Problem Set 7 (Key) II
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PROBLEM 7.1 Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.77.
SOLUTION ΣFx = 0: − F = 0
FBD JD:
F=0 ΣFy = 0: V − 20 lb − 20 lb = 0
V = 40.0 lb ΣM J = 0: M − ( 2 in.)( 20 lb ) − ( 6 in.)( 20 lb ) = 0 M = 160.0 lb ⋅ in.
PROBLEM 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.76.
SOLUTION FBD AJ:
ΣFx = 0: 60 lb − V = 0 V = 60.0 lb ΣFy = 0: − F = 0 F=0 ΣM J = 0: M − (1 in.)( 60 lb ) = 0 M = 60.0 lb ⋅ in.
PROBLEM 7.3 For the frame and loading of Prob. 6.80, determine the internal forces at a point J located halfway between points A and B.
SOLUTION FBD Frame:
ΣFy = 0: Ay − 80 kN = 0
A y = 80 kN
ΣM E = 0: (1.2 m ) Ax − (1.5 m )( 80 kN ) = 0
A x = 100 kN
0.3 m
θ = tan −1 = 21.801° 0.75 m FBD AJ:
ΣFx′ = 0: F − ( 80 kN ) sin 21.801° − (100 kN ) cos 21.801° = 0
F = 122.6 kN ΣFy′ = 0: V + ( 80 kN ) cos 21.801° − (100 kN ) sin 21.801° = 0
V = 37.1 kN ΣM J = 0: M + (.3 m )(100 kN ) − (.75 m )( 80 kN ) = 0
M = 30.0 kN ⋅ m
PROBLEM 7.4 For the frame and loading of Prob. 6.101, determine the internal forces at a point J located halfway between points A and B.
SOLUTION
FBD Frame:
ΣFy = 0: Ay − 100 N = 0
A y = 100 N
ΣM F = 0: 2 ( 0.32 m ) cos 30° Ax − ( 0.48 m )(100 N ) = 0
A x = 86.603 N
FBD AJ:
ΣFx′ = 0: F − (100 N ) cos 30° − ( 86.603 N ) sin 30° = 0
F = 129.9 N ΣFy′ = 0: V + (100 N ) sin 30° − ( 86.603 N ) cos 30° = 0
V = 25.0 N ΣM J = 0: ( 0.16 m ) cos 30° ( 86.603 N ) − ( 0.16 m ) sin 30° (100 N ) − M = 0
M = 4.00 N ⋅ m
PROBLEM 7.5 Determine the internal forces at point J of the structure shown.
SOLUTION FBD Frame:
AB is two-force member, so Ay Ax = 0.36 m 0.15 m
Ay =
5 Ax 12
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N Ay =
5 Ax = 260 N or A y = 260 N 12 ΣFx = 0: F − 624 N = 0
F = 624 N
FBD AJ: ΣFy = 0: 260 N − V = 0
V = 260 N ΣM J = 0: M − ( 0.2 m )( 260 N ) = 0
M = 52.0 N ⋅ m
PROBLEM 7.6 Determine the internal forces at point K of the structure shown.
SOLUTION FBD Frame:
ΣM C = 0: ( 0.3 m ) Ax − ( 0.48 m )( 390 N ) = 0
A x = 624 N
AB is two-force member, so Ay Ax 5 = → Ay = Ax 0.36 m 0.15 m 12
ΣFx = 0: − Ax + C x = 0
A y = 260 N
C x = A x = 624 N
ΣFy = 0: Ay + C y − 390 N = 0 C y = 390 N − 260 N = 130 N or C y = 130 N
FBD CK:
ΣFx′ = 0: F +
12 5 ( 624 N ) + (130 N ) = 0 13 13
F = −626 N ΣFy′ = 0:
F = 626 N
12 5 (130 N ) − ( 624 N ) − V = 0 13 13
V = −120 N
V = 120.0 N
ΣM K = 0: ( 0.1 m )( 624 N ) − ( 0.24 m )(130 N ) − M = 0
M = 31.2 N ⋅ m
PROBLEM 7.7 A semicircular rod is loaded as shown. Determine the internal forces at point J.
SOLUTION FBD Rod: ΣM B = 0: Ax ( 2r ) = 0
Ax = 0
ΣFx′ = 0: V − ( 30 lb ) cos 60° = 0
V = 15.00 lb
FBD AJ:
ΣFy′ = 0: F + ( 30 lb ) sin 60° = 0
F = −25.98 lb F = 26.0 lb ΣM J = 0: M − [ (9 in.) sin 60°] ( 30 lb ) = 0
M = −233.8 lb ⋅ in. M = 234 lb ⋅ in.
PROBLEM 7.8 A semicircular rod is loaded as shown. Determine the internal forces at point K.
SOLUTION FBD Rod:
ΣFy = 0: By − 30 lb = 0 ΣM A = 0: 2rBx = 0
B y = 30 lb Bx = 0
ΣFx′ = 0: V − ( 30 lb ) cos 30° = 0
FBD BK:
V = 25.98 lb V = 26.0 lb ΣFy′ = 0: F + ( 30 lb ) sin 30° = 0
F = −15 lb F = 15.00 lb ΣM K = 0: M − ( 9 in.) sin 30° ( 30 lb ) = 0
M = 135.0 lb ⋅ in.
PROBLEM 7.9 An archer aiming at a target is pulling with a 210-N force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at point J.
