∆A = 44.3 in.2
Σy∆A = 254.4 in.3
432 in.2
3x12.56 = 37.7 in.2
Ixc
3
4
48(9)
= 2916 in.
12
4
-3x(π)(4)
64
4
= -38 in.
ΣIxc = 2878 in.4
I x = ΣI xc + ΣAd 2y
ΣAdy 2 = 203 in.4
ΣAd 2y = 0
ΣIxc = 357.9 in.4
No transfer is necessary since the solid slab and the
three holes all have their component centroids on the
major entroidal x-axis.
I x = ΣI xc = 2878 in.4
ΣyA 254.1 in.3
y=
=
= 5.74"
44.3 in.2
A
Ix = ΣI xc + ΣAd 2y = 358 + 203 = 561 in.4
48”
∆A
€
€
6.6
6.10
6.12
6.12
6.11
Y
Y
xc1
dy1
X
X
X
x = -0.36”
y = 7.0”
dy2
xc2
y = 10.4”
Component
Ref.
Component
∆A
y
y∆A
Ixc
16
14.31+1
=15.31
245
5.33
dy
Ady2
y1 - y
=4.9”
384
(in. )
(in.)
x
x∆A3
(in. )
(in.)
y
y∆A
3
3.38
0
0
10+.22-.57
= 9.63
32.6
5
22.45
∆A2
-0.634
4.49
Σ∆A = 7.87 in.2
24.1
y - y2
= 3.2”
882
172.6
7.16
247
x=
−2.85 in.3
= −0.36"
7.87 in.2
Component
2
Σ∆A = 40.1 in.
y=
3
Σy∆A = 417.6 in.
4
ΣIxc = 887.3 in.
4
ΣAdy 2 = 631 in.
∆A
55.05 in.3
= 7.0"
3.38
7.87 in.2
-2.85
Σx∆A = -2.85 in.3
2
dy
Ady
1.32
2.65
23.7
67.4
ΣyA 417.6 in,3
y=
=
= 10.4"
40.1 in.2
A
I y = ΣI yc + ΣAd 2x = 34.9 + 0.8 = 36.7 in.4
d y 2 = y − y 2 = 7"−5"= 2"
I x = ΣI xc + ΣAd 2y = 887 + 631 = 1518 in.4
d x1 = x = −0.36"
€
Σy∆A = 55.05 in.3
Ixc
I x = ΣI xc + ΣAd 2y = 68.7 + 41.7 = 110.4 in.4
d y1 = y1 − y = 9.65"−7"= 2.65"
4.49
(in. )
4
ΣIxc = 68.7 in.
2
18
4
ΣAdy 2 = 41.7 in.
d x 2 = x 2 − x = 0.634"−0.36"= 0.27"
€
€
6.7
-0.634
4.49
6.12 cont’d
5
-2.85
6.13
6.13
Σy∆A = 55.05 in.3
Σx∆A = -2.85 in.3
Σ∆A = 7.87 in.2
22.45
Y
∆A
Ixc
dy
Ady
3.38
1.32
2.65
23.7
Component
yc
2
yc
xc
dx
dy
X
4.49
2
67.4
18
dy
6.12b
4
ΣAdy 2 = 41.7 in.
4
ΣIxc = 68.7 in.
Component
2
∆A
Iyc
dx
Ad x
3.38
32.6
-0.36
0.44
4.49
2.28
ΣIyc = 34.9 in.4
y = 6”
xc
-0.27
Component
2
dx
2
A
Ixc
dy
Ad y
Iyc
Adx
5.5
4.44
6 - 0.93
= 5.07
141.4
4.44
0.93+0.25
=1.18
6.0
72
0
0
0.13
0
0
5.5
4.44
5.07
141.4
4.44
1.18
7.66
7.66
0.33
ΣAdx 2 = 0.77 in.4
−2.85 in.
x=
= −0.36"
7.87 in.2
3
y=
ΣIxc = 80.9
55.05 in.3
= 7.0"
7.87 in.2
ΣAdy 2 = 282.8
ΣIyc = 9.0
ΣAdx 2 =15.3
I x = ΣI xc + ΣAd 2y = 80.9 + 282.8 = 363.7 in.4
I x = ΣI xc + ΣAd 2y = 68.7 + 41.7 = 110.4 in.4
I y = ΣI yc + ΣAd 2x = 9.0 + 15.3 = 24.3 in.4
I y = ΣI yc + ΣAd 2x = 34.9 + 0.8 = 35.7 in.4
€
€
6.8
6.14
6.14
6.14 cont’d
Component
yc Y
dy
xc
CG
yc
dy
12
0.56
7.5+.375
=7.88
Component
yc Y
xc
CG
7.88
CG
745
12
0.56
14.7
404
0
0
14.7
404
0
0
∴ 2299 in.4 = 534 in.4 + 29.4d 2x
d 2x =
Iyc
dx
Ad x 2
12
256
0
0
12
256
0
0
d x = 7.8"
w
= d x + x − t w = 7.8"+ 0.8" − 0.7"= 7.9"
 from AISC
2



 Tables
ΣAdy2 = 1490 in.4
∆A
1765
= 60.2 in.2
29.4
∴ w = 15.8"
€
yc
dx
xc
I y = 534 + 29.4d 2x
X
xc
Y
745
But : I x = I y
ΣIxc = 809in.4
yc
I x = 809 + 1490 = 2299in.4
X
CG
dy
Ad y 2
yc
dx
xc
Ixc
X
xc
Y
∆A
2
14.7
11
dx
14.7dx
14.7
11
dx
14.7dx
X
ΣIyc = 534in.4
2
ΣAdx 2 = 29.4 dx
2
6.9
Chapter 7 Problem Solutions
7.1
7.1
10k
E
A
5’
5’
x
G
G
ΣME = 0: -10k(x) = M; M = 10x
@ x = 0, M = 0; @ x = 5’, M = +50k-ft.
E
F
5’
5’
D
G
10k
5’
+10k
F
B
B
x
‘V’
-10k
ΣMF = 0: -10k(x) + 10k(x - 5’) + M = 0
M
M = +50k-ft. (constant for x = 5’ to x = 10’)
C
G
+
0
-
‘M’
Section G: x = 10’ to x = 15’
10k
5’
50k-ft.
ΣFy = 0: +10k - 10k - V = 0; V = 0 (no shear)
V
F
x
+
0
-
Section F: x = 5’ to x = 10’
G
10k
10k
ΣFy = 0: V = +10k (constant +shear)
M
5’
5’
10k
G
C
B
5’
10k
A
A
D
V
E
10k
F
Sect. E: x = 0 to x = 5’
10k
A
10k
10k
E
C
F
A
10k
F
E
E
7.1 cont’d
10k
B
10k
7.1b
ΣFy = 0: +10k - 10k - 10k + V = 0;
V
V = 10k (constant -shear)
M
ΣMG = 0: -10k(x) + 10k(x - 5’) + 10k(x - 10’) + M = 0
M = 150k-ft. - 10x
(varies with x; between x = 10’ to x = 15’)
7.1
7.2
7.2
C
ω = 1 k/ft
MB = 200k-ft.
A
Load
B
C
ω = 1 k/ft
Section cut C:
(for x = 0 to x = 20)
MC
A
VC
10k
'V'
Linear
1st degree
-20k
x
‘M’
Parabolic
2nd degree
-200k-ft.
ΣMD =+(10k)(15’)+(20k)(5’)-RB = 0
RB = +25k
Cut sections E, F and G and write equations
of equilibrium ΣMi = 0 and
ΣFy = 0 to determine the internal shear and
moment developed at each of the respective
sections.
VE
10k
VF
A
B
F
MF
25k
5'
x
ΣMC = 0: 1k/ft(x)(x/2) - M = 0
M = x2/2 (2nd degree function)
Solve for the external reactions at B and D.
RD = +5k
ME
A
@ x = 0, V = 0;
@ x = 20’, V = 20k (negative shear)
@ x = 0; M = 0
@ x = 20’, M = 200k-ft.
RD=5k
FBD
ΣFy = -10k - 20k + 25k + RD = 0
E
x
D
5'
5'
RB=25k
V = 1k/ft(x) (1st degree function)
G
B
5'
Shear varies as a function of x (linear)
+
0
-
C
F
E
A
ΣFy = 0: -1k/ft(x) + V = 0
x
20k
10k
VB = 20k
20’
+
0
-
7.3
7.3
20k
10k
C
A
B
5'
25k
5'
x
G
VG
MG
Section cut E:
ΣFy = 0: -10k + VE = 0;
VE = 10k
The shear between A and B remains
constant.
