Insstructor’s Manual to accompany
Statics and Strength of Materials For
Architecture and Building Construction
Fourth Edition
Barry S. Onouye
Upper Saddle River, New Jersey Columbus, Ohio
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10 9 8 7 6 5 4 3 2 1
ISBN-13: 978-0-13-511455-1 ISBN-10: 0-13-511455-1
Instructor’s Manual to Accompany
For Architecture and Building Construction
Fourth Edition
Barry Onouye
Seattle Public Library Architect: Rem Koolhaus
Pearson/Prentice HallUpper Saddle River, New Jersey Columbus, Ohio
Statics and Strength of Materials
Preface
This Instructor’s Manual is intended to accompany Statics and Strength of Materials for Architecture and Building Construction. It was initially developed as a study guide for students to practice on a va-riety of problems to enhance their understanding of the principles covered in the text. Solutions were developed in sufficient detail to allow students to use these problems as additional example problems.
Although the problem solutions contained in this Instructor’s Manual have been worked, re-worked, checked and scrutinized by my many stu-dents over the years, there are inevitably errors that remain to be discov-ered by others using the book. If you detect discrepancies, omissions and errors as you work through these problems, I would appreciate hearing from you so that I can incorporate the changes for any future editions of the Instructor’s Manual or book.
I realize that many instructors do not allow student’s access to the Instruc-tor’s Manual but I have personally found that my students appreciated having it as a study guide.
Fall, 2010
Barry Onouye, Senior LecturerDept. of ArchitectureCollege of Built EnvironmentsUniversity of Washingtone-mail: [email protected]
Table of Contents
Chapter 2 Statics • Graphical addition of vectors pg 2.1 - 2.2 • Resolution of forces: x and y components pg 2.2 - 2.3 • Vector addition by components pg 2.3 - 2.6 • Moment of a force pg 2.6 - 2.7 • Varignon’s theorem pg 2.7 – 2.8 • Moment couples pg 2.9 • Equilibrium of concurrent forces pg 2.10 – 2.13 • Equilibrium of rigid bodies pg. 2.13 – 2.16 • Supplementary problems pg 2.16 – 2.26
Chapter 3 Analysis of Determinate Systems • Cables with concentrated loads pg 3.1 – 3.3 • Equilibrium of rigid bodies with distributed loads pg 3.4 – 3.5 • Planar trusses – method of joints pg 3.6 – 3.8 • Truss analysis – method of sections pg 3.8 – 3.10 • Diagonal tension counters pg 3.10 – 3.12 • Zero-force members pg 3.12 • Pinned frames – multi-force members pg 3.13 – 3.15 • Supplementary problems pg 3.16 – 3.28 • Retaining walls pg 3.29 – 3.32
Chapter 4 Load Tracing • Gravity load trace pg 4.1 – 4.8 • Lateral load trace pg 4.8 – 4.11
Chapter 5 Strength of Materials • Tension, Compression and shear stress pg 5.1 – 5.2 • Deformation and strain pg 5.3 • Elasticity, strength and deformation pg 5.3 – 5.4 • Thermal stress and deformation pg. 5.4 – 5.5 • Statically indeterminate, axially loaded members pg 5.5 – 5.6
Chapter 6 Cross-Sectional Properties • Centroids pg 6.1 – 6.3 • Moment of inertia pg 6.3 – 6.7 • Moment of inertia for composite sections pg 6.7 – 6.9
Chapter 7 Bending and Shear Diagrams • Equilibrium method for shear and moment diagrams pg 7.1 – 7.4 • Semi-graphical method for shear and moment diagrams pg 7.5 – 7.10
Chapter 8 Bending and Shear Stress in Beams • Bending stress pg 8.1 – 8.5 • Bending and shear stresses pg 8.6 – 8.12 • Deflection in Beams pg 8.13 – 8.15
Chapter 9 Column Analysis and Design • Euler buckling loads and stresses pg 9.1 – 9.2 • Axially loaded steel columns - analysis pg 9.3 – 9.4 • Design of steel columns pg 9.5 – 9.6 • Axially load wood columns pg 9.6 – 9.9
Chapter 10 Structural Connections • Bolted steel connections pg 10.1 – 10.3 • Framed connections pg 10.3 • Welded connections pg 10.4 – 10.5
2.1
Chapter 2 Problem Solutions
2.1
40˚
20˚
50˚ x
y
R = 17
3#Fa = 100#
Fb = 200#
40˚
20˚50˚
x
y
R = 17
3#Fa = 100#
Fb = 200#
or
Prob. 2.2
R = 173#θ = 50˚ from horiz.φ = 40˚ from vert.
R = 10.2kNθ = 5˚
35˚10˚
30˚F1 = 3kN
F2 =6kN
F3 =5kN
x
y
2.2
40˚
20˚
50˚ x
y
R = 17
3#Fa = 100#
Fb = 200#
40˚
20˚50˚
x
y
R = 17
3#
Fa = 100#
Fb = 200#
or
Prob. 2.2
R = 173#θ = 50˚ from horiz.φ = 40˚ from vert.
R = 10.2kNθ = 5˚
35˚10˚
30˚F1 = 3kN
F2 =6kN
F3 =5kN
x
y
2.3
F1 = 600#
F2 = 72
0#
R =
930
#
40˚
40˚
x
yProb. 2.3
Prob. 2.43
4
3
4
5
12
T1 = 50
00#
T2 = 5910#
R =
10,
000#
T3 = 910#
2.4
F1 = 600#
F2 = 72
0#
R =
930
#
40˚
40˚
x
yProb. 2.3
Prob. 2.43
4
3
4
5
12
T1 = 50
00#
T2 = 5910#
R =
10,
000#
T3 = 910#
2.2
2.5
T1
T2 = 240 lb.
20°
10°
W =
1,200 lb.
x
y
Prob. 2.5
2.6
€
sinθ =35
and cosθ =45
∴ Fx = F cosθ = 1000#( ) 45( ) = 800#
Fy = F sinθ = 1000#( ) 35( ) = 600#
€
∴ Fx = 45 F = 4
5 1000#( ) = 800#
Fy = 35 F = 3
5 1000#( ) = 600#€
Fx4
=Fy
3=
F5
By similar triangles:
Prob. 2.6
3
4
F=1000 lb.
AB
x
Fy
Fx
Prob. 2.7
3
4
F=1000 lb.