SOLUTION FBD Point A:
By symmetry T1 = T2 3 ΣFx = 0: 2 T1 − 210 N = 0 5
T1 = T2 = 175 N
Curve CJB is parabolic: y = ax 2 FBD BJ:
At B :
x = 0.64 m,
y = 0.16 m
a=
0.16 m
( 0.64 m )
2
=
1 2.56 m
1 ( 0.32 m )2 = 0.04 m 2.56 m
So, at J : yJ =
Slope of parabola = tan θ =
dy = 2ax dx
2 At J : θ J = tan −1 ( 0.32 m ) = 14.036° 2.56 m
So
α = tan −1
4 − 14.036° = 39.094° 3 ΣFx′ = 0: V − (175 N ) cos ( 39.094° ) = 0
V = 135.8 N ΣFy′ = 0: F + (175 N ) sin ( 39.094° ) = 0 F = −110.35 N
F = 110.4 N
PROBLEM 7.9 CONTINUED 3 Σ M J = 0 : M + ( 0.32 m ) (175 N ) 5
4 + ( 0.16 − 0.04 ) m (175 N ) = 0 5
M = 50.4 N ⋅ m
PROBLEM 7.10 For the bow of Prob. 7.9, determine the magnitude and location of the maximum (a) axial force, (b) shearing force, (c) bending moment.
SOLUTION By symmetry
T1 = T2 = T 3 ΣFx = 0: 2T1 − 210 N = 0 5
FBD Point A:
4 (175 N ) = 0 5
FC = 140 N
3 (175 N ) − VC = 0 5
VC = 105 N
ΣFy = 0: FC − ΣFx = 0:
FBD BC:
T1 = 175 N
3 4 ΣM C = 0: M C − ( 0.64 m ) (175 N ) − ( 0.16 m ) (175 N ) = 0 5 5 M C = 89.6 N ⋅ m
Also: if y = ax 2 and, at B, y = 0.16 m, x = 0.64 m 0.16 m
Then
a=
And
θ = tan −1
( 0.64 m )
FBD CK:
2
=
1 ; 2.56 m
dy = tan −1 2ax dx
ΣFx′ = 0: (140 N ) cos θ − (105 N ) sin θ + F = 0
So
F = (105 N ) sin θ − (140 N ) cos θ dF = (105 N ) cos θ + (140 N ) sin θ dθ ΣFy′ = 0: V − (105 N ) cos θ − (140 N ) sin θ = 0
So
V = (105 N ) cos θ + (140 N ) sin θ
PROBLEM 7.10 CONTINUED And
dV = − (105 N ) sin θ + (140 N ) cos θ dθ ΣM K = 0: M + x (105 N ) + y (140 N ) − 89.6 N ⋅ m = 0 M = − (105 N ) x −
(140 N ) x 2 ( 2.56 m )
+ 89.6 N ⋅ m
dM = − (105 N ) − (109.4 N/m ) x + 89.6 N ⋅ m dx
Since none of the functions, F, V, or M has a vanishing derivative in the valid range of 0 ≤ x ≤ 0.64 m ( 0 ≤ θ ≤ 26.6° ) , the maxima are at the limits ( x = 0, or x = 0.64 m ) . Therefore,
(a)
Fmax = 140.0 N
at C
(b)
Vmax = 156.5 N
at B
(c)
M max = 89.6 N ⋅ m
at C
PROBLEM 7.11 A semicircular rod is loaded as shown. Determine the internal forces at point J knowing that θ = 30o.
SOLUTION FBD AB:
4 3 ΣM A = 0: r C + r C − 2r ( 70 lb ) = 0 5 5
C = 100 lb ΣFx = 0: − Ax +
4 (100 lb ) = 0 5
A x = 80 lb ΣFy = 0: Ay +
3 (100 lb ) − 70 lb = 0 5
A y = 10 lb
FBD AJ:
ΣFx′ = 0: F − ( 80 lb ) sin 30° − (10 lb ) cos 30° = 0
F = 48.66 lb F = 48.7 lb
60°
ΣFy′ = 0: V − ( 80 lb ) cos 30° + (10 lb ) sin 30° = 0 V = 64.28 lb V = 64.3 lb
30°
ΣM 0 = 0: ( 8 in.)( 48.66 lb ) − ( 8 in.)(10 lb ) − M = 0 M = 309.28 lb ⋅ in. M = 309 lb ⋅ in.
PROBLEM 7.12 A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.
SOLUTION 4 3 ΣM A = 0: r C + r C − 2r ( 70 lb ) = 0 5 5
FBD AB:
C = 100 lb
ΣFx = 0: − Ax +
4 (100 lb ) = 0 5
A x = 80 lb
ΣFy = 0: Ay +
3 (100 lb ) − 70 lb = 0 5
A y = 10 lb
ΣM J = 0: M − ( 8 in.)(1 − cosθ )(10 lb ) − ( 8 in.)( sin θ )( 80 lb ) = 0
FBD AJ:
M = ( 640 lb ⋅ in.) sin θ + ( 80 lb ⋅ in.) ( cos θ − 1) dM = ( 640 lb ⋅ in.) cos θ − ( 80 lb ⋅ in.) sin θ = 0 dθ
θ = tan −1 8 = 82.87° ,
for where
d 2M = − ( 640 lb ⋅ in.) sin θ − ( 80 lb ⋅ in.) cos θ < 0 dθ 2
M = 565 lb ⋅ in. at θ = 82.9° is a max for AC
So FBD BK:
ΣM K = 0: M − ( 8 in.)(1 − cos β )( 70 lb ) = 0 M = ( 560 lb ⋅ in.)(1 − cos β ) dM = ( 560 lb ⋅ in.) sin β = 0 dβ So, for β =
π 2
for β = 0, where M = 0
, M = 560 lb ⋅ in. is max for BC
∴ M max = 565 lb ⋅ in. at θ = 82.9°
PROBLEM 7.13 Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D. Determine the internal forces at point J.
