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
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