ΣME = 0: +10k(x) - ME = 0; ME = 10x
The moment M increases as a function of x,
between A and B; x = 0 to x = 5’.
Section F:
ΣFy = -10k + 25k - VF = 0; VF = 15k
The shear remains constant (positive)
between B and C.
ΣMF = (10k)(x) - (25k)(x - 5’) + MF = 0
MF = 15x - 125
The moment varies linearly from x = 5’ to
x = 10’.
Section G:
ΣFy = -10k + 25k - 20k + VG = 0;
VG = 5k
Shear is constant between C aqnd D.
ΣMG = (10k)(x) - (25k)(x - 5’) - (20k)(x - 10’+
MG = 0;
MG = -5x + 75
Moment varies linearly for x = 10’ to x = 15’.
7.2
7.3b
7.4
7.3 cont’d
7.4
20k
10k
C
A
B
5'
5'
5'
+15k
+
0
-
A
+15k
0
-10k
V=0
V
-5k (Shear)
-5k
20k
A
+25 k-ft.
M
(Moment)
A
20k
c
b
B
ω = 2 k/ft
x
ω = 2 k/ft
C
c
b
10’
10’
10’
a
V
a
B
b
M
V
x
b
ω = 2 k/ft
c
B
x
D
20k
Secction a-a: @ x= 0 to x = 10’
ΣFy = 0: +20k - ωx - V = 0
V = 20k - 2x
(as x increases, V decreases)
ΣMa-a = 0: -20(x) +(ωx)(x/2) + M = 0
M = 20x - x2
(M increases with x; 2nd degree curve)
M
ω = 2 k/ft
20k
0
-50 k-ft.
a
a
20k
RD=5k
RB=25k
+
0
-10k
D
Load
A
ω = 2 k/ft
M
C
V
c
Section b-b: @ x = 10’ to x = 20’
ΣFy = 0: +20k - 2(k/ft)(10’) - V = 0
V=0
(No shear between x = 10’ to x = 20’)
ΣMb-b = 0: -20x + 2(k/ft)(10’)(x - 5’) - M = 0
M = -100k-ft.
(assumed direction of M on the FBD is
incorrect; it should be counter-clock)
Section c-c: @ x = 20’ to x = 30’
ΣFy = 0: +20k - 2(k/ft)(10’) - 2(k/ft)(x - 20’) + V = 0
V = 2x - 40k (V increases with x)
ΣM = 0: -20(x) + 2(10)(x-5’) + 2(x-20’)(x - 20’)/2
+M=0
M = -x2 + 40x - 300
(M is a function of x ..... 2nd degree)
7.3
7.4b
7.5
7.4 cont’d
A
20k
+20k
ω = 2 k/ft
7.5
a
c
b
B
a
10’
C
b
10’
4k
ω = 2 k/ft
c
10’
D
-20k
100k-ft.
‘M’
C
6k
“V” Diagram:
‘V’
100k-ft.
B
A
5'
@ x = 10’; V = 0 (V = 20k - 2x )
Between x = 10’ to x = 20’; V = 0
Between x = 20’ to x = 30’;
V goes from 0 to -20k
2nd degree
2k
20k
1st degree
+
0
-
+
0
-
2k
‘M’ Diagram:
@x=0M=0
(no moment at the hinge)
Between x = 0 to x = 10’, M is
increasing.
@ x = 10’, M = 100 k-ft.
(positive bending)
Between x = 20’ to x = 30’ M is
decreasing.
@ x = 30’, M = 0.
5'
Load
2k
10'
5'
+4k
V=0
+
0
-2k
'V'
-2k
5'
+
0
-
+10 k-ft.
'M'
x
Inflection point
M=0
-10k-ft.
7.4
8.4.4
7.7
7.7
7.6
15k
ω=2k/ft
ω=3k/ft
A
20’
10’
30k
10.5k
45k
B
C
9.5k
8’
6’
8’
10.5k
15k
+4k
'V'
+
0
-
x= 5’
-17kN
300k-ft
D
A
C
B
30k
4k
15kN
337.5k-ft.
x=5.25’
+
0
-
1
'M'
-5.5k
27.6k-ft
2
2
'V'
20k-ft
'M'
1
-24k-ft
1
7.5
7.8
7.8
7.9
ω=240#/ft
1° curve
5k
A
C
Load
By=360#
Ay=360#
2'
A
B
ω=600lb/ft
2'
MA=10k-ft.
D
C
B
3k
5'
3’
3’
360#
+
(+)
0
Area=2bh/3=ωL2/12
V=0
C
-
(-)
2° curve
3° curve
+
0
-
A
5'
ω=600lb/ft
MA=10k-ft.
2° curve
3k
5k
5'
B
20k-ft.
3k
C
Load
D
3k
'V'
-360#
+3k
+3k
+3k
+
0
-
M=720 #-ft
'V'
+5k-ft
'M'
+
0
-
'M'
-7.5k-ft
-10k-ft
-15k-ft
7.6
7.10
7.11
7.10
7.11
5k
A
7.8k
+7.8
6’
B
C
6’
1°
-3.2
1°
2°
600#
x = 5’
14’
1080#
+600#
+4
+1.8
+
0
-
D
4’
13.2k
1°
M = 3360#-ft
ω = 120#/ft
ω = 1k/ft
‘V’
+
0
-
‘V’
-9.2
-1080#
+1500#-ft
28.8k-ft
+
0
-
2°
+
0
-
-8k-ft
2°
‘M’
‘M’
-3360#-ft
7.7
7.12b
7.12
7.12
7.12 cont’d
2
A= 1/2(x)(2/3x) = x /3
ω=6k/ft
6k/ft
1
A
B
9'
x
18k
9k
9'
2
+9k
k
From similar triangles: ω/x = 6 /ft/9' ; ω = 2x/3
V=0
+
0
-
2
∆V = 9k = x /3 ; x = 5.2'
'V'
x=5.2'
zero slope
2
-18k
3
A=2/3(5.2')(9')=31.2k-ft
9k
Mmax=31.2k-ft
x = 5.2'
+
0
-
'M'
C
A
0
E
B
x = 5.2'
D
AEBD = AACD - AACBE =
81k-ft - 49.8k-ft = 31.2k-ft
AACD = (1/3)(9')(27k) = 81k-ft
AACB0 = (9k)(9') = 81k-ft
AAE0 = (2/3)(5.2')(9k) = 31.2k-ft
AACBE = AACB0 - AAE0 =
81k-ft - 31.2kft. = 49.8k-ft.
7.8
7.13
7.14
7.13
ω=4k/ft
7.14
P = 10k
16k
A
26k
(a)
M = 6.67k-ft.
C
B
70k
ω=4k/ft
D
FBD's of beam components
16k
C
ω = 5k/ft
16k
(b)
H
B
ω=4k/ft
A
7.5k
2
D
B
26k
C ((hinge)
4'
16'
+
0
-
16k
70k
8'
A
D
10k
G
-2.5k E
A = (1/3)(4’)(10k)=13.33k-ft.
B
C
F
2
‘V’
B
A = (1/3)(2’)(2.5k)=1.67k-ft.
2’
+32k
A = 13.33 - 1.67 = 11.67k-ft
+
0
4'
+
0
-
84.5k-ft
M=0
B
3°
-16k
-38k
+
0
-
A
8.33k-ft.
'V'
4'
C
D D
V=16k
6.5'
2’
H
H
10k
+26k
0 slope
D
A
C
1.67k-ft
10k
3°
‘M’
A
2’
A = 20 - 11.67 = 8.33k-ft
H
D
A = 20k-ft
G
A
G
32k-ft
'M'
-96k-ft
7.9
7.15
7.15
ω2=3k/ft
C
B
30k
30k
ω1=2k/ft
MA=400k-ft
30k
30k
B
A
ω1=2k/ft
MD=400k-ft
D
C
50k
RA = 50k
FBDs of beam components
ω2=3k/ft
ω1=2k/ft
MA=400k-ft
A
B
ω1=2k/ft
C
hinges
50k
20'
10'
MD=400k-ft
D
10'
50k
50k
30k
+
0
-
V=0
'V'
-30k
-50k
150k-ft
+
0
-
'M'
M=0 @ hinge
-400k-ft
-400k-ft
7.10
Chapter 8 Problem Solutions
Prob. 8.2
8.2
8.18.1
P = 2,000 lb.
P=2k
M= -18 k-ft.