SOLUTION FBD Frame:
24 ΣM A = 0: ( 0.336 m ) C − ( 0.252 m )( 480 N ) = 0 25 C = 375 N
ΣFy = 0: Ax −
24 C =0 25
Ax =
24 ( 375 N ) = 360 N 25
A x = 360 N
ΣFy = 0: Ay − 480 N +
FBD CD:
7 ( 375 N ) = 0 24
A y = 375 N
ΣM C = 0: ( 0.324 m )( 480 N ) − ( 0.27 m ) B = 0 B = 576 N
ΣFx = 0: C x −
24 ( 375 N ) = 0 25
C x = 360 N
ΣFy = 0: −480 N +
FBD CJ:
7 ( 375 N ) + ( 576 N ) − C y = 0 25
C y = 201 N
ΣFx′ = 0: V − ( 360 N ) cos 30° − ( 201 N ) sin 30° = 0 V = 412 N
ΣFy′ = 0: F + ( 360 N ) sin 30° − ( 201 N ) cos 30° = 0 F = −5.93 N
F = 5.93 N
ΣM 0 = 0: ( 0.168 m )( 201 N + 5.93 N ) − M = 0 M = 34.76 N ⋅ m
M = 34.8 N ⋅ m
PROBLEM 7.14 Two members, each consisting of straight and 168-mm-radius quartercircle portions, are connected as shown and support a 480-N load at D. Determine the internal forces at point K.
SOLUTION FBD CD: ΣFx = 0: C x = 0 ΣM B = 0: ( 0.054 m )( 480 N ) − ( 0.27 m ) C y = 0 C y = 96 N
ΣFy = 0: B − C y = 0
FBD CK:
B = 96 N
ΣFy′ = 0: V − ( 96 N ) cos 30° = 0 V = 83.1 N
ΣFx′ = 0: F − ( 96 N ) sin 30° = 0 F = 48.0 N
ΣM K = 0: M − ( 0.186 m )( 96 N ) = 0 M = 17.86 N ⋅ m
PROBLEM 7.15 Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point J of the frame shown.
SOLUTION FBD Frame:
Note: Tension T in cord is 90 lb at any cut. All radii = 0.6 ft ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0 B x = 140 lb
ΣM E = 0: ( 5.4 ft )(140 lb ) − ( 7.2 ft ) By + ( 4.8 ft ) 90 lb − ( 0.6 ft ) 90 lb = 0
FBD BCE with pulleys and cord:
B y = 157.5 lb
ΣFx = 0: Ex − 140 lb = 0
E x = 140 lb
ΣFy = 0: 157.5 lb − 90 lb − 90 lb + E y = 0 E y = 22.5 lb
FBD EJ:
ΣFx′ = 0: −V +
3 4 (140 lb ) + ( 22.5 lb − 90 lb ) = 0 5 5
V = 30 lb
ΣFy′ = 0: F + 90 lb −
V = 30.0 lb
4 3 (140 lb ) − ( 90 lb − 22.5 lb ) = 0 5 5
F = 62.5lb
F = 62.5 lb
ΣM J = 0: M + (1.8 ft )(140 lb ) + ( 0.6 ft )( 90 lb ) + ( 2.4 ft )( 22.5 lb ) − ( 3.0 ft )( 90 lb ) = 0 M = − 90 lb ⋅ ft
M = 90.0 lb ⋅ ft
PROBLEM 7.16 Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point K of the frame shown.
SOLUTION FBD Whole:
Note: T = 90 lb ΣM B = 0: ( 5.4 ft ) Ax − ( 6 ft )( 90 lb ) − ( 7.8 ft )( 90 lb ) = 0
A x = 2.30 lb
FBD AE:
Note: Cord tensions moved to point D as per Problem 6.91 ΣFx = 0: 230 lb − 90 lb − Ex = 0 E x = 140 lb ΣM A = 0: (1.8 ft )( 90 lb ) − ( 7.2 ft ) E y = 0 E y = 22.5 lb
FBD KE:
ΣFx = 0: F − 140 lb = 0
F = 140.0 lb ΣFy = 0: V − 22.5 lb = 0
V = 22.5 lb ΣM K = 0: M − ( 2.4 ft )( 22.5 lb ) = 0
M = 54.0 lb ⋅ ft
PROBLEM 7.17 Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point J of the frame shown.
