PROBLEM 15.61
The cantilever beam BC is attached to the steel cable AB as
shown. Knowing that the cable is initially taut, determine the
tension in the cable caused by the distributed load shown. Use
SOLUTION
PROBLEM 15.62
Before the 2–kip/ft load is applied, a gap,
0
0.8
δ
=
in., exists
between the
W16 40×
beam and the support at C. Knowing
that
6
29 10E= ×
psi, determine the reaction at each support
after the uniformly distributed load is applied.
SOLUTION
Data:
3
0
63
4
62
32
0.8 in. 66.667 10 ft
29 10 psi 29 10 ksi
518 in
15.022 10 kip in
104.319 10 kip ft
E
I
EI
−
= = ×
=×=×
=
=×⋅
= ×⋅
δ
Loading I: Case 6 of Appendix C.
4
4
3
3
5
384
5(2)(24)
384(104.319 10 )
82.823 10 ft
C
wL
yEI
−
′= −
= − ×
=−×
Loading II: Case 4 of Appendix C.
33
3
3
(24)
48 48(104.319 10 )
2.7608 10
CC
C
C
RL R
yEI
R
−
′′ = = ×
= ×
Deflection at C.
0CCC
yyy
δ
′ ′′
=+=−
33 3
82.823 10 2.7608 10 66.667 10
5.8519 kips
C
C
R
R
−− −
− ×+ × =− ×
=
5.85 kips
C= ↑R
0: (2)(24)(12) (24) (5.8519)(12) 0
BA
MR
Σ= − − =
21.074 kips
A
R=
21.1kips
A= ↑R
0: 21.074 2(24) 5.8519 0
yB
FRΣ= − + + =
21.1kips
B
= ↑R
PROBLEM 15.61
The cantilever beam BC is attached to the steel cable AB as
shown. Knowing that the cable is initially taut, determine the
tension in the cable caused by the distributed load shown. Use
SOLUTION
PROBLEM 15.62
Before the 2–kip/ft load is applied, a gap,
0
0.8
δ
=
in., exists
between the
W16 40×
beam and the support at C. Knowing
that
6
29 10E= ×
psi, determine the reaction at each support
after the uniformly distributed load is applied.
SOLUTION
Data:
3
0
63
4
62
32
0.8 in. 66.667 10 ft
29 10 psi 29 10 ksi
518 in
15.022 10 kip in
104.319 10 kip ft
E
I
EI
−
= = ×
=×=×
=
=×⋅
= ×⋅
δ
Loading I: Case 6 of Appendix C.
4
4
3
3
5
384
5(2)(24)
384(104.319 10 )
82.823 10 ft
C
wL
yEI
−
′= −
= − ×
=−×
Loading II: Case 4 of Appendix C.
33
3
3
(24)
48 48(104.319 10 )
2.7608 10
CC
C
C
RL R
yEI
R
−
′′ = = ×
= ×
Deflection at C.
0CCC
yyy
δ
′ ′′
=+=−
33 3
82.823 10 2.7608 10 66.667 10
5.8519 kips
C
C
R
R
−− −
− ×+ × =− ×
=
5.85 kips
C= ↑R
0: (2)(24)(12) (24) (5.8519)(12) 0
BA
MR
Σ= − − =
21.074 kips
A
R=
21.1kips
A= ↑R
0: 21.074 2(24) 5.8519 0
yB
FRΣ= − + + =
21.1kips
B
= ↑R