1.07k
L = 9'
V = 2k
+
0
-
B
A
2.53k
4'
8'
4'
'V'
+1.6k
+1.07k
-2 k
-2 k
ω = 400lb/ft
+
0
-
'V'
-0.93k
+
0
-
111/4"
'M'
x
x
4.28k-ft
-18 k-ft.
3
W 8 x 18 (S x = 15.2 in. )
fb =
M (18k ft.) (12 in. ft.)
=
= 14.2 k in.2 < 22ksi
Sx
15.2in.3
O.K., the beam is not overstressed.
31/2"
+
0
-
'M'
4x12 S4S
Sx = 73.8 in.3
-3.16k-ft.
fb =
Mc M 4.28k ft. 12 in. ft.
=
=
= 0.696ksi
I
Sx
73.8in.3
fb = 696psi < 1,300psi
OK
The beam is safe.
8.1
Prob. 9.3
8.4
8.3
8.4
P = 6k
A
36k
4'
1000 lb.
w = 3k/ft.
ω = 500 lb./ft.
C
B
4'
8'
18k
A
B
C
4150 lb.
4350 lb.
9'
6'
+4150 lb.
+24k
12k
+
0
-12k
48k-ft
+
0
-
6k
'V'
-350 lb.
'V'
x = 8.3'
-1350
2'
-4350 lb.
-18k
17.2k-ft.
17.1k-ft
54k-ft.
'M'
-24k-ft.
W8x35; (Ix = 127 in.4, Sx = 31.2 in.3, d = 8.12”, c = 4.06”)
Mc M (54k ft.) (12in. ft.)
=
=
= 20.8ksi < 30ksi
31.2in.3
I x Sx
OK
fb =
+
0
-
+
0
-
'M'
Mmax = 17.2k-ft.
fb =
M
;
Sx
S min . =
M max (17.2k ft.) (12 in. ft.)
=
= 9.38in.3
Fb
22 k in.
Scan the W8x sections until you find one that has an Sx equal to
or greater than the Smin. value above.
Use: W8x13 (Sx = 9.91 in.3)
8.2
Prob. 8.6
8.5
8.6
ω = 2 k/ft
2400 lb.
ω = 400 lb./ft.
A
B
C
5k
5k
5'
4.4k
5.2k
6'
5'
12'
+5.2k
+5k
+2.8k
(zero slope)
0.4k
+
0
-
'V'
+
0
-
'V'
1'
-5k
-4.4k
24k-ft
M=2/3(5')(5k)=16.67k-ft.
24.2k-ft.
+
0
+
0
-
'M'
W8x18 (d = 8.14’, c = d/2 = 4.07”, Ix = 61.9 in.4, Sx = 15.2 in.3)
fb =
Mc M ( 24.2k ft.) (12 in. ft.)
=
=
= 19.1 k in.2
Ix
Sx
15.2in.3
fb = 19.1ksi < Fb = 30ksi
OK
'M'
a) W18x15 (Sx = 11.8 in.3)
f=
M (16.67k ft.)(12 in. ft.)
=
= 17ksi < Fb = 22ksi
Sx
11.8in.3
OK
b) Timber beam required:
S req' d =
M max (16.67k ft.)(12 in. ft.)
=
= 125in.3
Fb
1.6 k in.2
Use 8 x 12 S4S (Sx = 165.3 in.3)
8.3
8.8
Prob. 8.7
8.7
8.8
P = 1,600 lb.
ω = 475#/ft
ω = 400 lb/ft
24'
3.8k
10.6k
8'
8’
4’
V = 3800#
4’
4.8k
3.8k
M = 30,400 #-ft.
+
0
-
‘V’
-1900#
1.6k
'V'
-3800#
-3800#
x=9.5'
-5.8k
+
0
-
‘M’
18.1k-ft.
-3,800#-ft
'M'
-15,200#-ft
-25.6k-ft.
M@ 4' =
fb =
M ( 25.6k ft.)(12 in. ft.)
=
= 1.9ksi < 2.4ksi
Sx
162in.3
OK
fmax =
f 4'
1
2
(1900# )( 4') = 3800#
ft.
-30,400#-ft
M ( 30.4k ft )(12 in. ft.)
=
= 15ksi
24.3in.2
S
from
free end
=
M ( 3.8k ft )(12 in. ft.)
=
= 1.88ksi
24.3in.2
S
8.4
8.10
Prob. 8.9
8.9
8.10
ω = 1 k/ft
P
P
A
4'
4'
4'
4'
P
B
L = 20'
10k
3P/2
3P/2
10k
+10k
3P/2
+
0
-
'V'
10'
+P/2
+
0
-
'V'
-P/2
-10k
-3P/2
Mmax = 50k-ft.
8P
6P
6P
+
0
-
'M'
+
0
-
fb =
M max (50k ft.)(12in. ft.)
=
= 20in.3
Sx
30k in.2
Use : W14x18 (Sx = 21.1in.3 )
'M'
W18x40 (Sx = 68.4 in.3)
fb =
M
Sx
M max = Fb S x
M = (22k in.2 )(68.4in.3 ) = 1504.8k in. = 125.4k ft
M max = 8P = 125.4k ft.
P=
125.4k ft.
= 15.68k
8
8.5
8.11
Prob. 8.11
8.11
8.11b
10k
5k
10'
5'
A
5'
fb =
B
fb = 27.5kksi < Fb = 30ksi
8.75k
6.25k
Mc ( 43.75k ft.)(12 in. ft.)( 5.93")
=
= 27.5 k in.2
Ix
113.2in.4
OK.
+6.25k
+
0
-
'V'
y = 2.97"
N.A.
(
+
0
-
'M'
A
y
8
5
(in.2)
6
Σ=
1/2
14
yA
(in.3)
40
3
( 6)(1)
c = 5.93"
Ix =
3
12
= 42.7
= 0.5
fv =
dy
dy2
1.93
3.8
30.4
2.57
6.6
39.6
43.2
y=
N.A.
3
12
43
y
x
Ixc
(in.4)
(1)(8)
yA
A
I xc +
(
y = 1.93"
shear plane
)
(8.75k) 5.93in.2 2.97"
fv = VQ =
= 1.36 k in.2 < Fv = 20ksi
(@ N.A.) Ib
113.2in.4 (1")
+43.75k-ft.
+31.25k-ft.
N.A.
shear plane
-8.75k
Component
A = 8 in.2
A = 5.93 in.2
+1.25k
Ady2
(in.4)
)
(8.75k)(8in.2
) = 1.19ksi
1.93"
(113.2in. )(1")
4
70
=
43in.3
= 3.07"
14in.2
Ad 2y = 43.2in.4 + 70in.4 = 113.2in.4
y = 3.07"
Ref.
8.6
Prob. 8.12
8.12b
8.12
8.12b
ω = 0.4 k/ft )
4R
3π
N.A.
L = 32'
6.4k
Log beam
Shear:
6.4k
Fb = 1200psi; Fv = 100psi
+6.4k
Vmax =
+
0
-
'V'
L ( 400lb. ft.)( 32')
=
= 6400lb.
2
2
16'
-6.4k
fb =
R2 =
( 6400lb.)
R4
R2
2
(4R 3 )
4 ( 2R)
= 100 lb. in.2
( 4)( 6400) = 27.2
( 3)( 3.14)(100)
Bending governs the design. Use an 18” diameter log.
'M'
2
L2 ( 400lb. ft.)( 32')
=
= 51,200lb. ft.
8
8
D4
R4
I=
=
64
4
M max =
N.A.
VQ
fv =
=
Ib
R = 5.2"
Mmax = 51.2k-ft.
+
0
-
x
c = d/2 = R
x
Mc ( 51, 200lb. ft.)(12 in. ft.) R
=
= 1200 lb. in.2
Ix
R4
4
R3 =
( 51,200 12)( 4) = 653
3.14(1200)
Due to the bending requirement; R = 8.67” (say 9”)
8.7
Prob. 8.13
8.13b
8.13
8.13b
ω
Bending:
Vmax
L = 20'
ωL
2
ωL
2
L
=
;
2
M max =
1"x10" steel plate
(centered)
Mc
fb =
;
I
2
L
8
N.A.
xc
c = y= 6.35"
720 8
720 8
=
= 1.2 k ft.
2
12 L
12 20 2
Shear stress at the flange:
A = 10 in2
y= 2.15"
A
y
yA
Ixc
dy
Ady2
xc
10
8.5
85
0.833
2.15
46
9.13
4
36.4
110
2.35
50.3
121.4
111
19.13
y=
Ix =
yA
A
=
3
121.4in.
= 6.35"
19.13in.2
I xc +
)
X
Component
xc
)(
L2 720k in.