SOLUTION FBD Whole: ΣM A = 0: ( 5.4 ft ) Bx − ( 7.8 ft )( 90 lb ) = 0
B x = 130 lb
FBD BE with pulleys and cord:
ΣM E = 0: ( 5.4 ft )(130 lb ) − ( 7.2 ft ) By + ( 4.8 ft )( 90 lb ) − ( 0.6 ft )( 90 lb ) = 0
B y = 150 lb ΣFx = 0: Ex − 130 lb = 0
E x = 130 lb ΣFy = 0: E y + 150 lb − 90 lb − 90 lb = 0
E y = 30 lb
FBD JE and pulley: ΣFx′ = 0: − F − 90 lb +
4 3 (130 lb ) + ( 90 lb − 30 lb ) = 0 5 5
F = 50.0 lb ΣFy′ = 0: V +
3 4 (130 lb ) + ( 30 lb − 90 lb ) = 0 5 5
V = −30 lb
V = 30.0 lb
ΣM J = 0: − M + (1.8 ft )(130 lb ) + ( 2.4 ft )( 30 lb ) + ( 0.6 ft )( 90 lb ) − ( 3.0 ft )( 90 lb ) = 0
M = 90.0 lb ⋅ ft
PROBLEM 7.18 Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point K of the frame shown.
SOLUTION FBD Whole:
ΣM B = 0: ( 5.4 ft ) Ax − ( 7.8 ft )( 90 lb ) = 0
A x = 130 lb
FBD AE:
ΣM E = 0: − ( 7.2 ft ) Ay − ( 4.8 ft )( 90 lb ) = 0 Ay = −60 lb
A y = 60 lb
ΣFx = 0:
FBD AK:
F=0
ΣFy = 0: −60 lb + 90 lb − V = 0
V = 30.0 lb ΣM K = 0: ( 4.8 ft )( 60 lb ) − ( 2.4 ft )( 90 lb ) − M = 0
M = 72.0 lb ⋅ ft
PROBLEM 7.19 A 140-mm-diameter pipe is supported every 3 m by a small frame consisting of two members as shown. Knowing that the combined mass per unit length of the pipe and its contents is 28 kg/m and neglecting the effect of friction, determine the internal forces at point J.
SOLUTION FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N ΣM A = ( 0.6 m ) Cx − ( 0.315 m )( 824.04 N ) = 0
C x = 432.62 N
FBD pipe:
By symmetry: N1 = N 2 ΣFy = 0: 2 N1 =
21 N1 − W = 0 29
29 (824.04 N ) 42
= 568.98 N
Also note:
20 a = r tan θ = 70 mm 21 a = 66.67 mm
FBD BC: ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) C y + ( 0.06667 m )( 568.98 N ) = 0
C y = 532.42 N
PROBLEM 7.19 CONTINUED FBD CJ:
ΣFx′ = 0: F −
21 20 ( 432.62 N ) − ( 532.42 N ) = 0 29 29
F = 680 N ΣFy′ = 0:
21 20 ( 532.42 N ) − ( 432.62 N ) − V = 0 29 29
V = 87.2 N ΣM J = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 532.42 N ) + M = 0
M = 18.96 N ⋅ m
PROBLEM 7.20 A 140-mm-diameter pipe is supported every 3 m by a small frame consisting of two members as shown. Knowing that the combined mass per unit length of the pipe and its contents is 28 kg/m and neglecting the effect of friction, determine the internal forces at point K.
SOLUTION FBD Whole:
(
)
W = ( 3 m )( 28 kg/m ) 9.81 m/s 2 = 824.04 N ΣM C = 0: (.6 m ) Ax − (.315 m )( 824.04 N ) = 0
A x = 432.62 N
FBD pipe
By symmetry: N1 = N 2 ΣFy = 0: 2 N2 =
21 N1 − W = 0 29
29 824.04 N 42 = 568.98 N
Also note: FBD AD:
a = r tan θ = ( 70 mm )
20 21
a = 66.67 mm ΣM B = 0: ( 0.3 m )( 432.62 N ) − ( 0.315 m ) Ay − ( 0.06667 m )( 568.98 N ) = 0
A y = 291.6 N
PROBLEM 7.20 CONTINUED FBD AK:
ΣFx′ = 0:
21 20 ( 432.62 N ) + ( 291.6 N ) − F = 0 29 29
F = 514 N ΣFy′ = 0:
21 20 ( 291.6 N ) − ( 432.62 N ) + V = 0 29 29
V = 87.2 N ΣM K = 0: ( 0.15 m )( 432.62 N ) − ( 0.1575 m )( 291.6 N ) − M = 0
M = 18.97 N ⋅ m
PROBLEM 7.21 A force P is applied to a bent rod which is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at point J.
SOLUTION (a) FBD Rod:
ΣM D = 0: aP − 2aA = 0 A=
P 2
ΣFx = 0: V −
P =0 2 V =
FBD AJ:
ΣFy = 0:
F=0
ΣM J = 0: M − a
P =0 2 M =
(b) FBD Rod:
ΣM D = 0: aP −
A=
5P 2
P 2
a4 A = 0 25
aP 2
PROBLEM 7.21 CONTINUED FBD AJ:
ΣFx = 0:
3 5P −V = 0 5 2
V = ΣFy = 0:
3P 2
4 5P −F =0 5 2
F = 2P M =
(c) FBD Rod:
3 aP 2
3 4 ΣM D = 0: aP − 2a A − 2a A = 0 5 5 A=
5P 14
3 5P ΣFx = 0: V − =0 5 14
V = ΣFy = 0:
3P 14
4 5P −F =0 5 14
F=
2P 7
3 5P ΣM J = 0: M − a =0 5 14
M =
3 aP 14
PROBLEM 7.22 A force P is applied to a bent rod which is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at point J.