θ
5
5
y
x
T
10°
Tx
Ty
x
y
T
10°
Tx
Ty
2.3
2.7
Prob. 2.6
3
4
F=1000 lb.
AB
x
Fy
Fx
Prob. 2.7
3
4
F=1000 lb.
θ
5
5
y
x
T
10°
Tx
Ty
x
y
T
10°
Tx
Ty
€
Tx = T sin10°
Ty = T cos10°
∴ T =Ty
cos10°=
250N0.985
= 254N
2.8
Problem Solutions: 2.3.11-2.3.15
3
4
F=1000 lb.
AB
x
Prob. 2.3.11
Fy
Fx
y
x
T
10°
Prob. 2.3.12
Tx
Ty
412
θ
P=300 lb.
Rafter
Purlin Detail
Prob. 2.3.13
θPy
Px
9
12
16' 16'
P = 600 lb.
Q = 600 lb.
AB
Prob. 2.3.14
PyPx
Qy
Qx
3
4
34
x
3
4
y
12
5
F=2500N
Inclinedrafter
Problem 2.3.15
Fy
Fx
α
β
φ
α
90∞
€
θ = tan−1 412( ) = 18.43°
€
Px = P 412.65( ) = 300#( ) 0.316( ) = 94.9#
Py = P 1212.65( ) = 300#( ) 0.949( ) = 285#
2.9
Graphical solution using the tip-to-tail method
30
11
y F =10k1
F =12k2
F =18k3
xO
x
y
R = 7.03 k
Rx = +5.73 k
Ry
= -4
.07
k
Resultant
θ = 35.1°
1
1
30 F1 =
10k
F2 = 12k
F3 = 18k
R = 7.03k
x
y
θ = 35.1°
30
11
y F =10k1
F =12k2
F =18k3
xO
x
y
R = 7.03 k
Rx = +5.73 k
Ry
= -4
.07
k
Resultant
θ = 35.1°
1
1
30 F1 =
10k
F2 = 12k
F3 = 18k
R = 7.03k
x
y
θ = 35.1°
30
11
y F =10k1
F =12k2
F =18k3
xO
x
y
R = 7.03 k
Rx = +5.73 k
Ry
= -4
.07
k
Resultant
θ = 35.1°
1
1
30 F1 =
10k
F2 = 12k
F3 = 18k
R = 7.03k
x
y
θ = 35.1°
€
F1y = +F1 cos30° =10k 0.866( ) = 8.66k
F1x = +F1 sin30° =10k 0.50( ) = 5kF2 = −F2x = −12k
F3x = +12
F3( ) = +18k
2
F3y = −12
F3( ) = −18k
2
€
Rx = ΣFx = +5k −12k +18k
2= +5.73k
Ry = ΣFy = +8.66k −18k
2= −4.07k
€
tanθ =Ry
Rx
=4.075.73
= 0.710
θ = tan−1 0.710( ) = 35.4° from horizontal
€
sinθ =Ry
R
R =Ry
sinθ=
Ry
sin35.4°
∴ R =4.07k0.579( )
= 7.03k
2.4
2.10
2.10
x
y
TABx
TABy
TAB=600N
TACx
TAC=800N
TACy
60˚ 40˚
xy
TAC=
800N
TAB=600N
60˚ 40˚
R = 1079N
φ = 3.2˚
Scale 1mm = 10N
φ
x
y
θ
Rx = 59.6N
Ry = 1078N R = 1079N
φ = 3.2˚
θ = 86.8˚ €
−TACx= −TAC cos 60° = −0.5TAC
−TACy = −TAC sin 60° = −0.866TAC
+TABx= +TAB cos 40° = +0.766TAB
−TABy= −TAB sin 40° = −0.642TAB
€
Rx = ΣFx = − 0.5( ) 800N( ) + 0.766( ) 600N( ) = 59.6N
Ry = ΣFy = − 0.866( ) 800N( ) − 0.642( ) 600N( ) = −1078N
€
θ = tan−1 Ry
Rx
⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 1078
59.6⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 18.1( ) = 86.8°
φ = tan−1 RxRy
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = tan−1 59.6
1078⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 0.055( ) = 3.2°
R = 59.62 +10782 = 1079N
2.10
x
y
TABx
TABy
TAB=600N
TACx
TAC=800N
TACy
60˚ 40˚
xy
TAC=
800N
TAB=600N
60˚ 40˚
R = 1079N
φ = 3.2˚
Scale 1mm = 10N
φ
x
y
θ
Rx = 59.6N
Ry = 1078N R = 1079N
φ = 3.2˚
θ = 86.8˚
2.10
x
y
TABx
TABy
TAB=600N
TACx
TAC=800N
TACy
60˚ 40˚
xy
TAC=
800N
TAB=600N
60˚ 40˚
R = 1079N
φ = 3.2˚
Scale 1mm = 10N
φ
x
y
θ
Rx = 59.6N
Ry = 1078N R = 1079N
φ = 3.2˚
θ = 86.8˚
2.10 cont’d
Graphical Solution:
2.5
2.11y x
30˚
40˚F
W=200#
axis
of b
oom
Fy
Fx
Wx
Wy
2.11
axis
of b
oom
30˚
W=200#R =
293#
F = 156#
Tip-to-tail
Scale: 1” = 100#
y x
30˚
40˚F
W=200#
axis
of b
oom
Fy
Fx
Wx
Wy
2.11
axis
of b
oom
30˚
W=200#R =
293#
F = 156#
Tip-to-tail
Scale: 1” = 100#
€
−Wx = −W cos 30° = −0.866W−Wy = −Wsin 30° = −0.50W−Fx = −F cos 40° = −0.766F+Fy = +F sin 40° = +0.642F
€
Ry = ΣFy = 0; − 0.50 200#( ) + 0.642F = 0
∴ F =100#0.642
= 156#
R = Rx = ΣFx = −0.866 200#( ) − 0.766 156#( )R = −173# −120# = −293#
2.12
2.12
25°
y
R = resultant reaction
F1
F2
A
F1 = 7kN
F2 = 6.34kN
R =2.95kN
25˚
2.12
25°
y
R = resultant reaction
F1
F2
A
F1 = 7kN
F2 = 6.34kN
R =2.95kN
25˚
-F2x = -F2cos 25˚-F2y = -F2sin 25˚
Since the resultant must be vertical,Then: Rx = ΣFx = 0
-F2x + F1 = 0
F2cos25˚ = F1From this equation, it is seen that F1Is only a fraction of F2 , therefore, F2 = 7kN.