=
8
12 in. ft.
W8x31
(A36 steel)
Ref.
(
M max = M allow ;
=
xc
M allow
22 k in.2 207.3in.4
Fb I x
=
=
= 720k in.
c
6.35"
A 2y = 111in.4 + 96.3in.4 = 207.3in.4
X
Vmax =
fv =
VQ
Ib
L (1.2 k ft.)( 20')
=
= 12k
2
2
A
96.3
fv =
(12k) 10in.2
y
2.15in.
(207.3in. )(8in.)
4
= 0.156 k in.2 < Fv = 14.5ksi
Shear stress is not critical.
8.8
8.14
8.14b
8.14
8.14b
ω = 600 lb./ft
Shear stress:
X
6'
5.98"
1800 lb.
6'
1800 lb.
Component
6.73"
1800 lb.
3.37"
N.A.
(zero slope)
+
0
-
'V'
A
y
Q=yA
6.75in.2
3.37”
22.7in.3
5.25in.2
5.98”
31.4in.4
Q=54.1in.
-1800 lb.
Q = Ay
M=2/3(6')(1800 lb.)
=7200 lb.-ft.
+
0
-
3
fv =
'M'
(
)
3
VQ (1800lb.) 54.1in.
=
= 196 lb. in.2
Ib
496.9in.4 (1")
(
)
Y
Component
5.27"
dy
yA
Ixc
dy
Ady2
10.5
10.5
110.5
12
6
72
3.5( 3)
= 7.89
12
3.77
149.5
0.73
6.4
5.98
188
y = 6.75"
X
5.25
27.75
yA 186.5in.3
=
= 6.73"
A
27.75in.2
( ) = 144
2( 12) 12 3
12
dy
Ref.
y=
y
3
CG
12"
A
0.75
3.9
186.5
( ) = .99
3.5 1.53
12
152.9
344
I x = I xc + Ad 2y = 152.9 + 344 = 469.9in.4
c max = 6.73"
fb =
Mc ( 7, 200lb. ft )(12 in. ft.)( 6.73")
=
= 1170psi
Ix
469.9in.4
8.9
8.15
8.16
8.16
8.15
ω = (lb/ft)
2P
2'
P
2'
3ω
A
12"
L = 10'
3ω
6'
6'
6'
1.33P
6"
+3ω
M max = 6 + 1 2 ( 3)( 3
-3ω
3'
Ix =
Mmax = 10.5ω
c = 6”
Sx =
+
0
-
(
M allow = Fb S x = 1.6 k in.
M max = 10.5 = 19.2k ft.
bending
A = 36 in.2
y = 3"
x
x
=
= 864in.
Vallow. =
3
fv =
(@ N.A.)
Vmax =
'M'
1.5Vmax
A
Fv ( A)
1.5
(85 lb. in. )(39.4in. ) = 2, 230lb.
2
=
2
1.5
Equating the shear equations;
2230lb. = 1.67P;
P = 1340lb.
Bending governs: P = 985 lb.
2
) (144in. ) = 230k
3
in. = 19.2k ft.
19.2k ft.
= 1.83k ft.
10.5
VQ
fv =
Ix b
@ NA
10P
8P
Fv = 85 psi
(73.8in. )
10P = 9.85k ft.
P = 0.985k = 985lb.
4
+
0
-
M
Sx
Equating the two moment equations,
'V'
) = 10.5
I x 864in.4
=
= 144in.3
c
6in.
fb =
= 118k in. = 9.85k ft.
-1.67P
Fb = 1,600 psi
'M'
Mc M
fb =
=
Ix
Sx
12
4 x 12 S4S (Sx = 73.8 in.3, A = 39.4 in.2)
0.33P
A = 72 in.2
( 6)(12 3 )
1.67P
M max = Fb Sx = (1.6k in.2 )
+
0
-
Vmax = 3
'V'
a) Vmax = 1.67P
1.33P
Beam cross section
+
0
-
B
(
)(
b) @V4' = 1.33P = 1.33( 985lb.) = 1,315lb.
M = 1.33P( 4') = 5.33P = 5.33( 985lb.) = 5,250lb.
@ 4'
fv =
1.5V 1.5(1315lb.)
=
= 50.2psi
A
39.4in.2
fb =
M ( 5250lb. ft.)(12 in. ft.)
=
= 854psi
Sx
73.8in.3
)
4
k
Fv Ib .085 in.2 864in. ( 6")
=
= 4.08k
Q
36in.2 3"
(
)
Vmax = 3 = 4.08k
6"
(shear )
=
4.08k
= 1.36 k ft.
3ft.
Shear governs.
8.10
8.18
8.17
8.17
8.18
ω
L = 20'
ωL/2
4k
L2
8
M max =
ω = 2 k/ft
M max = Fb Sx = (22k in.2 )
ωL/2
M max =
Vmax = ωL/2
(79in. ) = 1735k
3
M max = 32k ft. = 384k in.
in.
1735k in.
= 145k ft.
12in. ft.
9.3k
4'
12'
S req' d =
26.7k
Equating the moment equations,
+
0
-
L2
= 145k ft.
8
145(8)
=
= 2.9 k ft.
20 2
'V'
10'
( )
-ωL/2
2
+
0
-
4k
+
0
-
'V'
x=4.7'
c = 5.5"
Mmax = ωL /8
1/2" x 10" steel
cover plate (top & bottom)
From the Appendix Tables;
12k
9.3k
(2) C10x20
Use: W12x19 (Sx = 21.3 in. 3)
does not account for the beam’s weight.
or W10x10 (S x = 18.8 in. 3) Same weight
as the W12x19 but shallower in depth.
f =
v
(ave)
-14.7k
N.A.
'M'
M ( 384k in.)
=
= 17.5in.3
Fb
22 k in.2
Vmax
14.7k
=
t w d (0.235")(12.16")
f = 5.14ksi < 14.5ksi
v
(ave)
OK
21.8k-ft.
Component
A
dy
Ady
0.1
5
5.25
138
157.8
11.8
0
0
0.1
5
5.25
138
158
I xc +
+
0
-
'M'
-32k-ft.
276
Ad 2y = 158 + 276 = 434in.4
S x = I x c = 434in. 5.5in. = 79in.3
L ( 2.9 k ft.)( 20')
V=
=
= 29k
2
2
b = 2 2.74"= 5.48"
Y
4
c = 5.5”
(
y = 5.25"
Ix =
2
Ixc
N.A.
)
Q = Ay = 5in.2 ( 5.25") = 26.3in.3
( 29k)( 26.3in.3 )
VQ
=
= 0.32 k
fv =
Ib ( 434in.4 )( 5.48in.)
in.2
8.11
8.20
8.19
8.19
8.20
ω = 80x
3’
4’
810#
2430#
ω = 720#/ft
Fb = 1200 psi
Fv = 100psi
2’
+1280
+
0
-
‘V’
-360
1.37’
-1150
+290
+
0
-
‘M’
-360
-980
2”x4”
rough cut
20”
L = 16’
N.A.
x
1/ ”
2
12”
h
+450
5k
p (pitch)
+2.5K
x
Plank Cross Sectio9n
3
bh 3 (12)h
Ix =
=
= h3
12
12
A = 12h
Bending :
Mc
;
fb =
Ix
1200 =
(980
h = 2.2"
Shear :
Ix =
( )
12) h 2 5800
= 2 = 4.9in.2
h3
h
(5")(20")
-2.5K
V = 2500#
+
0
-
‘M’
VQ
F = fv bp = (bp)
Ib
VQ
F = (p)
I
F = capacity of 2 nails at the flange
representing 2shear surfaces
(2 80# = 160# )
p = pitch or spacing
3
( 4")(16")
3
= 1965in.4
12
12
Q = Ay = (8in.2 )(9") = 72in.3
20k-ft.
1.5(1280)
1.5V
; 100 =
12h
A
1.5(1280)
h=
= 1.6"
1200
Bending controls : h = 2.2"
fv =
‘V’
+
0
-
2 shear
planes
y = 9”
N.A.
p=
x
4
FI (160# )(1965in. )
=
= 1.75"
VQ (2500# )( 72in.3 )
Use : 1 3 4 " spacing
Alternate method :
fv =
3
VQ ( 2500# )( 72in. )
=
= 22.8psi
Ib
(1965in.4 )(4")
P
F
A=
A
fv
A = 2" 2p = 4p
160#
F
;
4p = =
fv 22.8 # in 2
fv =
p = 1.75"
8.12
8.22
8.21
8.21
8.22
P = 2k
M = 48k-ft.