SOLUTION (a) FBD Rod:
ΣFx = 0: Ax = 0 ΣM D = 0: aP − 2aAy = 0
Ay =
P 2
ΣFx = 0: V = 0
FBD AJ:
ΣFy = 0:
P −F =0 2
F=
P 2
ΣM J = 0: M = 0
(b) FBD Rod:
ΣM A = 0 4 3 2a D + 2a D − aP = 0 5 5
D=
5P 14
ΣFx = 0: Ax −
4 5 P=0 5 14
Ax =
2P 7
ΣFy = 0: Ay − P +
3 5 P=0 5 14
Ay =
11P 14
PROBLEM 7.22 CONTINUED FBD AJ:
ΣFx = 0:
2 P −V = 0 7 V =
2P 7
11P −F =0 14
ΣFy = 0:
F= ΣM J = 0: a
2P −M =0 7 M =
(c) FBD Rod:
ΣM A = 0:
a 4D − aP = 0 2 5
4 5P =0 5 2
ΣFx = 0: Ax − ΣFy = 0: Ay − P −
FBD AJ:
11P 14
3 5P =0 5 2
D=
2 aP 7
5P 2
Ax = 2P Ay =
5P 2
ΣFx = 0: 2 P − V = 0 V = 2P ΣFy = 0:
5P −F =0 2 F=
5P 2
ΣM J = 0: a ( 2 P ) − M = 0 M = 2aP
PROBLEM 7.23 A rod of weight W and uniform cross section is bent into the circular arc of radius r shown. Determine the bending moment at point J when θ = 30°.
SOLUTION FBD CJ:
Note α =
180° − 60° π = 60° = 2 3 r =
r
α
sin α =
3r
3 3 3 r = π 2 2π
Weight of section = W ΣFy′ = 0: F −
120 4 = W 270 9
4 W cos 30° = 0 9
ΣM 0 = 0: rF − ( r sin 60° ) 2 3 3 3 3 − M = r 2π 2 9
F =
2 3 W 9
4W −M =0 9
2 3 1 4 − Wr W = π 9 9
M = 0.0666Wr
PROBLEM 7.24 A rod of weight W and uniform cross section is bent into the circular arc of radius r shown. Determine the bending moment at point J when θ = 120°.
SOLUTION (a) FBD Rod:
ΣFx = 0: Ax = 0 ΣM B = 0: rAy +
2r W 2r 2W − =0 π 3 π 3
Ay =
2W 3π
FBD AJ:
α =
Note:
60° π = 30° = 2 6
Weight of segment = W F =
r
α
sin α =
ΣM J = 0: ( r cos α − r sin 30° )
M =
2W 9
60 2W = 270 9
r 3r sin 30° = π /6 π 2W 2W + ( r − r sin 30° ) −M =0 9 3π
3r 3 r 3 1 3r 1 − + − + = Wr 2 2π 9 3π π 2 3π
M = 0.1788Wr
PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when
θ = 30o.
SOLUTION FBD Rod:
ΣFx = 0: A x = 0 ΣM B = 0:
2r
π
W − rAy = 0
α = 15°, weight of segment = W FBD AJ: r =
r
α
sin α =
ΣFy′ = 0:
2W
π F=
Ay =
2W
π
30° W = 90° 3
r sin15° = 0.9886r π /12
cos 30° −
W cos 30° − F = 0 3
W 3 2 1 − 2 π 3
2W W ΣM 0 = M + r F − =0 + r cos15° π 3 M = 0.0557 Wr
PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when
θ = 30o.
SOLUTION FBD Rod:
ΣM A = 0: rB − B=
r =
r
π /12
π
π
π 12
sin15° = 0.98862r
Weight of segment = W ΣFy′ = 0: F −
W =0
2W
α = 15° =
FBD BJ:
2r
30° W = 90° 3
W 2W cos 30° − sin 30° = 0 π 3
3 1 + W F= 6 π
ΣM 0 = 0: rF − ( r cos15° )
W −M =0 3
3 1 cos15° M = rW + − 0.98862 Wr 6 π 3
M = 0.289Wr
PROBLEM 7.27 For the rod of Prob.7.26, determine the magnitude and location of the maximum bending moment.
SOLUTION FBD Bar:
ΣM A = 0: rB −
α =
2r
π
θ
W =0
r =
r
α
=
F = =
2W
π 2W
π
4α
π
( sin 2α
M =
or
W
2W
π
sin 2α = 0
+ θ cos θ ) 4α
π
W −M =0
r 4α Wr ( sin θ + θ cos θ ) − sin α cos α W π α π 2
But, so
π
+ 2α cos 2α )
ΣM 0 = 0: rF − ( r cos α )
FBD BJ:
4
2α π /2
4α
W cos 2α −
( sin θ
π
sin α ,
Weight of segment = W
ΣFx′ = 0: F −
2W
π
0≤α ≤
so
2
B=
sin α cos α = M =
2Wr
π
1 1 sin 2α = sin θ 2 2
( sin θ M =
+ θ cosθ − sin θ ) 2
π
Wrθ cosθ
dM 2 = Wr ( cos θ − θ sin θ ) = 0 at θ tan θ = 1 dθ π
PROBLEM 7.27 CONTINUED
Solving numerically
θ = 0.8603 rad
and
M = 0.357Wr
at θ = 49.3° (Since M = 0 at both limits, this is the maximum)
PROBLEM 7.28 For the rod of Prob.7.25, determine the magnitude and location of the maximum bending moment.
