6.5 Cables: Concentrated Loads - Civil Engineering · 6.5 Cables: Concentrated Loads Example 1,...
Transcript of 6.5 Cables: Concentrated Loads - Civil Engineering · 6.5 Cables: Concentrated Loads Example 1,...
6.5 Cables: Concentrated Loads Example 1, page 1 of 3
1.4 kN
D
B
C
A
AA
x
1.4 kN
T BC
A y
B
2 kN
4 m3 m
2 m
2 m
1. For the cable system shown, determine the
reactions at support A and the distance yC.
1 Strategy: Note that both
horizontal and vertical
distances between A and B are
known. Thus summing
moments about B, for a
free-body AB, would give an
equation involving Ax and Ay
only.
2 Free-body diagram of AB
+Equilibrium equation
MB = 0: Ax(2.5 m) Ay(2 m) = 0 (1)
3
2 m
2.5 m yC
2.5 m
6.5 Cables: Concentrated Loads Example 1, page 2 of 3
1.4 kN
D
BC
A
2 kN
D x
D yA
y
A x
2 m 3 m 4 m
Equilibrium equation
MD = 0: (1.4 kN)(3 m + 4 m)
+ (2 kN)(4 m)
y(2 m + 3 m + 4 m) = 0 (2)
Solving gives
y = 1.978 kN Ans.
Using this result in Eq. 1 gives
Ax(2.5) y(2) = 0 ( Eq. 1 repeated)
1.978 kN
Solving gives
Ax = 1.582 kN Ans.
6
We can get another equation involving the reaction at
A as the only unknown by summing moments about
D for a free body of the entire cable.
Free-body diagram of entire cable
4
5
+
6.5 Cables: Concentrated Loads Example 1, page 3 of 3
1.4 kN
BC
A
2 kN
A y 1.978 kN
A x 1.582 kN
T CD
2 m 3 m
Now that the values of Ax and Ay are known,
summing moments about C, for a free body ABC,
will give an equation with a single unknown, yc.
Free-body diagram of ABC
+
Equilibrium equation
MC = 0: (1.4 kN)(3 m)
(1.978 kN)(2 m + 3 m)
+ (1.582 kN) yc) = 0
Solving gives
yc = 3.60 m Ans.
9
7
8
yC
6.5 Cables: Concentrated Loads Example 2, page 1 of 3
15 m
3.5 kN
B
2 kN
P A
C
4 m
2. The horizontal force P is applied to end A of the
cable as shown. Determine the value of P and the
distance d required to keep the cable system in the
configuration shown. Also determine the total
length of the cable.
Strategy: Note that we know the
horizontal distance between points A
and B, and we can compute the vertical
distance between these points. Thus
summing moments about B for a free
body AB, will give an equation from
which P can be found.
1
d
12 m
6.5 Cables: Concentrated Loads Example 2, page 2 of 3
Free-body diagram of ABC5
Equilibrium equation for ABC
MC = 0: (2.667 kN)(15 m) + (2 kN)(4 m + d)
+ (3.5 kN)(d) = 0
Solving gives
d = 5.819 m Ans.
Free-body diagram of AB
6
+
+
15 m 12 m = 3 m
C
15 m
d
3.5 kN
B
2 kN
P = 2.667 kNA
4 m
C y
C x
P
2 kN
T BC
3.5 kNA
B
4 m
2
Equilibrium equation for AB
MB = 0: (2 kN)(4 m) P(3 m) = 0
Solving gives
P = 2.667 kN Ans.
3
Now that the value of P is known, we can sum
moments about C, for a free-body of the whole
cable, to obtain an equation for d.
4
6.5 Cables: Concentrated Loads Example 2, page 3 of 3
B
A
C
4 m
12 m
d 5.819 m
15 m 12 m 3 m
The total length of the cable can be
found by applying the Pythagorean
Theorem to segment AB and to BC
7
Geometry8
LTotal = LAB + LBC
= (4 m)2 + (3 m)2 + (5.819 m)2 + (12 m)2
= 5 m + 13.34 m
= 18.34 m Ans.