ωDL = 300 lb/ft
w = 1 k/ft
ωLL = 400 lb/ft
A
E = 29 x 103 ksi
F b = 22 ksi
L = 8'
Vmax = 10 k;
V = 10k
S req' d =
+10k
F v = 14.5 ksi
B
5600 lb.
M max = 48 k-ft.
M ( 48k ft.)(12 in. ft.)
=
= 26.2in.3
Fb
22 k in.2
Try: W8x31 (Sx = 27.5 in.3,
+2k
+
0
-
'V'
d = 8”, tw = 0.285”, Ix = 110 in.4)
M add. =
bm
(31lb. ft.)(8')
=
L2
2
M add. = 992 lb. ft.
+
0
-
'M'
Sadd. =
2
2
M add. (992lb. ft.)(12in. ft.)
=
Fb
22k in.2
Sadd. = 0.54in.3
Stotal = Sreq' d + Sadd. = 26.2 + 0.54
Stotal = 26.74in.3 < 27.5in.3
-48k-ft.
OK
V
10k
=
fv =
t w d (0.285")(8")
(ave)
fv = 4.38k in.2 < Fv = 14.5k in.2
(ave)
Deflection:
total =
L4 PL3
+
8EI 3EI
L ( 700lb.)(16ft.)
=
= 5, 600lb.
2
2
A req' d =
1.5V 1.5( 5,600lb.)
=
= 98.8in.2
Fv
85lb. in.2
= 25.5 lb. ft.*)
2
=
( 25.5 lb. ft.)(16') = 204lb.
2
1.5Vadd. 1.5( 204lb.)
=
= 3.6in.2
A add. =
Fv
85 lb. in.2
A total = A req' d + A add. = 98.8in.2 + 3.6in.2 = 102.4in.2 > 101.3in.3
The beam is overstressed in shear.
Try: 8 x 16 S4S
= 29.3 lb. ft. )
(A = 116.25 in.2, Sx = 300.3 in.3, Ix = 2327 in.4,
= L 360 =
(16ft.)(12 in. ft.) = 0.53"
360
=
(
)
3
3
5 L4 5( 400 lb. ft.)(16ft.) 1728 in. ft.
=
= 0.16"< 0.53"
384EI
384 1.6 10 6 lb. in.2 2327in.4
(
)(
)
Use: 8 x 16 S4S
(1728 in. ft. ) + (2k)(8ft.) (1728 in. ft. )
k in. )(110in. )
3( 29 10 k in. )(110in. )
4
3
total
=
total
= 0.286"+0.185"= 0.47"
(
bm L
Vadd. =
OK for deflection.
(1.031k ft.)(8ft.)
(8) 29 10 3
Vmax =
= 0.252 A (for Douglas Fir and Southern Pine)
*Note:
actual
(LL)
+
M ( 22,400lb. ft )(12 in. ft.)
=
= 207in.3
Fb
1300 lb. in.2
(A = 101.25 in.2, Sx = 227.8 in.3, Ix = 1538 in.4,
4
P
w
S req' d =
Try: 8 x 14 S4S
allow.
OK
L2 ( 700 lb. ft.)(16')
=
= 22,400lb.ft.
8
8
5600 lb.
L = 16'
2
M max =
2
3
3
4
3
3
2
3
4
8.13
8.24
8.23
8.23
8.24
2k
4" concrete slab
1k
1k
4'
4'
A
ω = 99 lb/ft2 x 8' = 792 lb/ft
4'
A
4'
L = 16'
2k
B
2k
8712 lb.
Vmax = 2000lb.
A req' d =
1.5V 1.5( 2000lb.)
=
= 27.3in.2
Fv
110 lb. in.2
M max = 12, 000lb. ft.
+2k
S req' d =
+1k
+
0
-
'V'
-1k
-2k
8k-ft.
8k-ft.
+
0
-
M (12, 000lb. ft.)(12 in. ft.)
=
= 92.9in.3
Fb
1550 lb. in.2
'M'
(A = 46.4 in.2, S = 102.4 in.3,
Ix = 678.5 in.4,
= 12 lb./ft.)
=
(16')(12 in. ft.) = 0.8"
L
=
240
240
P
P
actual
P
=
OK
PL3
PL3
5 L4
+
+
20.1EI 48EI 384EI
P
ωbeam
+
+
3
2
Stotal = Sreq' d + Sadd. = 92.9in.3 + 3in.3
Stotal = 95.9in.3 < 102.4in.3
allow
3
(1k)(16') (1728) + (1k)(16') (1728) + 5(.012)(16') 4 (1728)
act. =
20.1(1.6 10 3 )( 678.5) 48(1.6 10 3 )( 678.5) 384 (1.6 10 3 )( 678.5)
act
= 0.324"+0.136"+0.016"= 0.48"< 0.80"
Loads:
Metal
decking
8712 lb.
L = 22'
Beam B-1
Section A-A
(
) (150 lb. ft. ) = 50 lb. ft.
Conc. = 4 12 ft.
3
2
Metal deck = 4 lb. ft.2
Plaster ceiling = 5lb. ft.2
Try: 4 x 14 S4S
L2 (12 lb. ft.)(16ft.)
M add. = bm =
= 384lb. ft.
8
8
M
( 384lb. ft.)(12 in. ft.) = 3in.3
S add. = add. =
Fb
1550 lb. in.2
12k-ft.
B
Dead Load = 59 psf
DL
Live Load = 40 psf
LL
(
)
= ( 40 lb. ft. )(8') = 320 lb. ft.
= 59 lb. ft.2 (8') = 472 lb. ft.
2
DL + LL = 99 psf
= 99 lb. ft.2 8' = 792 lb. ft.
M max =
Sreq' d =
2
L2 ( 792 lb. ft.)( 22')
=
= 48,000lb. ft.
8
8
M ( 48k ft.)(12in. ft.)
=
= 26.2in.3
22k in.2
Fb
Try: W14 x 22 (Sx = 29 in.3, A = 6.49 in.2, Ix = 199 in.4)
Deflection check:
allow
(LL)
act.
=
=
L
( 22')(12 in. ft.) = 0.73"
=
360
360
4
5 LL L4 5( 320 lb. ft.)( 22') (1728)
=
= 0.29"< 0.73"
384EI
384 29 10 3 (199)
(
)
OK
OK
Check the bearing stress.
fp =
P
=
A brg
2000lb.+ 96lb.
( 5.5"
(bm.wt )
3.5")
= 109psi < Fc = 410psi
OK
Use: 4x14 S4S
8.14
8.24b
8.24b
8.95k (Beam B1 react.
incl. beam wt.)
8.95k
ω = 0.3 k/ft
8'
8'
8'
SB1:
Beam
12.55k
12.55k
M max = 95.6k ft.
12.55k
Sreq' d =
10.41k
1.2k
+
0
-
'V'
-10.41k
-12.55k
fv
( ave)
4
allow =
(D+L)
act.
=
act.
= 0.193"+0.584"= 0.78"< 1.20"
=
(
=
L
( 24')(12 in. ft.) = 1.2"
=
240
240
5 L4
PL3
+
384EI 28.2EI
3
(8.95)( 24') (1728)
3
10 )( 448) 28.2( 29 10 3 )( 448)
act.
384 29
= 52.1in.3
3
2
4
(S x = 56.5in. , A = 10.6in. , I x = 448in. )
'M'
5(.336)( 24') (1728)
22
12)
Try: W16 x 36
Mmax = 95.6 k-ft.
+
0
-
(95.6
+
OK
V
(12.55k) = 2.7ksi < F = 14.5ksi
=
v
t w d ( 0.295")(15.86")
∴ OK in shear
Use : W16x36 for SB1
€
8.15
9.3
Chapter_9_Problem Solutions
9.3
9.1
9.1
Pcr = 250k
Imin. = 103 in.4
W10x54;
Pcr
2
EI
L2
Pcr =
W8x31
I y min . = 37.1in.4
L=?
2
L = 20'
A = 9.13 in.
(
)(
)
2
3
2
4
EI 3.14 29 10 k in. 37.1in.
=
= 184.2k
2
L2
( 20' 12 in. ft.)
P
184.2k
fcr = cr =
= 20.2ksi
A 9.13in.2
Pcr =
2
L2 =
2
EI
Pcr
=
(
)(
3.14 2 29 10 3 k in.2 103in.4
250k
) = 117,800in.
2
L = 343” = 28.6’
Pcr
Pcr
9.49.4
9.2
9.2
Pcr = 25k
Y
2-31/2”
X
φ standard pipe
8" Dia.