Then, F1 = F2cos25˚ = (7kN)(0.706)
F1 = 6.34kN and F2 = 7kN
R = F2y = (7kN)(sin25˚) = 2.95kN.
Graphical solution
2.6
2.13
€
−T1x = −T1 cos 30° = −0.866T1
−T1y = −T1 sin 30° = −0.50T1
−T2x = −T2 cos 60° = −0.50T2
−T2y = −T2sin 60° = −0.866T2
€
+Fx = +F cos 45° = +0.707 8k( ) = +5.65k
−Fy = −F sin 45° = −.707 8k( ) = −5.65k
But T1 = T2
For the resultant to be vertical,
€
Rx = ΣFx = 0∴ − 0.866T − 0.50T + 5.65k = 0T = 4.14kR = Ry = ΣFy = −0.50 4.14k( ) − 0.866 4.14k − 5.65k( ) = −11.3k
2.14
W = 25#
F = 20#8’
5’
A
4’
2.14
2.15
A
800N
W=700N
x
2m 1m
2.15
€
ΣM A = 0( )+800N 1m( ) − 700N x( ) = 0
x =800N( ) 1m( )
700N( )= 1.14m
Since x > 1m, the man is OK.
W = 25#
F = 20#8’
5’
A
4’
2.14
2.15
A
800N
W=700N
x
2m 1m
2.13
F = 8k
T1
T2
y
x30°
60°45°
A
A B C
AB = BC so that x-components cancel
T
T
TR
8k
R = 11.3k
xy
Using the parallelogram law
2.13
F = 8k
T1
T2
y
x30°
60°45°
A
A B C
AB = BC so that x-components cancel
T
T
TR
8k
R = 11.3k
xy
Using the parallelogram law
MA = -20#(5’) + 25#(4’) = -100 #-ft + 100 #-ft = 0
The box is just on the verge of tipping over.
2.7
2.16
€
ΣM A = −36# 15"( ) +15# 8"( ) =
− 540# −in( ) + 120# −in( ) = −420# −in.
€
ΣM A = −W 18.8"( ) = −100# 18.8"( ) = −1880# −in. clockwise( )ΣM A = 0[ ] −100# 18.8"( ) + P 45.1"( ) = 0
P =1880# −in.
45.1 in.= 41.7#
€
ΣM A = 0[ ] − 200# 12"( ) = F 26"( ) = 0; ∴ F = 92.3#
ΣM B = 0[ ] + F 4"( ) − P 36"( ) = 0; ∴ P = 10.3#
2.17
2.18
5
12
8”
17”
15”
5
12
Fy = 15#
Fx = 36#
A
F = 39#
2.16
W = 100# P18.8” 26.3”
6”20”
28”
20˚
A
12” 14”
4”
36”
W = 200# P
F
FA
B
2.19
24m
5kN
10kN
9kN
8kN
8kN
8kN
4m
4m
4m
4m
4m
4m
A
2.19
€
MA = − 5kN( ) 24m( ) − 10kN( ) 20m( ) − 9kN( ) 16m( ) − 8kN( ) 12m( )− 8kN( ) 8m( ) − 8kN( ) 4m( )MA = − 120kN − m( ) − 200kN − m( ) − 144kN − m( )− 96kN − m( ) − 64kN − m( ) − 32kN − m( )MA = −656kN − m
2.8
€
Fx = 1213 1300#( ) =1200#
Fy = 513 1300#( ) = 500#
€
MB = −Fx 5'( ) + Fy 0( ) = −1200# 5'( ) = −6000#−ft
MC = −Fx 5'( ) + Fy 12'( ) = −1200# 5'( ) + 500# 12'( )MC = − 6000#−ft( ) + 6000#−ft( ) = 0
2.202.20
5’
12’ 12’
F = 1300# Fy
Fx
512
AB
C
D
2.21
2.21
6”
6”
4”
25˚
25˚
A
BF = 30#Fy
Fx
€
Fx = Fcos25° = 30#( ) 0.906( ) = 27.2#
Fy = Fsin25° = 30#( ) 0.423( ) =12.7#
€
MA = Fy 6"( ) = 27.2#( ) 4"( ) =108.8#− in CCW
€
MB = −Fy 6"( ) − Fx 2"( ) = − 12.7#( ) 6"( ) − 27.2#( ) 2"( ) = −130.6#− in.
2.22
2.22
A
Fx
Fy F = 1.5kN
60˚
30˚
dx
dy
120mm
200m
m
€
Fx = Fcos30° = 1.5kN( ) 0.866( ) =1.3kN
Fy = Fsin30° = 1.5kN( ) 0.50( ) = 0.75kN
dx = dcos60° = 200mm( ) 0.50( ) =100mm
dy = dsin60° = 200mm( ) 0.866( ) =173mm
€
MA = −Fx dy( ) + Fy 120mm + dx( )MA = −1.3kN 173mm( ) + 0.75kN 220mm( )MA == −60kN − mm = −0.06kN − m
2.23
2.23
60˚
dx1
dx2
dyT = 2000#
Tx
Ty
W
A
B
3’
7’
30˚
€
Tx = T cos 30° = 2000# .866( ) = 1732#
Ty = T sin 30° = 2000# 0.50( ) = 1000#
dx1 = 7' cos 60° = 7' 0.50( ) = 3.5'
dx2 = 10' 0.50( ) = 5'
dy = 7' sin 60° = 7' .866( ) = 6.06'
€
ΣM C = 0[ ] Tx dy( ) − Ty dx1( ) − W dx2( ) = 0
1732#( ) 6.06'( ) − 1000#( ) 3.5'( ) − W 5'( ) = 0
5W = 10,500#−ft − 3500#−ft
W =7000#−ft
5'= 1400#
2.9
2.24
€
MA = − 10k( ) 11.3'( ) − 4k( ) 14'( ) = −113k−ft − 56k−ft = −169k−ft
MB = − 10k( ) 11.3'( ) − 4k( ) 14'( ) = −169k−ft
2.252.25
RAx = 25#
RAy = 150#
RBx = 25#
B
A
2’ 2’
6’
6’
150#
€
RAx and RBx form a coupleMcouple 1 = 25# 12'( ) = +300#−ft
€
RAy and 150# man form a couple
Mcouple2 = − 150#( ) 2'( ) = −300#−ft
Since the moment due to a couple is constant,
MA = MB = MC = +300#-ft - 300#-ft = 0
2.26
2.26
CL
90kN
125mm 180mm
CL
90kN
M = 90kN(0.305m) = 27.45kN-m
2.27
2.27
30°
55°
C
4"
3" 12"
A B
Fy = 69.6#
Fx = 48.7#
F = 85#
€
Fx = 85 lb.( ) cos 55° = 48.8 lb.