SOLUTION FBD Rod:
ΣFx = 0: Ax = 0 2r
ΣM B = 0:
W − rAy = 0
π
α =
θ 2
r =
,
r
α
F =
2W
π
4α
π
2W ΣM 0 = 0: M + F − π
FBD AJ:
M =
2W
π
(1 + θ cosθ
M =
so
2r
π
2W
π
2W
π
cos 2α = 0
(1 − θ ) cosθ
4α W =0 r + ( r cos α ) π
− cosθ ) r −
sin α cos α =
But,
=
π
2α 4α W = π /2 π
W cos 2α +
(1 − 2α ) cos 2α
2W
sin α
Weight of segment = W ΣFx′ = 0: − F −
Ay =
4αW r
π
α
sin α cos α
1 1 sin 2α = sin θ 2 2
W (1 − cosθ + θ cosθ − sin θ )
dM 2rW = ( sin θ − θ sin θ + cosθ − cosθ ) = 0 dθ π
(1 − θ ) sin θ
for dM =0 dθ
for
=0
θ = 0, 1, nπ ( n = 1, 2,")
Only 0 and 1 in valid range At
θ = 0 M = 0, at θ = 57.3°
at θ = 1 rad M = M max = 0.1009 Wr
PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD beam:
(a) By symmetry: Ay = D =
1 L ( w) 2 2
Ay = D =
wL 4
Along AB:
ΣFy = 0: ΣM J = 0: M − x
wL −V = 0 4
wL =0 4
M =
V =
wL x (straight) 4
Along BC:
ΣFy = 0:
wL − wx1 − V = 0 4
V =
ΣM k = 0: M + M =
wL − wx1 4 V =0
straight with
wL 4
at
x1 =
L 4
x1 L wL wx1 − + x1 =0 2 4 4
w L2 L + x1 − x12 2 8 2
PROBLEM 7.29 CONTINUED
Parabola with
M =
3 L wL2 at x1 = 32 4
Section CD by symmetry (b) From diagrams: V
max
M
=
max
wL on AB and CD 4 =
3wL2 at center 32
PROBLEM 7.30 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) Along AB:
ΣFy = 0: −wx − V = 0 straight with
V =−
ΣM J = 0: M +
wL 2
L 2
x=
at
x wx = 0 2
M =−
parabola with
V = − wx
1 M = − wx 2 2
wL2 L at x = 8 2
Along BC:
ΣFy = 0: − w
L −V = 0 2
1 V = − wL 2
L L ΣM k = 0: M + x1 + w = 0 4 2 M =− straight with (b) From diagrams:
wL L + x1 2 4 3 L M = − wL2 at x1 = 8 2 V M
max
max
=
wL on BC 2
=
3wL2 at C 8
PROBLEM 7.31 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) Along AB:
ΣFy = 0: P − V = 0
V = P
ΣM J = 0: M − Px = 0
straight with M =
M = Px
PL at B 2
Along BC:
ΣFy = 0: P − P − V = 0
V =0
L ΣM K = 0: M + Px1 − P + x1 = 0 2 M =
PL 2
(constant)
(b) From diagrams:
V M
max
max
=
= P along AB PL along BC 2
PROBLEM 7.32 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION ΣM C = 0: LAy − M 0 = 0
(a) FBD Beam:
Ay =
M0 L
ΣFy = 0: − Ay + C = 0
M0 L
C=
Along AB:
ΣFy = 0: −
M0 −V = 0 L
ΣM J = 0: x
V =−
M0 +M =0 L
M0 L
M =−
straight with M = −
M0 x L
M0 at B 2
Along BC:
ΣFy = 0: − ΣM K = 0: M + x
M0 −V = 0 L
M0 − M0 = 0 L
straight with M = (b) From diagrams:
V =−
M0 L
x M = M 0 1 − L M0 at B 2 V
max
M = 0 at C
= P everywhere M
max
=
M0 at B 2
PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam: ΣM B = 0:
(.6 ft )( 4 kips ) + ( 5.1 ft )(8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay A y = 12.625 kips
ΣFy = 0: 12.625 kips − 10 kips − 8 kips − 4 kips + B = 0 B = 9.375 kips
Along AC:
ΣFy = 0: 12.625 kips − V = 0 V = 12.625 kips ΣM J = 0: M − x (12.625 kips ) = 0 M = (12.625 kips ) x M = 22.725 kip ⋅ ft at C
Along CD:
ΣFy = 0: 12.625 kips − 10 kips − V = 0 V = 2.625 kips ΣM K = 0: M + ( x − 1.8 ft )(10 kips ) − x (12.625 kips ) = 0 M = 18 kip ⋅ ft + ( 2.625 kips ) x M = 29.8125 kip ⋅ ft at D ( x = 4.5 ft )
=0
PROBLEM 7.33 CONTINUED Along DE:
Along DE: ΣFy = 0: (12.625 − 10 − 8 ) kips − V = 0
V = −5.375 kips