I y = 2 I = 2( 4.79in.4 ) = 9.58in.4
( min )
L = 24'= 288"
Pcr =
EI
=
L2
2
(3.14 )
2
(29 10
(288)
3
2
)( 9.58 .in
)
2
A=
πD2 π (8")
=
= 50.2"
4
4
I=
D4
=
64
(8")
64
4
= 201in.4
Pcr = 25 k
K = 0.7
4
8” diameter pole
Pcr =
= 33k
( KL)
2
EI
( KL)
2
=
2
;
( KL)
(
2
=
)(
2
EI
Pcr
3.14 2 1 10 3 k in.2 201in.4
Pcr
KL = 282” ;
25k
L=
) = 7.93
10 4 in.2
282"
= 403"= 33.6'
0.7
9.1
9.6
9.6
9.5
Pcr
8" 6"
3
Pcrit.
8 " rectangular tube.
L = 38'
W8x28 (W200x42)
(A = 9.58in. ; ry = 2.36"; I y = 53.5in. )
(A = 8.25in.2 ; I y = 21.7in.4 ; rx = 3.45"; ry = 1.62")
KL (1)( 38' 12 in. ft.)
=
= 193.2
2.36"
ry
Weak Axis:
2
Pcr =
2
EI y
( KL)
2
=
4
(
Le = 16’
)(
3.14 2 29 10 3 k in.2 53.5in.4
( 456")
2
) = 73.64k
Bracing
Leff.= 16'
L e 16' 12 in. ft.
=
= 118.5
ry
1.62"
Strong Axis:
P
73.64k
fcr = cr =
= 7.7ksi
A 9.58in.2
L = Le = 26’
L e 26' 12 in. ft.
=
= 90.4
rx
3.45"
Pcr
Weak axis buckling
Pcrit.
The weak axis governs.
Pcr =
fcr =
2
EI y
L2e
=
(
)(
3.14 2 29 10 3 k in.2 21.7in.4
(16'
12 in. ft.)
2
) = 168k
Pcr
168k
=
= 20.4ksi
A 8.25in.2
Leff.= 26'
9.5
Strong axis buckling
9.2
9.7
9.7
9.8
Pcr
9.8
W12x65 (A = 19.1in.2 ; ry = 3.02")
Pa
Y
Case a):
yc
KL ( 0.65)(18' 12 in. ft.)
=
= 46.5
3.02in.
ry
Fa = 18.66 ksi
Pa = Fa
(a)
(
A = 18.66 k in.2
yc
2-C12x20.7 (A36 steel )
) (19.1in. ) = 356 k
2
K = 1.0
dx = bf –x =2.94” – 0.70” = 2.24”
L = 20'
L = 18'
(K = 0.65)
X
Pcr
dx dx
Pcr
A
Ixc
Iyc
dx
Adx2
6.09
129
3.88
2.24
30.6
6.09
129
3.88
2.24
30.6
12.18 in.2
258 in.4
7.76 in.4
Component
Case b):
KL (0.8)(18' 12in. ft.)
=
= 57.2
3.02"
ry
L = 18'
(K = 0.80)
Fa =17.69 ksi
(
Pa = 17.69 k in.2
(b)
) (19.1in. ) = 338 k
2
Pcr
Pcr
Case c):
KL (1)(18' 12 in. ft.)
=
= 71.5
3.02"
ry
L = 18'
(K = 1.0)
Fa = 16.28 ksi
(
Pa = 16.28 k in.2
(c)
) (19.1in. ) = 311k
2
Ix =
I xc = 258in.4
Iy =
I yc +
Ad 2x = 7.8 + 61.2 = 69in.4
KL (1)( 20' 12 in. ft.)
=
= 100.8
ry
2.38"
Pa = Fa
(
A = 12.88k in.2
ry =
61.2 in.4
Iy
A
=
69in.4
= 2.38"
12.2in.2
Fa =12.88 ksi
) (12.2in. ) = 157 k
2
Pcr
9.3
9.9
9.11
9.9
9.11
Compression member
L = 7’
Weak Axis:
L5 3 1 2
1
2"
From the Table 10.1; Fa = 17.99ksi
(A = 4in.2 , rz = 0.755")
Pa = 17.99 k in.2 31.2in.2 = 561k > 500k
Since trusses are assumed to be pin connected,
KL (1)( 7' 12 in. ft.)
=
= 111
0.755"
rz
(
A = 11.54 k in.2
Ptotal = 500k
) (4in. ) = 46k
at the top of the column. This is a bit conservative.
KL (1)( 26' 12 in. ft.)
=
= 57
5.47"
rx
From the Table 10.1;
9.10
9.10
Pa = 17.71k in.
P = 60k
5”
Std. wt.pipe
A = 4.3in.2; r = 1.88"; P = 60k
Fa =
L=?
KL = 0.8L
Strong Axis:
Assumes that the 2nd floor beam loads are applied
2
KL=26'
Pa = Fa
OK
Weak axis
it is reasonable to assume K = 1.0.
Fa = 11.54ksi
W12x106; A = 31.2in.2; rx = 5.47"; ry = 3.11"
KL = 14’
Bracing
(2nd flr beams)
KL 14' 12 in. ft.
=
= 54
ry
3.11"
KL = 14'
12'
Ptotal = 500k
2
Fa = 17.71ksi
31.2in.2 = 553k > 500k
OK
Strong axis
P
60k
=
= 13.95ksi
A 4.3in.2
KL
= 92.2 (from Table 10.1)
r
1.88"
r
92.2 =
92.2 = 216.7"= 18'
L=
0.80
K
9.4
9.12
9.14
9.12
9.14
P = 30k
L = 20’
KL = 20’
P = 30k
P = 350k
L = 12’; KL = 12’
Enter Table C-36;
KL (1)(12' 12 in. ft.)
=
= 58
ry
2.48"
12'
Bracing Strong Axis:
(2nd flr L = 24’; KL = 24’
beams) KL (1)( 24' 12 in. ft.)
KL = 12'
KL (1)(20' 12 in ft )
=
= 127.7
1.88"
r
Try a W14x74 (A = 21.8 in.2; rx = 6.04”; ry = 2.48”)
Weak Axis:
Try a 5”φ pipe: A = 4.3 in.2 ; r = 1.88”
KL = 20'
Fy = 36 ksi
P = 350k
rx
=
6.04"
Fallow = 9.17 ksi
The weak axis governs the design.
Pallow = Fallow x A = 9.17 x 4.3 in.2 = 39.4k
Enter the slenderness ratio table with
Weak axis
P = 350k
A = 17.62 k in.2
Pa = Fa
9.13
W14x68: (A = 20.0 in.2; rx = 6.01”; ry = 2.46”)
Weak Axis:
KL = 12’;
Try: W8 x 18 (A = 5.26 in.2, ry = 1.23”)
KL 20' 12 in ft
=
= 195
1.23"
r
KL = 20'
Fy = 36 ksi
Enter Table C-36;
Fallow = 3.93 ksi
Pallow = Fallow x A = 3.93 ksi x 5.26 in.2 = 20.7k
Pallow < P = 30k; Inadequate design
Try W8 x 24 (A = 7.08in.2; ry = 1.61”)
KL (20x12)"
=
= 149
1.61"
r
Fallow = 6.73 ksi
OK
350k
100% =
100% = 91%
384k
Try for a more efficient column section.
KL=24'
L = 20 ; KL = 20 ; P = 30k
21.8in.2 = 384k
Pa = 384k > Pact. = 350k
Pact.
Efficiency:
Pallow
9.13
KL
= 58.
ry
Fa = 17.62ksi
Pallow > Pactual
P = 30k
= 47.7
KL (1)(12' 12 in. ft.)
=
= 58.5
ry
2.46"
Strong Axis:
Strong axis
KL = 24’;
KL (1)( 24' 12 in. ft.)
=
= 47.9
rx
6.01"
The weak axis governs.
KL
= 58.5; Fa = 17.57 ksi
ry
Pa = Fa
A = 17.57 k in.2 20.0in.2 = 351.4k
Pa = 351.4k > Pact. = 350k
OK
P
350k
100% = 99.6%
Efficiency: act. 100% =
351.4k
Pa
Use: W14x68
Pallow = 6.73ksi x7.08in.2 = 47.6k
Pallow >Pactual = 30k ;
Use: W8x24
9.5
9.15
9.16
9.16
9.15
P = 397.5k
P=?
Roof Load: DL = 80 psf
6 x 6 S4S Southern Pine
A = 30.25 in.2; E = 1,600 ksi; Fc = 975 psi
LL = 40 psf
le (14' 12 in. ft.)