€
Fy = 85 lb.( ) sin 55° = 69.6 lb.
€
M A = −Fy 15"( ) + Fx 4"( ) = − 69.6 lb.( ) 15"( ) + 48.8 lb.( ) 4"( ) = −849 lb.− in.
€
M B = −Fy 12"( ) + Fx 4"( ) = − 69.6 lb.( ) 12"( ) + 48.8 lb.( ) 4"( ) = −640 lb.− in.
2.24
A
B
7.07 k
7.07 k4 k
7.07 k
7.07 k
6’ 8’ 8’
8’8’ 10k
10k11.3’
2.10
2.28
2.28
B
Ax
AyRA RC
Cx
Cy
1000#
60º 45º
y
x
1000#
RC RA
Cx
Cy Ay
Ax
B
2.29
F = 500N
ACBC
20˚
30˚10˚
x
y
Free-body diagram of joint C
C
2.29
€
ΣFx = 0[ ] − Cx + Ax = 0−0.707C + 0.50A = 0
C =0.50A0.707
= 0.707A
€
ΣFy = 0[ ] + Cy + Ay −1000#= 0
0.707 0.707A( ) + 0.866A =1000#∴ 1.37A =1000#A = 732#C = 0.707 732#( ) = 518#
€
ΣFx = 0[ ] − 470N + 0.174AC + 0.50BC = 0 ... Eq 1( )ΣFy = 0[ ] −171N + 0.984AC − 0.866BC = 0 ... Eq 2( )
Solving equations (1) and (2) simultaneously,
€
0.866 × 0.50BC + 0.174AC = 470N[ ] ... Eq 1( )0.50 × −0.866BC − 0.984AC =171N[ ] ... Eq 2( )
Therefore,
€
+0.433BC + 0.15AC = 407N ... Eq 1( )−0.433BC + 0.492AC = 86N ... Eq 2( )
Adding the two equations;
AC = +768N (compression)
Substitute and solve for BC;
BC = 672N (tension)
€
Ax = Acos60° = 0.50AAy = Asin60° = 0.866ACx = Ccos45° = 0.707CCy = Csin45° = 0.707C
Force Magnitude Fx__________________ Fy____________________
F 500N -500cos20º = -470N -500sin20º = -171N
AC ? +ACsin10º = +0.174AC +ACcos10º = +0.984AC
BC ? +BCsin30º = +0.50BC -BCcos30º = -0.866BC
2.11
2.30
2.30
F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ α x
y
P F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ x
y
α =61.1˚
P = 925#
3
4
F1 = 50#
F2 = 150#W = 60#
P = 925#
α =61.1˚25˚
Graphical check
2.30
F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ α x
y
P F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ x
y
α =61.1˚
P = 925#
3
4
F1 = 50#
F2 = 150#W = 60#
P = 925#
α =61.1˚25˚
Graphical check
2.30
F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ α x
y
P F1 = 50#
F2 = 150#
W = 60#
3
4
25˚ x
y
α =61.1˚
P = 925#
3
4
F1 = 50#
F2 = 150#W = 60#
P = 925#
α =61.1˚25˚
Graphical check
Force Magnitude Fx__________________ Fy__________________
F1 50# -50#cos25° = −45.3# +50#sin25° = +21.1#
F2 150# +150#(3/5) = +90# +150#(4/5) = +120#
W 60# 0 -60#
P ? +Pcosα +Psinα
€
ΣFx = 0[ ] − 45.3# +90# +P cosα = 0 ..... 1( )ΣFy = 0[ ] + 21.1# +120# −60# +P sinα = 0 .... 2( )
€
P =−44.7#cosα
.... 1( )
P =−81.1#sinα
.... 2( )
€
Equating : −44.7#cosα
=−81.1#sinα
sinαcosα
= tanα =−81.1#−44.7#
= +1.81
α = tan−1 1.81( ) = 61.1°
€
P =−44.7#
cos 61.1°=
−44.7#0.483
= −92.5#
Note that the negative sign for Pindicates that it was initially assumedin the wrong direction.
Final Free Body Diagram
2.12
2.31
2.31
A
B
W = 2.5kN
75˚
30˚
FBD of the sphere
15˚ 60˚
A = 2.24kN
B = 0.67kN
Forces exerted bt the sphere onto the smooth surface.
90˚
90˚
2.31
A
B
W = 2.5kN
75˚
30˚
FBD of the sphere
15˚ 60˚
A = 2.24kN
B = 0.67kN
Forces exerted bt the sphere onto the smooth surface.
90˚
90˚
€
Tx = Tsin5° = 0.087TTy = Tcos5° = 0.996TPx = Pcos20° = 0.940PPy = Psin20° = 0.342P
€
ΣFx = 0[ ] − 0.087T + 0.94P = 0
T =0.94P0.087
=10.8P
ΣFy = 0[ ] + 0.996T − 0.342P − 2000#= 0
substituting;0.996 10.8P( ) − 0.342P = 2000#P =192#AB = T =10.8 192#( ) = 2074#
2.32
€
ΣFx = 0[ ] Acos75° − Bsin60° = 0
A =B 0.866( )
0.259= 3.346B
€
ΣFy = 0[ ] Asin75° + Bcos60° − 2.5kN = 0
3.346B 0.966( ) + 0.50B = 2.5kN3.732B = 2.5kNB = 0.67kNA = 2.24kN
2.32
y
x
W = 2000#
P
T5˚
20˚
2.33
34
60°
y
x
CA
CD
BC
34
DC
y
W = 200 lb.
DE
15°FBD(b)
FBD(a)
C
D
2.13
2.33
Solving FBD(a) first:Force _________Fx_________ __________Fy__________
DC
€
−45
DC = −0.80DC
€
+35
DC = +0.60DC
DE
€
+DE cos15° = +0.966DE
€
+DE sin15° = +0.259DE
W 0 -200 lb. ___________________ _____________________
€
Fx∑ = − 0.80DC + 0.966DE = 0 ; DC = 1.21DE
€
Fy∑ = + 0.60(1.21DE)DCx( )
+ 0.259DE − 200lb. = 0
€
0.985DE = 200lb. ; DE = +203lb.; DC = 1.21 203lb.( ) = 246lb.