6.5 Cables: Concentrated Loads Example 3, page 1 of 5
xC
2 m
4 m
2 m
C500 N
150 NB
6 m
3 m
150 NB
A
A y
A x
T BC
2
A
D
3. The cable supports the 150 N and 500 N
loads shown. Determine the distance xC and
the tension in each segment of the cable.
1 Strategy: Note that both the horizontal
and vertical distances between A and B
are known. Thus summing moments
about B, for a free-body AB would give
an equation involving the reaction
components at A (Ax and Ay), only; no
other unknowns are present.
Free-body diagram of AB
Equilibrium equation
MB = 0: Ay (4 m) x (2 m) = 0 (1)
3
+
4 m
6.5 Cables: Concentrated Loads Example 3, page 2 of 5
C500 N
150 N B
A
D
A y
A x
6 m
D y
D
x
2 m
Now summing moments about D, for a free body of the entire
cable, will give another equation for Ax and Ay.
4
Free-body diagram of entire cable5Equilibrium equation
MD = 0: x (2 m + 6 m + 3 m) (150 N)(6 m + 3 m) (500 N)(3 m) = 0
Solving gives
Ax = 259.09 N (2)
Substituting this in Eq. 1 gives
4 Ay 2 Ax = 0 (Eq. 1 repeated)
259.09 N
Solving gives
Ay = 129.55 N (3)
6
+
3 m
6.5 Cables: Concentrated Loads Example 3, page 3 of 5
9
We can get the tension in AB by considering a
free body of support A.
10
Free-body diagram of A for calculating tension TAB
Summing horizontal and vertical forces gives
TABx = 259.09 N (4)
TABy
= 129.55 N (5)
and the magnitude of the tension in AB is
TAB = (TABx)2 + (TABy
)2
= (259.09 N)2 + (129.55)2
= 290 N Ans.
12
x C
+
C500 N
150 NB
A
11
A y 129.55 N
A x 259.09 N
T ABy
T ABx
T CD
A y 129.55 N
A x 259.09 N
6 m
T AB
2 m
Now that Ax and Ay are known, we can calculate xC by
summing moments about C for a free-body ABC.
Free-body diagram of ABC
7
8
Equilibrium equation
MC = 0: (150 N)(6 m) (259.09 N)(2 m + 6 m)
+ (129.54 N)(xC) = 0
Solving gives
xC = 9.05 m Ans.
6.5 Cables: Concentrated Loads Example 3, page 4 of 5
B
TABy
129.55 N
T BCy
TABx 259.09 N
T BC
T BCx
14
150 N
Free-bodies of B and C will yield the
values of the tension in BC and CD
13
Free-body diagram of B for
calculating tension TBC
The magnitude of the tension in BC is
TBC = (TBCx)2 + (TBCy
)2
= (109.09 N)2 + (129.55 N)2
= 169.4 N Ans.
+
Summing vertical forces gives
TBCy
= 129.55 N
Summing horizontal forces gives
Fx = 0: 150 N TBCx + 259.09 N = 0
Solving gives
TBCx = 109.09 N
15
16
6.5 Cables: Concentrated Loads Example 3, page 5 of 5
C
500 N
17
T BCy
129.55 N
T CDy
T CD
TBCx 109.09 NT
CDx
Free-body diagram of C for
calculating tension TCD
Summing vertical forces gives
TCDy
= 129.55 N
Summing horizontal forces gives
Fx = 0: 500 N TCDx + 109.09 N = 0
Solving gives
TCDx = 390.91 N
The magnitude of the tension in CD is
TCD = (TCDx)2 + (TCDy
)2
= (390.91 N)2 + (129.55)2
= 412 N Ans.
+
C500 N
150 NB
6 m
3 m
A
D
4 m
xC
2 m
6.5 Cables: Concentrated Loads Example 4, page 1 of 6
2.2 kN
E
B C
A
1.8 kN
1.2 kN
2 m
3 m
D
y D
0.5 m0.25 m
4. For the cable system shown, determine the distance yC for
which segment BC will be horizontal. Also determine yD.