=
= 30.5
d
5.5"
DL + LL = 120psf x 500 ft.2 = 60k
KL = 16'
LL = 125 psf
DL + LL = 225psf x 500 ft.2 = 112.5k
The third floor column supports the 4th, 5th, and 6th floors plus the
roof. The column load is therefore: P = 3 x 112.5k + 60k = 397.5k
2
Try: W12x79 (KL = 16’; A = 23.2 in. ; ry = 3.05”)
3rd Floor Column
KL (16' 12 in. ft.)
=
= 63;
3.05"
ry
Pa = Fa
P = 622.5k
(
A = 17.14 k in.
2
Fa = 17.14 ksi
) (23.2in. ) = 397.6k
2
Pa = 397.6k > Pact = 397.5k
OK
The ground floor supports an additional two floors of load.
KL = 14'
Floor Load: DL = 100 psf
6x6 S4S
So. Pine
(Dense No. 1)
Fc E =
0.3E
le
d
2
=
(
0.3 1.6 10 6 lb. in.2
( 30.5)
(
2
) = 516psi
)
Fc* = FcC D = 975lb. in.2 (1.25) = 1219psi
FcE
516psi
=
= 0.423
Fc* 1219psi
Enter Appendix Table 14.
Cp = 0.377
Fc' = Fc* C p = 1219psi 0.377 = 460psi
(
Pa = Fc A = 460 lb. in.2
) (30.25in. ) = 13,900lb.
2
P = 5 x 112.5k + 60k = 622.5k
KL = 20'
KL = 20’
Use a W12x section for a better transition to the W12x79 above.
Try: W12x136 (A = 39.9 in.2; ry = 3.16”)
KL ( 20' 12 in. ft.)
=
= 76 ;
ry
3.16"
Pa = Fa
1st Floor Column
A = 15.79 k in.2
Fa = 15.79 ksi
39.9in.2 = 630k
Pa = 630k > Pact. = 622.5k
OK
9.6
9.17
9.18
9.18
9.17
P = 32k
8 x 8 S4S Douglas Fir
2
P
6
(A = 56.25 in. ; E = 1.6 x 10 psi; Fc = 1000 psi)
le 13.5' 12 in. ft.
=
= 21.6
d
7.5in.
0.30E
le
d
2
=
(
0.30 1.6 10 6 lb. in.2
( 21.6)
2
) = 1029psi
Fc* = FcC D = (1000psi)(1.0) = 1000psi
FcE 1029psi
=
= 1.03
Fc* 1000psi
8x8 S4S
Douglas Fir
(No. 1)
le 22' 12 in. ft.
=
= 25.15
d
10.5"
The strong axis governs the design.
Bracing
@ mid-height
6
2
0.418E 0.418 1.8 10 lb. in.
FcE =
=
= 1190psi
2
2
le
( 25.15)
d
Strong Axis:
(
)
FcE 1190psi
=
= 0.72
Fc* 1650psi
Cp = 0.701
Fc' = Fc*C p = (1000psi)( 0.701) = 701psi
Pa = 39.4k > 32k;
le 11' 12 in. ft.
=
= 19.55
d
6.75"
Fc* = FcC D = (1650psi)(1.00) = 1650psi
From Appendix Table 14;
Pa = Fc' A = ( 701psi)
(A = 70.88 in.2; E = 1.8 x 106 psi; Fc = 1650 psi)
Weak Axis:
KL = 11'
KL = 13.5'
FcE =
Glu-Lam Column: 6 3 4 " 10 1 2 "
(56.25in. ) = 39, 400lb.
2
OK
From Appendix Table 14; Cp = 0.619
Fc' = Fc*C p = (1650psi)( 0.619) = 1021psi
Pa = Fc' A = (1021psi)
(70.88in. ) = 72, 370lb. = 72.4k
2
Using Appendix Table 12, interpolate between 13’ and 14’;
Pa =39.4k
9.7
9.19
9.20
4x8 S4S Douglas − Fir
Hem − fir : E = 1400ksi; Fc = 1050psi; A = 5.25in.
2
(A = 25.38in. , E = 1600ksi, F
2
L e ( 7.5')(12 in. ft.)
=
= 25.7
d
3.5"
6
0.3E 0.3(1.4x10 psi)
=
= 636psi
FcE =
2
 L e 2
25.7)
(
 
d
Slenderness ratio =
ω = 2778# × 1216 = 2084 # ft.
Bearing stress :
Fc⊥ = 405psi
Slenderness ratio = L e
Weak axis governs.
(per
stud, every 16")
FcE =
PAllow < PAllow
( compression )
d
=
76.8"
= 21.9
3.5"
d
=
0.3E 0.3(1.6 ×10
=
2
 L e 2
(21.9)
 
d
6
120"
= 16.6
7.25"
) = 1001psi
Fc* = FcC D = (1000psi)(1.0) = 1000psi
FcE 1001psi
=
= 1.00
Fc* 1000psi
Enter Table 9 − Ed
C p = 0.691
Pallow = Fc⊥ × A = ( 405psi)(5.25in.2 ) = 2126#
( bearing)
Le
Strong Axis : L e = KL = 1.0(10')(12 in. ft.) = 120"
FcE 636psi
=
= 0.606
Fc* 1050psi
C p = 0.504
Pallow = Fc' × A = (529.2)(5.25) = 2778#
= 1000psi)
Weak Axis : L e = KL = 0.8(8')(12 in. ft.) = 76.8"
Fc* = FcC D = (1050psi)(1.0) = 1050psi
Fc' = Fc*C p = (1050psi)(0.504 ) = 529.2psi
c
∴ Bearing stress governs
Fc' = Fc* × C p = (1000psi)(0.691) = 691psi
€
Pallow = Fc' × A = 691psi × 25.38in.2 = 17,540#
Pallow
17,540#
=
= 351 ft.2
Atrib. =
50 # ft 2
DL + LL
€
9.8
9.21
9.22
Southern Pine : E = 1600ksi, Fc = 975psi
Try : 6x6 S4S
Le
d
FcE =
=
Glu − Lam : 6 3 4 "× ______; E = 1.8x10 6 psi; Fc = 1650psi
(A = 30.25in. )
Le
2
(16')(12 in. ft.) = 34.9
5.5"
FcE =
0.3(1.6x10 psi)
0.3E
= 394psi
2 =
2
L e 
34.9
(
)


 d
6
C p = 0.363
Areq. =
Insufficient capacity in the 6x6
(16')(12 in. ft.) = 25.6
€
0.3(1.6x10 6 )
0.3E
=
= 732psi
2
L e 2
(25.6)


 d
FcE 732psi
=
= 0.75
C p = 0.585
Fc* 975psi
FcE =
P 15,000#
=
= 21.6in.2
Fc'
693psi
Use : 6 3 4 "x7 1 2 " Glu − Lam; A = 50.63in.2
Try : 8x8 S4S ( A = 56.25in.2 )
7.5"
6
0.418E 0.418(1.8x10 psi)
=
= 735psi
2
L e 2
32
(
)


 d
Fc' = Fc* × C p = (1650psi)(0.420) = 693psi
Pallow = Fc' × A = ( 354psi)( 30.25in.2 ) = 10.7k < 25k
d =
6.75"
FcE 735psi
=
= 0.45
Fc* 1650psi
C p = 0.420
Fc' = Fc* × C p = (975psi)(0.363) = 354psi
Le
(18')(12 in. ft.) = 32
Fc* = FcC D = (1650psi)(1.0) = 1650psi
Fc* = FcC D = (975psi)(1.0) = 975psi
FcE 394psi
=
= 0.404
Fc* 975psi
d
=
Fc' = Fc* × C p = (975psi)(0.585) = 570psi
Pallow = Fc' × A = (0.570ksi)(56.25in.2 ) = 32k > 25k
∴OK
€
9.9
Chapter 10 Problem Solutions
10.1
10.1
10.2
Shear : (double) Pv = 10.4 k bolt 2 bolts = 20.8k
(Table 10.1)
Bearing : Thickness = 3 4 "
Ap = 5 8 "
3
4
"= 0.469in.2
Fp = 1.2Fu = 1.2(58ksi) = 69.6ksi
(0.469in. ) (69.6ksi) = 65.2k
Pp = 2 bolts
2
Net Tension : d = 5 8 "+ 116 "= 1116 "= 0.688"
Anet = 3 4 " ( 4" 0.688") = 2.48in.2
Ft = 0.5Fu = 0.5(58ksi) = 29ksi
Pt = 29 k in 2 2.48in.2 = 71.9k
3
4
Pplate = 22 k in.2
P = 28k;
A325 X bolts in double shear
a) Based on shear : 14 k bolt requirement
2
5
8
"
A325 X bolts (Pv = 18.4 k bolt ) (Table 10.1)
Capacity of 2
Pp = 2
(5 8"
7
8
5
8
"
"= 3.0in.2 ; Ft = 0.6Fy = 22ksi
3.0in.2 = 66k
bolts in bearing :
")(69.6 k in.2 ) = 32.6k;
OK
b) Based on net tension : d = 5 8 "+ 116 "= 0.688"
Anet = t (W d) = 3 8 " (W 0.688")
Pt net = Ft
Anet = 29 k in 2
W = 3.26"
Agross = t W = 3 8 " W;
Ft gross = 0.6Fy = 22ksi
Plate tension :
Agross = 4"
10.2
Pgross = Ft
W = 3.4"
Agross = 22 k in 2
3
8
" (W 0.688") = 28k
(3 8"
W ) = 28k
this condition governs
Shear governs the design; Pallow = 20.8k
10.1
10.3
10.4
10.3
10.4
Group A bolts : 3 3 4 " A325 X in double shear
Shear : Pv = 3 26.5 k bolt = 79.5k
( Table 10.1)
Bearing : Pp = 3
( Table 10.2)
(26.1 k bolt ) = 78.3k
( t= 1 2 ")
Net tension : Ft = 0.5Fu = 0.5(58ksi) = 29ksi
Anet = (2 plates
Pt = Ft
5
16
") 8" 2
13
16"
( d + 116")
Anet = 29 k in 2
= 3.98in.2
3.98in.2 = 115.5k
Group B bolts : 2 7 8 " A325 X in double shear
Shear : Pv = 2 36.1 k bolt = 72.2k
( Table 10.1)
Bearing : Pp = 2 30.5 k bolt = 61k
( t= 1 2")
( Table 10.2)
Net Tension : Ft = 29ksi; d +
Pt = 29 k in 2
( 1 2 ") (3"
15
16
Tension capacity of the
Pt = Ft
( 0.6Fy )
A = 22 k in 2
Net tension in the
Pmax = 29.9k
2
= 1516 "
") = 29.9k
1
( 12 "
1
1
16"
2
governs
" 3" bar :
3") = 33k
" 3" bar is critical;
Each member will be checked for shear and bearing only.