Writing equations of equilibrium for FBD(b);
Force Fx Fy
CD
€
+45
246lb.( ) = +197lb.
€
−35
246lb.( ) = −148lb.
CA
€
−CA cos 60° = −0.50CA
€
−CAsin 60° = −0.866CA
BC 0 +BC
€
Fx∑ = − 0.50CA +197lb. = 0 ; ∴ CA = 394lb. T( )
€
Fy∑ = − 0.866 394lb.( ) + BC −148lb. = 0 ; BC = +489lb. C( )
2.34
€
ΣMA = 0[ ] +100# 4'( ) − 300# 3'( ) + Bx 10'( ) = 0
∴ Bx = +50#
ΣFx = 0[ ] + Ax −100# − Bx = 0
+Ax −100# − 50# = 0∴ Ax = +150#
ΣFy = 0[ ] + Ay − 300# = 0
∴ Ay = +300#
2.34
Ax
Ay
Bx
300#
100#
3’ 2’
6’
4’
2.35
€
ΣMA = 0[ ] − 40kN 2.5m( ) − 50kN 5.0m( ) + By 7.5m( ) = 0∴ By = 46.7kN
ΣFy = 0[ ] + A − 40kN − 50kN + 46.7kN = 0
∴ = 43.3KN
No horizontal reaction is necessary for this load case.
2.35
40kN 50kN
A BC D
2.5m2.5m 2.5m
2.36
20’ 20’ 20’20’
2k 3k 4k
15’15’ 20’
Ax
Ay By
2.32
y
x
W = 2000#
P
T5˚
20˚
2.33
34
60°
y
x
CA
CD
BC
34
DC
y
W = 200 lb.
DE
15°FBD(b)
FBD(a)
C
D
2.14
2.36
€
ΣFx = 0[ ] No force to balance Ax, ∴ Ax = 0
ΣMA = 0[ ] − 2k 20'( ) − 3k 40'( ) − 4k 60'( ) + By 80'( ) = 0
By =40k−ft +120k−ft + 240k−ft
80'= +5k
ΣMB = 0[ ] + 4k 20'( ) + 3k 40'( ) + 2k 60'( ) − Ay 80'( ) = 0∴ Ay = +4k
2.35
40kN 50kN
A BC D
2.5m2.5m 2.5m
2.36
20’ 20’ 20’20’
2k 3k 4k
15’15’ 20’
Ax
Ay By
2.37
2.37
30° 30°
90° 90°1500#
1500#
3000#
Ay Dy
Dx8.66’
8.66’
30’
60°
2.38
5
126m
6m
2.5m
12
5
5
12
1kN
1kN
Ax
Ay
By
Bx
B
€
ΣMA = 0[ ] −1500# 17.33'( ) − 3000# 8.67'( ) + Dy 30'( ) = 0
∴Dy = +1733#
ΣFy = 0[ ] −1500# cos30° − 3000# cos30° −1500# cos30° + Ay +1733# = 0
∴ Ay = +3463#
ΣFx = 0[ ] +1500sin 30° + 3000sin30° +1500sin30° − Dx = 0
∴ Dx = +3000#
€
513( ) 1kN( ) = 0.385kN1213( ) 1kN( ) = 0.923kN
2.37
30° 30°
90° 90°1500#
1500#
3000#
Ay Dy
Dx8.66’
8.66’
30’
60°
2.38
5
126m
6m
2.5m
12
5
5
12
1kN
1kN
Ax
Ay
By
Bx
B
2.38
€
ΣMA = 0[ ] + B 12m( ) − 0.923kN( ) 2.5m( ) + 0.385kN( ) 2.5m( ) −
0.385kN + 0.923kN( ) 6m( ) = 0
Solving for B; B = 0.767kNBx = 5
13( ) 0.767kN( ) = 0.295kN
By = 1213( ) 0.767kN( ) = 0.708kN
Reverting back to the unresolved forces;
€
ΣFx = 0[ ] + Ax +1kN − 0.295kN( )Bx( )
= 0
∴ Ax = +0.705kN
ΣFy = 0[ ] + Ay −1kN + 0.708kN( )By( )
= 0
∴ Ay = +0.292kN
2.15
2.39 2.40
Left beam:
Right beam:
€
ΣM E = 0[ ] + 4k 12'( ) −16k 24'( ) + Fy 48'( ) − 4k 60'( ) = 0∴ Fy = +12k
ΣFy = 0[ ] − 4k + Ey −16k +12k − 4k = 0
∴ Ey = +12k
Upper beam:Cy +Dy =8k; Cy = Dy = 4k Upper beam:
Lower beam:€
ΣMA = 0[ ] − 20k 24'( ) − 4k 60'( ) + By 48'( ) = 0∴ By = +15k
ΣFy = 0[ ] + Ay − 20k +15k − 4k = 0
∴ Ay = +9k
€
ΣMD = 0[ ] + 400# 12'( ) − Cy 10'( ) = 0
Cy = +480#
ΣFy = 0[ ] − 400# + Cy + Dy = 0
Dy = −480# + 400# = −80# ↓( )
ΣFx = 0[ ] + 300# + Dx = 0
Dx = −300# ←( )
€
ΣMB = 0[ ] − Cy 5'( ) − Ay 10'( ) = 0
Ay = −240# ↓( )
ΣFy = 0[ ] − 240# + By − 480# = 0
By = +720# ↑( )
Σx = 0[ ] Bx = 0
2.39
Ay By
Cy Dy
Ey Fy
8k
16k20k
4k
24’ 36’
48’
12’ 12’
12’ 12’ 12’
36’ 36’
2.40500# 400#
300#Dx
DyCy
2’ 10’
Bx
ByAy
Cy
10’ 5’
4k 4k
2.39
Ay By
Cy Dy
Ey Fy
8k
16k20k
4k
24’ 36’
48’
12’ 12’
12’ 12’ 12’
36’ 36’
2.40500# 400#
300#Dx
DyCy
2’ 10’
Bx
ByAy
Cy
10’ 5’
4k 4k
2.16
2.41
2.41
Ax
Ay
FD FDy
FDx
2k 8k
x
y
DCDF = 18.9k
BD
DFx = 15.2k
DFy = 11.3k
2.41
Ax
Ay
FD FDy
FDx
2k 8k
x
y
DCDF = 18.9k
BD
DFx = 15.2k
DFy = 11.3k
€
FDx = 45( )FD FDy = 3
5( )FD
€
ΣM A = 0[ ] + FDx 4'( ) + FDy 20'( ) − 2k 16'( ) − 8k 32'( ) = 04
5( )FD 4'( ) + 35( )FD 20'( ) − 32k−ft − 256k−ft = 0
∴ FD = 18.9kFDx = 4
5( ) 18.9k( ) = 15.2k
FDy = 35( ) 18.9k( ) = 11.3k
€
ΣFx = 0[ ] + Ax − FDx = 0
∴ Ax = +15.2k →( )
ΣFy = 0[ ] + Ay − 2k − 8k +11.3k = 0
∴Ay = −1.3k ↓( )
€
ΣFx = 0[ ] − 3DC3.16
⎛
⎝ ⎜
⎞
⎠ ⎟
DCx( )
+BD5.1
⎛
⎝ ⎜
⎞
⎠ ⎟
BDx( )
+15.2kDFx( )