ΣM L = 0: M + x1 ( 8 kips ) + ( 2.7 ft + x1 )(10 kips ) − ( 4.5 ft + x1 )(12.625 kips ) = 0 M = 29.8125 kip ⋅ ft − ( 5.375 kips ) x1 M = 5.625 kip ⋅ ft at E
Along EB:
( x1 = 4.5 ft )
Along EB:
ΣFy = 0: V + 9.375 kips = 0
V = 9.375 kips
ΣM N = 0: x2 ( 9.375 kip ) − M = 0 M = ( 9.375 kips ) x2 M = 5.625 kip ⋅ ft at E
(b) From diagrams:
V
max
M
= 12.63 kips on AC
max
= 29.8 kip ⋅ ft at D
PROBLEM 7.34 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam: ΣM C = 0:
(1.2 m )( 4 kN ) − (1 m )(16 kN ) + ( 2 m )(8 kN ) + ( 3.2 m ) B = 0 B = −1.5 kN ΣFy = 0: −4 kN + C y − 16 kN + 8 kN − 1.5 kN = 0 C y = 13.5 kN
Along AC:
ΣFy = 0: −4 kN − V = 0 V = −4 kN ΣM J = 0: M + x ( 4 kN ) = 0
M = −4 kN x
M = −4.8 kN ⋅ m at C
Along CD:
ΣFy = 0: −4 kN + 13.5 kN − V = 0 V = 9.5 kN ΣM K = 0: M + (1.2 m + x1 )( 4 kN ) − x1 (13.5 kN ) = 0 M = −4.8 kN ⋅ m + ( 9.5 kN ) x1 M = 4.7 kN ⋅ m at D ( x1 = 1 m )
PROBLEM 7.34 CONTINUED
Along DE:
ΣFy = 0: V + 8 kN − 1.5 kN = 0 V = −6.5 kN ΣM L = 0: M − x3 ( 8 kN ) + ( x3 + 1.2 m ) (1.5 kN ) = 0 M = −1.8 kN ⋅ m + ( 6.5 kN ) x3 M = 4.7 kN ⋅ m at D ( x3 = 1 m )
Along EB:
ΣFy = 0: V − 1.5 kN = 0 V = 1.5 kN ΣM N = 0: x2 (1.5 kN ) + M = 0 M = − (1.5 kN ) x2
(b) From diagrams:
M = −1.8 kN ⋅ m at E V M
max max
= 9.50 kN ⋅ on CD = 4.80 kN ⋅ m at C
PROBLEM 7.35 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) Along AC:
ΣFy = 0: −3 kip − V = 0
V = −3 kips
ΣM J = 0: M + x ( 3 kips ) = 0
M = ( 3 kips ) x
M = −9 kip ⋅ ft at C
Along CD:
ΣFy = 0: −3 kips − 5 kips − V = 0
V = −8 kips
ΣM K = 0: M + ( x − 3 ft )( 5 kips ) + x ( 3 kips ) = 0 M = +15 kip ⋅ ft − ( 8 kips ) x M = −16.2 kip ⋅ ft at D ( x = 3.9 ft )
Along DE:
ΣFy = 0: −3 kips − 5 kips + 6 kips − V = 0 V = −2 kips ΣM L = 0: M − x1 ( 6 kips ) + (.9 ft + x1 )( 5 kips ) + ( 3.9 ft + x1 )( 3 kips ) = 0 M = −16.2 kip ⋅ ft − ( 2 kips ) x1 M = −18.6 kip ⋅ ft at E
( x1 = 1.2 ft )
PROBLEM 7.35 CONTINUED Along EB:
ΣFy = 0: −3 kips − 5 kips + 6 kips − 4 kips − V = 0
V = −6 kips
ΣM N = 0: M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips ) + ( 5.1 ft + x2 )( 3 kips ) − (1.2 ft + x2 )( 6 kips ) = 0 M = −18.6 kip ⋅ ft − ( 6 kips ) x2 M = −33 kip ⋅ ft at B
(b) From diagrams:
( x2
= 2.4 ft ) V
max
M
max
= 8.00 kips on CD = 33.0 kip ⋅ ft at B
PROBLEM 7.36 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION ΣM E = 0:
(a) FBD Beam:
(1.1 m )( 540 N ) − ( 0.9 m ) C y + ( 0.4 m )(1350 N ) − ( 0.3 m )( 540 N ) = 0 C y = 1080 N ΣFy = 0: −540 N + 1080 N − 1350 N −540 N + E y = 0
E y = 1350 N
Along AC:
ΣFy = 0: −540 N − V = 0
V = −540 N ΣM J = 0: x ( 540 N ) + M = 0
M = − ( 540 N ) x
Along CD:
ΣFy = 0: −540 N + 1080 N − V = 0
V = 540 N
ΣM K = 0: M + ( 0.2 m + x1 )( 540 N ) − x1 (1080 N ) = 0 M = −108 N ⋅ m + ( 540 N ) x1 M = 162 N ⋅ m at D ( x1 = 0.5 m )
PROBLEM 7.36 CONTINUED Along DE:
ΣFy = 0: V + 1350 N − 540 N = 0
V = −810 N
ΣM N = 0: M + ( x3 + 0.3 m )( 540 N ) − x3 (1350 N ) = 0 M = −162 N ⋅ m + ( 810 N ) x3 M = 162 N ⋅ m at D ( x3 = 0.4 )
Along EB:
ΣFy = 0: V − 540 N = 0
V = 540 N
ΣM L = 0: M + x2 ( 540 N ) = 0
( x2
M = −162 N ⋅ m at E
(b) From diagrams
M = −540 N x2 = 0.3 m ) V
M
max
max