Member a : P = 105k 63k = 42k
3 " A325
X(NSL) A36 steel
4
42k
= 1.6 ~ 2 bolts
Shear : n =
( double)
26.5 k bolt
42k
Bearing : n =
= 2.14 ~ 3 bolts
19.6 k bolt
( t= 3 8")
Member b : P = 26k
26k
= 1.95 ~ 2 bolts
Shear : n =
( sin gle)
13.3 k bolt
26k
Bearing : n =
= 2.65 ~ 3 bolts
3
9.8 k bolt
( t= 16")
Member c : P = 26k
26k
= 1.95 ~ 2 bolts
Shear : n =
( sin gle)
13.3 k bolt
26k
Bearing : n =
= 1.98 ~ 2 bolts
1
13.1 k bolt
( t= 4 ")
Member d : P = 42k
42k
= 3.16 ~ 4 bolts
Shear : n =
( sin gle)
13.3 k bolt
42k
Bearing : n =
= 3.21 ~ 4 bolts
13.1 k bolt
( t= 1 4 ")
governs
governs
governs
governs
10.2
10.7
10.5
10.5
10.7
4 3 4 " A325 SC(STD) w / A36 steel
governs
Shear : Pv = 15 k bolt 4bolts = 60k
( double)
( Table 10.1)
g : Pp = 26.1 k bolt
Bearin
1
(t
2"
)
4 bolts = 104.4k
( Table 10.2)
Net tension : Anet = (8" 2
Ft = 0.5Fu = 29ksi
Pt = Ft
Anet = (29 k in.2 )
13
16
")( 1 2 ") = 3.19in.2
Beam reaction = 210k
5 A490X bolts
U sing Table 10.3;
5 7 8 " A490X bolts carry 242k in shear.
10.7 angle thickness is 5 "
Clip
8
Angle length = 14 1 2 " (it fits within the beam flanges)
(3.19in. ) = 89.9k
2
Beam reaction = 210k
5 A490X bolts
U sing Table 10.3;
10.8
10.6
5
" A490X bolts carry 242k in shear.
Clip angle thickness is 5 8 "
10.8
10.1
Angle length = 14 1 2 " (it fits within the beam flanges)
10.6
1 1 3 8 " A490X
Shear : Pv = 119k
( double)
( t=1")
1
Pp = 69.6
k
2
Lmax
n = 6 bolts
Pv = 90.1k
1.375in. = 95.7k
2
Pt = Ft
1
2
0.93”
")(5 1 2 " 1 716 ") = 4.06in.2
Anet = (29 k in.2 )
(4.06in. ) = 117.8k
Bearing governs the design.
21.62”
6
3
4
"
A325
0.93”
SC
A325 SC @3" o.c.
a) Maximum clearance :
3
Net tension :
Hole diameter = 1 3 8 "+ 116 "= 1 716"
Anet = (2
"
From Table 10.3;
1”
10.8
using L = 17 1 2 "
")(1 3 8 ") = 1.375in.2
in. 2
4
1”
L max = 21.62"
2(0.93") 2(1") = 17.76"
Ap
Fp = 1.2Fu = 1.2(58ksi) = 69.6ksi
Ap = (2
8
A325 SC
@3"
W21
x 93 o.c.
a) Maximum clearance :
3
( Table 10.1)
Bearing : Pp = Fp
7
2
Pmax = 95.7k
4
"
L max = 21.62" 2(0.93") 2(1") = 17.76"
From Table 10.3;
using L = 17 1 2 "
n = 6 bolts
Pv = 90.1k
6
3
4
"
A325 SC
10.3
10.11
10.9
10.9
10.11
Plate capacity : PPL = Ft
A = (22 k in.2 )
(6"
3
) = 49.5k
8"
U sing the maximum weld size : t
= L 3.71 k in. = 3.71L
Pallow
Plate capacity : PPL = FT
L total = 12"
49.5k
= 4.13 k in.
12"
Use : 516 " weld (s = 4.64 k in.) Table 10.4
"= 516 "
1
16
"=
1
4
"
( Table 10.4 )
( based on weld )
Minimum weld size = 316 "
Maximum weld size = 516 "
1
16
Equating : 3.71L = 20.6k
L = 5.6"
A = 22 k in.2
( 516 "
3") = 20.6k
10.12
10.12
A = 3.59 in.2
10.10
10.10
PPL = 22 k in.2
4”
(5
5
16
") = 34.88k
Minimum weld size = 316 "
Maximum weld size =
1
4
"
(s = 2.78 k in.)
(s = 3.71 k in.)
L min = 5" (distance between longitudinal welds)
1/ ”
4
Fillet weld : 1 4 " weld = 3.71 k in.
Total weld length = 4 4"= 16"
Weld capacity = 3.71 k in. 16"= 59.4k
Fillet Weld
Total minimum weld length = 15"
34.88k
= 2.29 k in.
Required strength : s =
15"
Use : 316" weld with L = 5"
Full penetration groove weld :
Capacity is equal to the tensile capacity
of the square tube.
Pt = Ft
A = 22 k in.2 3.59in.2 = 79k
Groove Weld
10.4
10.13
10.13
Y
R1
Re
0.84”
P
O
a = 3.34”
x
b = y = 1.66”
R2
[
Fx = 0] R1 + R 2 + R e = P
[
M O = 0] R1 ( 3.34") + R e (0.84") R 2 (1.66") = 0
P = Ft
A = 22 k in.2
Minimum weld :
Maximum weld :
4.0in.2 = 88k
3
16
"
"
Try 16 " weld : s = 2.78 k in.
R e = s L e = 2.78 k in. 5"= 13.9k
7
16
3
Returning to the moment equilibrium equation;
R1 ( 3.34 ) + (13.9k )(0.84") = R 2 (1.66")
R2 =
R1 ( 3.34 ) + 11.68
= 2.01R1 + 7.04
1.66
Substituting int o the
Fx = 0 equation;
R1 + (2.01R1 + 7.04 ) + 13.9 = 88k
R1 = 22.28k
R 2 = 51.82k
R e = 13.9k
R1 = s L1
R2 = s L2
22.28k
= 8.01"
2.78 k in.
51.82k
L2 =
= 18.64"
2.78 k in.
L1 =
10.5