= 0
ΣFy = 0[ ] − DC3.16
⎛
⎝ ⎜
⎞
⎠ ⎟
DCy( )
+5BD5.1
⎛
⎝ ⎜
⎞
⎠ ⎟
BDy( )
−11.3kDFy( )
= 0
Solving the two equations simultaneously;
BD = 17.9k
DC = 19.7k
2.42
2.42
θ1 = 75° θ2 = 30°
F1 = 500#
F2 = 400#
R =
720#
θR = 72.5°
Parallelogram Method
x
y
F2 = 400#
F1 = 500#
R =
720#
θR = 72.5°θ1 = 75°
θ2 = 30°
Tip-to-Tail Method
2.17
2.432.43
512
11
30°
A = 2kN
B =
1.8k
N
C = 1kN
R = 3.4
kN
x
y
θ = 52°
O
Tip-to-Tail Method
2.442.44
45°
75°
30°
P = 16k
Q = 22k
S = 20.5k
R = 42.5k(vertical)
O x
2.452.45
y
x
6k
3k
A
60° 30°
R = 6.71k
Rx = -0.4k
Ry = -6.7k
θ = 86.6°
30°3k
6k
Tip-to-tail
xy
θ = 86.6°
R = 6.71k
60°
€
Rx = ΣFx = − 6k( ) cos 60° + 3k( ) cos 30°
Rx = −3k + 2.6k = −0.4k
Ry = ΣFy = − 6k( ) sin 60° − 3k( ) sin 30°
Ry = −5.2k −1.5k = −6.7k
€
θ = tan−1 Ry
Rx
⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 6.7
0.4⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 16.75( ) = 86.6°
€
R = Rx2 + Ry
2 = 0.40( )2+ 6.7( )2
= 6.71k
2.45
y
x
6k
3k
A
60° 30°
R = 6.71k
Rx = -0.4k
Ry = -6.7k
θ = 86.6°
30°3k
6k
Tip-to-tail
xy
θ = 86.6°
R = 6.71k
60°
2.45
y
x
6k
3k
A
60° 30°
R = 6.71k
Rx = -0.4k
Ry = -6.7k
θ = 86.6°
30°3k
6k
Tip-to-tail
xy
θ = 86.6°
R = 6.71k
60°
2.18
2.462.46
x
y
F1 = 800#
F2 = 1200#
30°
30°
F1x
F1y
F2x
F2y
P = 500#
Rx = 1732#
Ry = 700#
R = 1868#
θ = 22°
2.46
x
y
F1 = 800#
F2 = 1200#
30°
30°
F1x
F1y
F2x
F2y
P = 500#
Rx = 1732#
Ry = 700#
R = 1868#
θ = 22°
Force Fx______________________ Fy __________________________
P=500# 0 -500#
F1 +F1cos30°=(800#)(0.866)=692.8# + F1sin30°=(800#)(0.50)=400#
F2 + F2cos30°=(1200#)(0.866)=1039.2# - F2sin30°=-(1200#)(0.50)=-600#
Alternate way to find the resultant R:
€
Rx = ΣFx = +692.8#+1039.2#= +1732# (→)Ry = ΣFy = −500#+400#−600#= −700# (↓)
R = 1732#( )2+ −700#( )2
=1868#
€
tanθ =Ry
Rx
;
θ = tan−1 7001732
⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 0.404( ) = 22°
€
sinθ =Ry
R;
R =Ry
sinθ=
700sin22°
=700
0.375=1867#
2.47
Component Fx Fy
AD = 90kN
€
−12
⎛
⎝ ⎜
⎞
⎠ ⎟ 90kN( ) = −
90kN2
= −63.6kN
€
−12
⎛
⎝ ⎜
⎞
⎠ ⎟ 90kN( ) = −63.6kN
BD = 45kN
€
− 35 45kN( ) = −27kN
€
− 45 45kN( ) = −36kN
CD = 110kN
€
+ 110kN( ) cos 30° = +95.3kN
€
− 110kN( )sin 30° = −55kN
€
Rx = Fx∑ = +4.7kN
€
Ry = Fy∑ = −154.6kN
€
R = Rx2 + Ry
2 = 4.7( )2+ 154.6( )2
= 154.7kN
€
θ = tan−1 Ry
Rx
⎛
⎝ ⎜
⎞
⎠ ⎟ = tan−1 154.6
4.7⎛
⎝ ⎜
⎞
⎠ ⎟
θ = tan−1 32.9( ) = 88.3°
D
AD = 90kN k BD = 45kN CD = 110kN
34
11
30°
x
y
2.47
x
y
θ = 88.3˚
R = 154.7kN
Ry = -154.6kN
Rx = +4.7kN
Resultant
D
AD = 90kN k BD = 45kN CD = 110kN
34
11
30°
x
y
2.47
x
y
θ = 88.3˚
R = 154.7kN
Ry = -154.6kN
Rx = +4.7kN
Resultant
2.19
€
d2x = d2 cos 45° = 14' 0.707( ) = 9.9'
d2y = d2 sin 45° = 14' 0.707( ) = 9.9'
d1x = d1 cos 20° = 10' 0.94( ) = 9.4'
€
ΣM C = 0[ ] + 250# d1x( ) −100# d2y( ) − F d2x( ) = 0
F =250# 9.4'( ) −100# 9.9'( )
9.9'= 137.4#
€
ΣFx = 0[ ] + Ax − 3.83kNFx( )
= 0 Ax = 3.83kN
ΣFy = 0[ ] + Ay − 3.21kNFy( )
= 0 Ay = 3.21kN
2.48
2.48
C
250#100#
F
20° 45°
d1x d2x
d2y
d2 = 14
’
d1 = 10’
2.49
2.50
T2 = 700#
T2x
T2y6’
4’
5’
30’
T1 = 500#
T1x
T1y
A
MA = +T2x(30’) + T2y(6’) – T1y(35’) – T1y(4’) =0
MA = 689.5#(30’) + 121.8#(6’) – 483#(35’) – 129.5#(4’) = +3990#-ft
Force Fx_________________ Fy_________________
T1 = 500# (500#)cos15o = 483# (500#)sin15o =129.5#
T2 = 700@# (700#)cos10o = 689.5# (700#)sin10o = 121.8#2.49
4
3
Ax
Ay
MRA
1.33m
1.1m 1m 1m
F = 5kNFy = 3.21kN
Fx = 3.83kN
40°
€
MA = −3.21kN 3.1m( ) − 3.83kN 1.33m( )MA = −9.95kN − m − 5.09kN − m = −15.04kN − m
2.20
2.51
R = ΣFy = 10# + 7# + 6# - 18# = +5#
MO = +(7#)(4”) + (6#)(9”) – (18#)(17” = - 224#-in.