Strategy: We have to make use of the fact
that segment BC is horizontal. One way to
do this is to pass a section through BC and
then consider the portion of the cable to the
left of the section.
1
2 m
yC
6.5 Cables: Concentrated Loads Example 4, page 2 of 6
B
A
2.2 kN
A y
A x
2
T BC (horizontal)
Free-body diagram of AB
Equilibrium equation for AB (Because BC is
horizontal, we use the sum of vertical forces
so that the unknown tension TBC will not
appear in the equation.):
Fx = 0: Ay 2.2 kN = 0
Solving gives
Ay = 2.2 kN (1)
3
+2.2 kN
E
B C
A
1.8 kN
1.2 kN
2 m
3 m
D
y D
0.5 m0.25 m 2 m
yC
6.5 Cables: Concentrated Loads Example 4, page 3 of 6
2.2 kN
E
B C
A
1.8 kN
D
1.2 kN
2 m 2 m
y C
A y 2.2 kN
A x
E x
E y
3 m
0.5 m
Now that Ay is known, we can solve for
Ax by summing moments about E for a
free-body diagram of the entire cable.
4
Free-body diagram of entire cable5
+
6 Equilibrium equation for entire cable
ME = 0: (1.2 kN)(0.5 m)
+ (1.8 kN)(0.5 m + 2 m)
+ (2.2 kN)(0.5 m + 2 m + 2 m)
(2.2 kN)(0.5 m + 2 m + 2 m + 0.25 m)
x(3 m) = 0
Solving gives
Ax = 1.517 kN (2)
0.25 m
6.5 Cables: Concentrated Loads Example 4, page 4 of 6
yB
A y 2.2 kN
2.2 kN
0.25 m
B
A
T BC
A x 1.517 kN y
B y C (because BC is horizontal)
Summing moments about B for a free body AB
will now give us the value of yC.
7
Free-body diagram of AB8
Equilibrium equation for AB
MB = 0: (1.517 kN)(yC) (2.2 kN)(0.25 m) = 0
Solving gives
yC = 0.36 m Ans.
9
+
2.2 kN
E
B C
A
1.8 kN
1.2 kN
2 m
3 m
D
y D
0.5 m0.25 m 2 m
yC
6.5 Cables: Concentrated Loads Example 4, page 5 of 6
2.2 kN
B C
A
1.8 kN
D
1.2 kN
2 m 2 m0.25 m
A y 2.2 kN
A x 1.517 kN
T DE
y D
Finally, summing moments about D
for a free body ABCD will give an
equation for yD.
Free-body diagram of ABCD
10
11
Equilibrium equation
MD = 0: (1.517 kN)(yD)
+ (2.2 kN)(2 m + 2 m)
+ (1.8 kN)(2 m)
(2.2 kN)(0.25 m + 2 m + 2 m) = 0
Solving gives
yD = 2.01 m Ans.
12
+
6.5 Cables: Concentrated Loads Example 4, page 6 of 6
2.2 kN
E
B C
A
1.8 kN
D
1.2 kN0.36 m
The minus sign in yD = 2.01 m
indicates point D lies above point A,
not below it, as was assumed in
drawing the free-body diagram.
13
2.01 m
3 m
6.5 Cables: Concentrated Loads Example 5, page 1 of 4
4 m2 m
P B
B
A
C
200 N
P D
D
E
7 m
5 m
2 m
2 m
3 m
5. For the cable system shown,
determine the value of the forces PB
and PD necessary to maintain the
given configuration.
Strategy: Equilibrium equations for a free
body of the entire cable would involve
six unknowns (the components of the
support reactions: Ax, Ay, Ex, and Ey;
plus PB and PD). Thus using a free body
of the entire cable does not look like a
good place to start.
1
4 m
6.5 Cables: Concentrated Loads Example 5, page 2 of 4
P B
B
C
200 N
P D
D
5 m
A y
A x
E y
E x
4 m 4 m2 m 3 m
2 m
2 m
D
C
B
C 2 m + 3 m 5 m
2
BC
DC
BC
DC
A better place to start is to observe that we
know the horizontal and vertical distances
between points B, C and D. Thus we can
calculate the angles that segments BC and DC
make with the horizontal, and then we can use
the equilibrium equation for connector C to find
the tension in BC and in CD. Once these
tensions are known, we can use the equilibrium
equations for segments AB and ED to determine
PB and PD.
