978-0073398167 Chapter 15 Solution Manual Part 4

subject Type Homework Help
subject Pages 17
subject Words 1183
subject Authors David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf

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page-pf1
page-pf2
PROBLEM 15.22
Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
consent of McGraw-Hill Education.
page-pf3
PROBLEM 15.22 (Continued)
34
11 1 1
2 6 48 384
A
R L wL
 
−=
 
 
17
3 384
A
R wL=
7
128
A
wL= ↑R
From (1), with
,
2
L
x=
2
7
2 256
CA
L
M R wL

= =


2
0.0273
C
M wL=
From (4), with
,xL=
22
1 71 9
2 2 128 8 128
BA
L
M R L w wL wL
 
=− =−−
 
 
2
0.0703
B
M wL= −
Location of maximum positive M:
2
LxL<<
0
2
mA m
L
V R wx

=− −=


7
2 128
A
mLR
xL
w
−= =
7 71
2 128 128
mL
x LL=+=
From (4), with
,
m
xx=
2
1
22
m Am m
L
M Rx wx

=−−


2
7 71 1 7
128 128 2 128
wL L w L
 
= −
 
 
2
0.0288
m
M wL=
consent of McGraw-Hill Education.
page-pf4
PROBLEM 15.23
Determine the reaction at the roller support and the deflection at point D
if a is equal to L/3.
SOLUTION
0:xa≤≤
A
M Rx=
2
2
21
312
1
2
1
6
A
A
A
dy
EI M R x
dx
dy
EI R x C
dx
EIy R x C x C
= =
= +
= ++
:ax L≤≤
()
A
M Rx Px a= −−
2
2
22
3
33
34
()
11
()
22
11
()
66
A
A
A
dy
EI M R x P x a
dx
dy
EI R x P x a C
dx
EIy R x P x a C x C
== −−
= − −+
= −+ +
2
[ 0, 0] : 0 0 0xy C= = ++ =
2
0C∴=
,:
dy dy
xa
dx dx

= =


22
13
11
0
22
AA
Ra C Ra C+ = −+
13
CC∴=
[ , ]:x ay y= =
33
1 14
11
00
66
AA
R a Ca R a Ca C++= −++
4
0C∴=
, 0:
dy
xL
dx

= =


22
3
11
() 0
22
A
RL PL a C − +=
22
3
11
()
22
A
C PL a RL∴ = −−
page-pf5
PROBLEM 15.23 (Continued)
[ , 0] :x Ly= =
3 3 22
11 1 1
() () ()00
66 2 2
AA
RL PL a PL a RL L

− + +=


3
3 23 33
33
(2 3 ) 2 9
22
A
P PL
R L aL a L L
LL

= + = −+



14
27
A
P
= ↑R
Deflection at D.
at 3
L
yxa

= =


3
1
32
2
3
11
63 3
1 1 14 1 1 14
6 27 3 2 3 2 27 3
20
2187
DA
LL
y RC
EI
LL L
P PL PL
EI
PL
EI


 
= +

 
 





    
= + −−


    
    




= −
3
20
2187
D
PL
yEI
= ↓
consent of McGraw-Hill Education.
page-pf6
PROBLEM 15.24
Determine the reaction at the roller support and the deflection at point D
if a is equal to L/3.
SOLUTION
0: 0Σ= + = =−
y AB A B
F RR R R
00
0: 0Σ= − + = = +
A AB A B
M M M RL M RL M
0:xa≤≤
AA
M M Rx= +
2
2
21
23
12
1
2
11
26
AA
AA
AA
dy
EI M M R x
dx
dy
EI M x R x C
dx
EIy M x R x C x C
= = +
=++
= + ++
:ax L≤≤
0AA
M M Rx M
=+−
2
0
2
203
23 2
0 34
1
2
1 11
2 62
AA
AA
AA
dy
EI M M R x M
dx
dy
EI M x R x M x C
dx
EIy M x R x M x C x C
==+−
= + −+
= + − ++
1
0, 0 : 0 0 0
dy
xC
dx

= = ++ =


1
0C∴=
2
[ 0, 0]:000 0
xy C= = +++ =
2
0C∴=
,:
dy dy
xa
dx dx

= =


22
03
11
22
AA AA
Ma Ra Ma Ra Ma C+ = + −+
30
C Ma∴=
[ , ]:x ay y= =
23 23 2
00 4
11111
( )( )
26262
AA AA
Ma Ra Ma Ra Ma Ma a C+= +− + +
2
40
1
2
C Ma∴=
page-pf7
PROBLEM 15.24 (Continued)
[ , 0] :x Ly= =
23 2 2
00 0
1 11 1
( )( ) 0
2 62 2
AA
ML RL ML Ma L Ma+− + =
2 32 2
0 00 0
1 11 1
( ) () 0
2 62 2
BB
RL M L R L ML MaL Ma+ +− − + =
00 0
33
33 5
( 2) 2
33 6
22
BMa M L L M
R aL L L
LL
 
= −= − =
 
 
0
5
6
B
M
L
= ↓R
Deflection at D.
at 3
L
yxa

= =


23
23
00
0
2
0
11 1
26
11 5 1 5
26 3 66 3
7
486
D AA
y Mx Rx
EI
M L ML
LM
EI L L
ML
EI

= +




   
= − + ++

   
   