R(x) = 224 #-in;
€
x =224#−in
5# = 44.8"
ω = 30N/m
Assume the member weight is located at the center of the length.
Weight of wood member:
R = ΣFy = -100N – 90N + 150N = -40N
MO = - (100N)(1m) – (90N)(1.5m) + (150N)(3m) = +215 N-m
R(x) = MO
€
∴ x =215N − m
40N= 5.4m
For a 40N force to produce a moment directed counter-clockwise, the R = 40N force will be at an imaginary location where x = 5.4m to the left of the origin.
2.52
2.51
y
x
10#7# 6#
18#
4” 5” 8”origin
O
origin
y
x
R = 5#
44.8”O
x
2.51
y
x
10#7# 6#
18#
4” 5” 8”origin
O
origin
y
x
R = 5#
44.8”O
x
2.52
x
y
origin1m 2m
100N
150N
30N/m
1m
x
y
origin
100N
150N
W = 90N (beam wt.) 1.5m
yR = 40N
x = 5.4m 3morigin
O
O
O
2.52
x
y
origin1m 2m
100N
150N
30N/m
1m
x
y
origin
100N
150N
W = 90N (beam wt.) 1.5m
yR = 40N
x = 5.4m 3morigin
O
O
O
2.52
x
y
origin1m 2m
100N
150N
30N/m
1m
x
y
origin
100N
150N
W = 90N (beam wt.) 1.5m
yR = 40N
x = 5.4m 3morigin
O
O
O
2.21
2.53
origin
y
x
A B200#100#
400#
4’ 4’ 4’ 4’
20#/ft
originx
A B200#100#
400#
W = 320#
y
Total beam weight equals (20#/ft)(16’) = 320#at the center of the beam length.
For the beam to remain stationary and horizontal,the moments taken about points A and B shouldbe balanced by the opposing moments due to Band A respectively, resulting in no resultant moment.
€
ΣMA = 0[ ] + 200# 4'( ) +100# 8'( ) − 320# 8'( ) − 400# 12'( ) − B 16'( ) = 0
∴ B = 360#
ΣMB = 0[ ] + 400# 4'( ) + 320# 8'( ) −100# 8'( ) − 200# 12'( ) − A 16'( ) = 0
∴ A = 60#
Ry = ΣFy = +A +100# +100# − 320# − 400# + B = 0
= +60# + 200# +100# − 320# − 400# + 360# = 0
Force Fx_______ Fy_________
AB
€
−1213
AB
€
+5
13AB
AC
€
−45
AC
€
−35
AC
W 0 -5k
€
ΣFx = 0[ ] − 1213
AB−45
AC = 0; 1213
AB = −45
AC = 0
∴ AB = −1315
AC
ΣFy = 0[ ] + 513
AC −35
AC − 5k = 0; + 513
−1315
AC⎛
⎝ ⎜
⎞
⎠ ⎟ −
35
AC = +5k
−AC3
− 3AC5
= 5k; AC = −5.36k compression( )
AB = −1315
−5.36k( ) = +4.64k tension( )
2.54
2.54
125
34
AB
AC W = 5k
ABy
ABx
ACx
ACy
2.22
2.55
2.55
3
4
4
3
60°
y
x
F = 2kN
Fx = 1.2kN
Fy = 1.6kN
AC
ACx
ACy
BC
BCx
BCy
3
4
60°
4
3
F = 2kN
BC = 2.1kN
AC = 0.27kN
Tip-to-tail
2.55
3
4
4
3
60°
y
x
F = 2kN
Fx = 1.2kN
Fy = 1.6kN
AC
ACx
ACy
BC
BCx
BCy
3
4
60°
4
3
F = 2kN
BC = 2.1kN
AC = 0.27kN
Tip-to-tail
Force Fx_______________ Fy__________________
AC
€
−35
AC
€
−45
AC
BC
€
−BC cos 60° = −0.50BC
€
+BC sin 60° = +0.866BC
F=2kN
€
+35
2kN( ) = +1.2kN
€
−45
2kN( ) = −1.6kN
€
ΣFx = 0[ ] − 35
AC − 0.50BC +1.2kN = 0; 35
AC + 0.50BC = 1.2kN
ΣFy = 0[ ] − 45
AC + 0.866BC −1.6kN = 0; − 45
AC + 0.866BC = 1.6kN
Solving simul tan eously;BC = 2.1kN (compression)AC = 0.27kN (tension)
Force Fx______________ Fy_______________
BA
€
−1213
BA
€
−5
13BA
DB
€
+DBsin 30° = +0.50DB
€
+DB cos 30° = +0.866DB
W 0 -800#
€
Rx = ΣFx = 0[ ] − 1213
BA + 0.50DB = 0; DB =2413
BA
Ry = ΣFy = 0[ ] − 513
BA + 0.866DB − 800# = 0
−5
13BA + 0.866 24
13BA
⎛
⎝ ⎜
⎞
⎠ ⎟ = 800#
BA = 658.2# DB =2413
⎛
⎝ ⎜
⎞
⎠ ⎟ 658.2#( ) = 1215.2#
2.56
x
y
BA
DB W = 800#
5
12
30°
2.56
2.23
2.57
2.57
x
y
W
CACB
CAx CBx
CBy
CAy