Geometry3
3 m5 m1
BC = tan ( ) = 59.04° (1)
1 7 m5 mDC = tan ( ) = 54.46° (2)
7 m
5 m
3 m
7 m
6.5 Cables: Concentrated Loads Example 5, page 3 of 4
4 m
P B
B
C
A y
A x
2 m
C 200 N
T BC
T DC
T BC 177.46 N
A
4
6
BC 59.04°
DC 54.46°
BC 59.04°
Free-body diagram of C
Equilibrium equations for C
Fx = 0: TBC cos 59.04° TDC cos 54.46° + 200 = 0
Fy = 0: TBC sin 59.04° TDC sin 54.46° = 0
Solving gives
TBC = 177.46 N (3)
TDC = 187.02 N (4)
+
+
5
Free-body diagram of AB
Equilibrium equation for AB
MA = 0: PB(4 m) + (177.46 N)(cos 59.04°)(2 m)
(177.46 N)(sin 59.04°)(4 m) = 0
Solving gives
PB = 106.5 N Ans.
7
+
6.5 Cables: Concentrated Loads Example 5, page 4 of 4
C
P D
D
E
E y
2 m E x
T DC 187.02 N
DC 54.46°
8
8 m
Free-body diagram of ED
ME = 0: PD(2 m) (187.02 N)(cos 54.46 )(2 m)
+ (187.02 N)(sin 54.46 )(8 m) = 0
Solving gives
PD = 500 N Ans.
9 +
6.5 Cables: Concentrated Loads Example 6, page 1 of 8
80 lb
50 lbB
C
1.5 ftA
D
6. For the cable system shown, determine
distance yB and the tension in each segment.
Strategy: If we can compute the reactions
at supports A and D, then we can compute
the tensions in AB and CD. Let's start
with support D. Note that both horizontal
and vertical distances between C and D
are known. Thus summing moments
about C, for the free body CD, would give
an equation involving Dx and Dy only.
1
5 ft3.5 ft2 ft
3.5 ft
yB
6.5 Cables: Concentrated Loads Example 6, page 2 of 8
80 lb
50 lbB
C
A x
A y
D y
D x
2 ft
1.5 ft
50 lb
C
D x
D y
3.5 ft
D
T BC
5 ft
2
5
Free-body diagram of CD
Equilibrium equation
MC = 0: Dy(5 ft) Dx(3.5 ft) = 0 (1)
+
3
4 We can get another equation involving reactions at Dx and
Dy as the only unknowns by summing moments about A
for a free body of the entire cable.
Free-body diagram of entire cable
6
+
Equation of equilibrium
MA = 0: Dy(2 ft + 3.5 ft + 5 ft) Dx(1.5 ft)
(50 lb)(2 ft + 3.5 ft) 80 lb(2 ft) = 0 (2)
3.5 ft 5 ft
6.5 Cables: Concentrated Loads Example 6, page 3 of 8
yB
80 lb
50 lbB
C
D
D y 34.41 lb
D x 49.15 lb
T AB
9
Solving Eqs. 1 and 2 simultaneously gives
Dx = 49.15 lb (3)
Dy = 34.41 lb (4)
Now that the values of Dx and Dy are
known, summing moments about B, for a
free body BCD, will give an equation with
a single unknown, yB.
8
Free-body diagram BCD
MB = 0: (50 lb)(3.5 ft) + (34.41 lb)(3.5 ft + 5 ft) (49.15 lb)(yB) = 0
Solving gives
yB = 2.39 ft Ans (5)
+10
7
5 ft3.5 ft
3.5 ft
6.5 Cables: Concentrated Loads Example 6, page 4 of 8
We still must find the tension in each cable segment.
A free-body consisting of support D will give the
tension in CD.