= −
2
0
7
486
D
ML
yEI
= ↑
consent of McGraw-Hill Education.
page-pf8
PROBLEM 15.25
Determine the reaction at A and draw the bending moment diagram for
consent of McGraw-Hill Education.
page-pf9
PROBLEM 15.26
Determine the reaction at A and draw the bending moment diagram for
the beam and loading shown.
SOLUTION
page-pfa
PROBLEM 15.22
Determine the reaction at the roller support and draw the bending moment
diagram for the beam and loading shown.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 15.22 (Continued)
34
11 1 1
2 6 48 384
A
R L wL
 
−=
 
 
17
3 384
A
R wL=
7
128
A
wL= ↑R
From (1), with
,
2
L
x=
2
7
2 256
CA
L
M R wL

= =


2
0.0273
C
M wL=
From (4), with
,xL=
22
1 71 9
2 2 128 8 128
BA
L
M R L w wL wL
 
=− =−−
 
 
2
0.0703
B
M wL= −
Location of maximum positive M:
2
LxL<<
0
2
mA m
L
V R wx

=− −=


7
2 128
A
mLR
xL
w
−= =
7 71
2 128 128
mL
x LL=+=
From (4), with
,
m
xx=
2
1
22
m Am m
L
M Rx wx

=−−


2
7 71 1 7
128 128 2 128
wL L w L
 
= −
 
 
2
0.0288
m
M wL=
consent of McGraw-Hill Education.
PROBLEM 15.23
Determine the reaction at the roller support and the deflection at point D
if a is equal to L/3.
SOLUTION
0:xa≤≤
A
M Rx=
2
2
21
312
1
2
1
6
A
A
A
dy
EI M R x
dx
dy
EI R x C
dx
EIy R x C x C
= =
= +
= ++
:ax L≤≤
()
A
M Rx Px a= −−
2
2
22
3
33
34
()
11
()
22
11
()
66
A
A
A
dy
EI M R x P x a
dx
dy
EI R x P x a C
dx
EIy R x P x a C x C
== −−
= − −+
= −+ +
2
[ 0, 0] : 0 0 0xy C= = ++ =
2
0C∴=
,:
dy dy
xa
dx dx

= =


22
13
11
0
22
AA
Ra C Ra C+ = −+
13
CC∴=
[ , ]:x ay y= =
33
1 14
11
00
66
AA
R a Ca R a Ca C++= −++
4
0C∴=
, 0:
dy
xL
dx

= =


22
3
11
() 0
22
A
RL PL a C − +=
22
3
11
()
22
A
C PL a RL∴ = −−
PROBLEM 15.23 (Continued)
[ , 0] :x Ly= =
3 3 22
11 1 1
() () ()00
66 2 2
AA
RL PL a PL a RL L

− + +=


3
3 23 33
33
(2 3 ) 2 9
22
A
P PL
R L aL a L L
LL

= + = −+



14
27
A
P
= ↑R
Deflection at D.
at 3
L
yxa

= =


3
1
32
2
3
11
63 3
1 1 14 1 1 14
6 27 3 2 3 2 27 3
20
2187
DA
LL
y RC
EI
LL L
P PL PL
EI
PL
EI


 
= +

 
 





    
= + −−


    
    




= −
3
20
2187
D
PL
yEI
= ↓
consent of McGraw-Hill Education.
PROBLEM 15.24
Determine the reaction at the roller support and the deflection at point D
if a is equal to L/3.
SOLUTION
0: 0Σ= + = =−
y AB A B
F RR R R
00
0: 0Σ= − + = = +
A AB A B
M M M RL M RL M
0:xa≤≤
AA
M M Rx= +
2
2
21
23
12
1
2
11
26
AA
AA
AA
dy
EI M M R x
dx
dy
EI M x R x C
dx
EIy M x R x C x C
= = +
=++
= + ++
:ax L≤≤
0AA
M M Rx M
=+−
2
0
2
203
23 2
0 34
1
2
1 11
2 62
AA
AA
AA
dy
EI M M R x M
dx
dy
EI M x R x M x C
dx
EIy M x R x M x C x C
==+−
= + −+
= + − ++
1
0, 0 : 0 0 0
dy
xC
dx

= = ++ =


1
0C∴=
2
[ 0, 0]:000 0
xy C= = +++ =
2
0C∴=
,:
dy dy
xa
dx dx

= =


22
03
11
22
AA AA
Ma Ra Ma Ra Ma C+ = + −+
30
C Ma∴=
[ , ]:x ay y= =
23 23 2
00 4
11111
( )( )
26262
AA AA
Ma Ra Ma Ra Ma Ma a C+= +− + +
2
40
1
2
C Ma∴=
PROBLEM 15.24 (Continued)
[ , 0] :x Ly= =
23 2 2
00 0
1 11 1
( )( ) 0
2 62 2
AA
ML RL ML Ma L Ma+− + =
2 32 2
0 00 0
1 11 1
( ) () 0
2 62 2
BB
RL M L R L ML MaL Ma+ +− − + =
00 0
33
33 5
( 2) 2
33 6
22
BMa M L L M
R aL L L
LL
 
= −= − =
 
 
0
5
6
B
M
L
= ↓R
Deflection at D.
at 3
L
yxa

= =


23
23
00
0
2
0
11 1
26
11 5 1 5
26 3 66 3
7
486
D AA
y Mx Rx
EI
M L ML
LM
EI L L
ML
EI

= +




   
= − + ++

   
   


= −
2
0
7
486
D
ML
yEI
= ↑
consent of McGraw-Hill Education.
PROBLEM 15.25
Determine the reaction at A and draw the bending moment diagram for
consent of McGraw-Hill Education.
PROBLEM 15.26
Determine the reaction at A and draw the bending moment diagram for
the beam and loading shown.
SOLUTION

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