45° 30°
CBy
2.58
3
45
12
BE
BCAB = 1560#
y
xABx
ABy
BCx
BCy
3
4
45°
CB = 1800#CD
W
CDy
CDx CBx
CBy
y
x
Force Fx_______________ Fy__________________
CA
€
−CA cos 45° = −0.707CA
€
CAsin 45° = +0.707CA
CB
€
+CB cos 30° = +0.866CB
€
+CBsin 30° = +0.50CB
W 0 -W
€
ΣFx = 0[ ] − 0.707CA + 0.866CB = 0; CA =0.866CB
0.707= 1.22CB
This relationship indicates the CA > CB, therefore, CA = 1.8kNThen, CB = (1.8kN)/1.22 = 1.48kN
€
ΣFy = 0[ ] 0.707CA + 0.50CB − W = 0
W = 0.707 1.8kN( ) + 0.50 1.48kN( ) = 2.0kN
2.58
2.57
x
y
W
CACB
CAx CBx
CBy
CAy
45° 30°
CBy
2.58
3
45
12
BE
BCAB = 1560#
y
xABx
ABy
BCx
BCy
3
4
45°
CB = 1800#CD
W
CDy
CDx CBx
CBy
y
x
Force Fx______________ Fy______________
AB=1560#
€
+1213
1560#( ) = 1440#
€
−5
131560#( ) = −600#
BE 0 +BE
BC
€
−45
BC
€
−35
BC
€
ΣFx = 0[ ] +1440# −45
BC = 0; BC = 1800#
ΣFy = 0[ ] − 600# + BE −35
1800#( ) = 0; BE = 1680#
Joint B:
2.24
2.58 cont’d
2.57
x
y
W
CACB
CAx CBx
CBy
CAy
45° 30°
CBy
2.58
3
45
12
BE
BCAB = 1560#
y
xABx
ABy
BCx
BCy
3
4
45°
CB = 1800#CD
W
CDy
CDx CBx
CBy
y
x
Force Fx_____________ Fy______________
CD -0.707CD +0.707CD
CB = 1800#
€
+45
1800#( ) = 1440#
€
+35
1800#( ) = +1080#
W 0 -W
€
ΣFx = 0[ ] − 0.707CD +1440# = 0; CD = 2037#
ΣFy = 0[ ] + 0.707 2037#( ) +1080# − W = 0; W = 2520#
Joint C:
2.59
€
ΣM A = 0[ ] − 500# 10'( ) +5B13
10'( )Bx( )
+12B13
24'( )By( )
= 0
50B13
+288B
13= 5000#−ft ; B = 192.3#
Bx = 74#; By = 177.5#
€
ΣFx = 0[ ] + Ax − 74#Bx( )
= 0; Ax = 74#
ΣFy = 0[ ] + Ay − 500# +177.5#
By( )= 0; Ay = 322.5#
2.59
34
B3
4 10’
500#10’
24’
Ax
Ay
By
Bx
26’
2.25
€
ΣFx = 0[ ] Bx = 0
ΣM B = 0[ ] − Ay 4.5m( ) +1.8kN 2.5m( ) = 0; Ay = 1kN
ΣFy = 0[ ] +1kN −1.8kN + By = 0; By = 0.8kN
€
ΣFx = 0[ ] Cx = 0
ΣFy = 0[ ] − 0.8kN − 2.7kN + Cy = 0; Cy = 3.5kN
ΣM C = 0[ ] − M RC + 2.7kN 3m( ) + 0.8kN 6m( ) = 0 M RC = 8.1kN − m+ 4.8kN − m = 12.9kN − m
2.60
A
1.8kN 2.7kN
hinge
B C
AyCy
Cx
MRC
2.5m2m 3m 3m
1.8kN
A
AyBy
Bx2m 2.5m
Bx
By
2.7kN
MRC
Cx
Cy
3m3m
2.60
2.60
A
1.8kN 2.7kN
hinge
B C
AyCy
Cx
MRC
2.5m2m 3m 3m
1.8kN
A
AyBy
Bx2m 2.5m
Bx
By
2.7kN
MRC
Cx
Cy
3m3m
Beam AB:
Beam BC:
2.61
2.61
5k
4k
3k
2k 2k3k
Ay
Ax
By
20’
20’
20’ 20’ 20’
10’ 10’
2.62
Ay
Ax300#240#
180#
By
4’ 6’ 7’
4’
5’
5’
200#
80#
60#
By
Dy
Dx
Cy
€
ΣFx = 0[ ] − Ax + 4k = 0; Ax = +4k ←( )ΣM A = 0[ ] − 2k 20'( ) − 3k 40'( ) − 2k 60'( ) − 3k 40'( ) + 4k 20'( ) + By 80'( ) = 0
∴ By = +4k ↑( )ΣFy = 0[ ] + Ay − 2k − 3k − 2k − 3k + 4k
By( )= 0
∴ Ay = +6k ↑( )
2.26
2.61
5k
4k
3k
2k 2k3k
Ay
Ax
By
20’
20’
20’ 20’ 20’
10’ 10’
2.62
Ay
Ax300#240#
180#
By
4’ 6’ 7’
4’
5’
5’
200#
80#
60#
By
Dy
Dx
Cy
€
ΣM A = 0[ ] − 300# 5'( ) + By 8'( ) = 0; By = +187.5# ↑( )ΣFx = 0[ ] + Ax −180# = 0; Ax = +180# →( )ΣFy = 0[ ] +187.5# − 240# + Ay = 0; Ay = +52.5# ↑( )
€
ΣM D = 0[ ] + 200# 4'( ) + Cy 6'( ) −187.5# 9'( ) − 80# 13'( ) = 0
∴ Cy = +322# ↑( )ΣFx = 0[ ] + Dx − 60# = 0; Dx = +60# →( )ΣFy = 0[ ] − 200# + Dy + 322# −187.5# − 80# = 0
∴ Dy = +145.5# ↑( )
2.62
Upper Beam:
Lower Beam:
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