11
Summing x forces and then y forces gives
TCDx = 49.15 lb (6)
TCDy
= 34.41 lb (7)
The magnitude of the tension in CD is then
TCD = (TCDx)2 + (TCDy
)2
= (49.15 lb)2 + (34.41 lb)2
= 60.0 lb Ans.
D y 34.41 lb
D x 49.15 lb
T CDy
T CDx
T CD
12
D
Free-body diagram of D for determining tension TCD
80 lb
50 lbB
C
1.5 ftA
D
5 ft3.5 ft2 ft
3.5 ft
yB
6.5 Cables: Concentrated Loads Example 6, page 5 of 8
50 lb
C
14
TCDx 49.15 lb (Eq. 7)
T CDy
34.41 lb (Eq. 6)
T BCx
T BCyT
BC
A free-body diagram of connection C will give
the tension in cable segment BC.13
Free-body diagram of C for determining tension TBC
Summing horizontal forces gives
TBCx = 49.15 lb (8)
Summing vertical forces gives
Fy = 0: 34.41 lb TBCy
50 lb = 0
Solving gives
TBCy
= 15.59 lb (9)
+
80 lb
50 lbB
C
1.5 ftA
D
5 ft3.5 ft2 ft
3.5 ft
yB
6.5 Cables: Concentrated Loads Example 6, page 6 of 8
15 Why did TBCy
turn out to be negative? Because
in drawing the tension
TBC on the free-body
diagram of C, we
assumed that B was
below C:
But this assumption was
wrong, as we should have
realized, since we have
already calculated yB and
found yB = 2.39 ft (Eq. 5).
The corrected free-body
diagram shows TBC pulling
up on C, which would lead
to a positive value of TBCy
16
Free-body diagram of C
Corrected free-body diagram of C
B lies
above C
80 lb50 lb
B
C
A
D
y B
3.5 ft
80 lb 50 lb
B
C
3.5 ft
A
D
2.39 ft
C
CD y
CD x
50 lb
T BC (Pulls down on C)
50 lb
C
CD y
CD x
T BC (Pulls up on C)
6.5 Cables: Concentrated Loads Example 6, page 7 of 8
80 lb
B
T ABy
T BCx 49.15 lb
T ABx
T BCy
+15.59 lb
T AB
The magnitude of the tension in BC is
TBC = (TBCx)2 + (TBCy
)2
= (49.15 lb)2 + ( 15.59 lb)2
= 51.56 lb Ans.
17
The tension in AB can be found from a free-body
diagram of connector B.18
Free-body diagram of B for calculating TAB19
Since TBCy
points down, we are assuming correctly
that point C lies below B.
Summing x forces gives
TABx = 49.15 lb
Summing y forces gives
TABy
= 15.59 lb + 80 lb
= 95.59 lb
The magnitude of the tension in AB is then
TAB = (TABx)2 + (TABy
)2
= (49.15 lb)2 + (95.59 lb)2
= 107.5 lb Ans.
20
6.5 Cables: Concentrated Loads Example 6, page 8 of 8
80 lb
B
A
A y
A x
TBC (unknown)
B (known)
The tension in the cable segments could have
been calculated by other approaches, for
example, by calculating the angle of inclination
of each segment and then summing moments
about one end of the segment.
Example:
21
MA = 0 gives an equation that will give us the
value of TBC. Proceeding to the next cable segment,
BC, and summing moments about end B of that
segment would give the value of the tension in that
segment, etc.
80 lb
50 lbB
C
1.5 ftA
D
5 ft3.5 ft2 ft
3.5 ft
yB
6.5 Cables: Concentrated Loads Example 7, page 1 of 7
2 kip2 kip
2 kip
2 kip
A
F
18 ft 18 ft 18 ft 18 ft 18 ft
B
CD
E
15 ft
35 ft
7. The cable supports the four forces shown.
Determine the maximum tension in the cable.
Strategy: To find the maximum tension, Tmax in
the cable, we could find the tension in each of
the five segments and then pick the largest.
However, this is a tedious and time-consuming
approach, and we can find Tmax more easily if we
note two facts: 1) The horizontal component of
tension is the same in all cable segments, and 2)
the maximum tension occurs in the cable
segment with the maximum slope.
1
6.5 Cables: Concentrated Loads Example 7, page 2 of 7
2 kip
B T BCxT
ABx
T BCy
T ABy
T BC
T AB
T horizontal
T
Demonstration that horizontal components
of tension are equal:2
Free-body diagram of B
Fx = 0: TABx + TBCx = 0
Thus
TABx = TBCx
That is, the horizontal components are
equal. A similar argument holds for points
C, D, and E.
+
Demonstration that Tmax occurs in the segment
with the greatest slope:3
For each cable segments,
T =
Since Thorizontal is the same for all segments, it
follow that T will be largest where cos is
smallest, that is, where is the largest the
steepest slope.
Thorizontal
cos
6.5 Cables: Concentrated Loads Example 7, page 3 of 7
15 ft
2 kip2 kip2 kip
2 kip
A
B
EC
D
F
A x
A y
A
F y
F x
F
18 ft 18 ft 18 ft 18 ft 18 ft
35 ft
Thus the problem of finding Tmax has now
been reduced to determining whether the slope
is greater at A than at B.
Free-body diagram of entire cable
4
5 To determine A, we can find Ax and Ay
first. To do this, sum moments about F for
a free-body consisting of the entire cable.
Equilibrium equation for cable
MF = 0: Ax(15 ft) y(5 18 ft)
+ (2 kip)(4 18 ft)
+ (2 kip)(3 18 ft)
+ (2 kip)(2 18 ft)
+ (2 kip)(1 18 ft) = 0 (1)
6
+
6.5 Cables: Concentrated Loads Example 7, page 4 of 7
2 kip2 kip
2 kip
A
B
CD
A x
T DE
A y
18 ft 18 ft 18 ft
8
35 ft
Equilibrium equation for ABCD
MD = 0: Ax(35 ft) y(3 18 ft)
+ (2 kip) (2 18 ft)
+ (2 kip) (18 ft) = 0 (2)
Solving Eqs. 1 and 2 simultaneously gives
Ax = 4.154 kip (3)
y = 4.692 kip (4)
9
+
To obtain another equation for Ax and y, pass a
section through segment DE and consider a free
body ABCD.
7
Free-body diagram of ABCD
6.5 Cables: Concentrated Loads Example 7, page 5 of 7
2 kip2 kip
2 kip
2 kip
A
FB
C
DE
F y
F x
A y 4.692 kip
A x 4.154 kip
Free-body diagram of entire cable
10 To find the slope at F, consider a free body of the
entire cable.
11
12
+
+
Equilibrium equations for entire cable
Fx = 0: 4.154 kip + Fx = 0
Fy = 0: 4.692 kip + Fy (4 2 kip) = 0
Solving gives
Fx = 4.154 kip (5)
Fy = 3.308 kip (6)
6.5 Cables: Concentrated Loads Example 7, page 6 of 7
A F
13
A y 4.692 kip
A x 4.154 kip
T AB
A
A
F y 3.308 kip
F x 4.154 kip
F
F
T FE
Compare slopes at A and F
Thus A is greater than F and it follows that Tmax occurs at A.
Tmax = magnitude of the reaction force at A
= (Ax)2 + (Ay)2
= (4.154 kip)2 + (4.692 kip)2
= 6.27 kip Ans.
14
A = tan ( ) = 48.5°1 4.692
4.154 4.1543.3081
F = tan ( ) = 38.5°
6.5 Cables: Concentrated Loads Example 7, page 7 of 7
15 Of course, once we had calculated the components of the
reaction at F,
Fx = 4.154 kip (Eq. 5 repeated)
Fy = 3.308 kip (Eq. 6 repeated)
we could have computed the tension in EF:
TEF = (4.154 kip)2 + (3.308 kip)2
= 5.31 kip
and then we could have compared TEF with the tension in
TAB. That is, in this particular example, we didn't have to
compare slopes, but we did it to illustrate the principle
that Tmax occurs where the slope is a maximum.