= 810 N on DE W
= 162.0 N ⋅ m at D and E W
PROBLEM 7.37 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam: ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0 A y = 2 kips
ΣM A = 0: M A + ( 3 ft )( 6 ft )( 2 kips/ft ) − (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0 M A = 114 kip ⋅ ft
Along AC:
ΣFy = 0: 2 kips + x ( 2 kips/ft ) − V = 0 V = 2 kips + ( 2 kips/ft ) x V = 14 kips at C ( x = 6 ft ) ΣM J = 0: 114 kip ⋅ ft − x ( 2 kips ) −
x x ( 2 kips/ft ) + M = 0 2
M = (1 kip/ft ) x 2 + ( 2 kips ) x − 114 kip ⋅ ft M = −66 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: V − 12 kips − 2 kips = 0
V = 14 kips
ΣM k = 0: −M − x1 (12 kips ) − (1.5 ft + x1 )( 2 kips ) = 0
PROBLEM 7.37 CONTINUED M = −3 kip ⋅ ft − (14 kips ) x1
Along DB:
M = −66 kip ⋅ ft at C
ΣFy = 0:
( x1 = 4.5 ft )
V − 2 kips = 0
V = + 2 kips
ΣM L = 0: −M − 2 kip x3 = 0 M = − ( 2 kips ) x3 M = −3 kip ⋅ ft at D ( x = 1.5 ft )
(b) From diagrams:
V
max
M
max
= 14.00 kips on CD W = 114.0 kip ⋅ ft at A W
PROBLEM 7.38 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam: ΣM A = (15 ft ) B − (12 ft )( 2 kips/ft )( 6 ft ) − ( 6 ft )(12 kips ) = 0
B = 14.4 kips ΣFy = 0: Ay − 12 kips − ( 2 kips/ft )( 6 ft ) + 14.4 kips
A y = 9.6 kips
Along AC:
ΣFy = 0: 9.6 kips − V = 0 V = 9.6 kips ΣM J = 0: M − x ( 9.6 kips ) = 0 M = ( 9.6 kips ) x M = 57.6 kip ⋅ ft at C ( x = 6 ft )
Along CD:
ΣFy = 0: 9.6 kips − 12 kips − V = 0 V = −2.4 kips ΣM K = 0: M + x1 (12 kips ) − ( 6 ft + x1 )( 9.6 kips ) = 0 M = 57.6 kip ⋅ ft − ( 2.4 kips ) x1 M = 50.4 kip ⋅ ft at D
PROBLEM 7.38 CONTINUED Along DB:
ΣFy = 0: V − x3 ( 2 kips/ft ) + 14.4 kips = 0 V = −14.4 kips + ( 2 kips/ft ) x3 V = −2.4 kips at D ΣM L = 0: M +
x3 ( 2 kips/ft )( x3 ) − x3 (14.4 kips ) = 0 2
M = (14.4 kips ) x3 − (1 kip/ft ) x32 M = 50.4 kip ⋅ ft at D ( x3 = 6 ft )
(b) From diagrams:
V M
max max
= 14.40 kips at B W = 57.6 kip ⋅ ft at C W
PROBLEM 7.39 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) By symmetry: Ay = B = 8 kN +
1 ( 4 kN/m )( 5 m ) 2
A y = B = 18 kN
Along AC:
ΣFy = 0: 18 kN − V = 0
V = 18 kN
ΣM J = 0: M − x (18 kN )
M = (18 kN ) x
M = 36 kN ⋅ m at C ( x = 2 m )
Along CD:
ΣFy = 0: 18 kN − 8 kN − ( 4 kN/m ) x1 − V = 0 V = 10 kN − ( 4 kN/m ) x1 V = 0 at x1 = 2.5 m ( at center ) ΣM K = 0: M +
x1 ( 4 kN/m ) x1 + (8 kN ) x1 − ( 2 m + x1 )(18 kN ) = 0 2
M = 36 kN ⋅ m + (10 kN/m ) x1 − ( 2 kN/m ) x12 M = 48.5 kN ⋅ m at x1 = 2.5 m
Complete diagram by symmetry (b) From diagrams:
V
max
M
= 18.00 kN on AC and DB W max
= 48.5 kN ⋅ m at center W
PROBLEM 7.40 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a)
ΣM D = 0: ( 2 m )( 2 kN/m )( 2 m ) − ( 2.5 m )( 4 kN/m )( 3 m ) − ( 4 m ) F − ( 5 m )( 22 kN ) = 0
F = 22 kN ΣFy = 0: − ( 2 m )( 2 kN/m ) + D y − ( 3 m )( 4 kN/m ) − 22 kN + 22 kN = 0
D y = 16 kN
Along AC:
ΣFy = 0: − x ( 2 kN/m ) − V = 0 V = − ( 2 kN/m ) x
V = −4 kN at C
ΣM J = 0: M +
x ( 2 kN/m )( x ) ≠ 0 2
M = − (1 kN/m ) x 2
M = −4 kN ⋅ m at C
Along CD:
ΣFy = 0: − ( 2 m )( 2 kN/m ) − V = 0
V = −4 kN
ΣM K = 0: (1 m + x1 )( 2 kN/m )( 2 m ) = 0 M = −4 kN ⋅ m − ( 4 kN/m ) x1
M = −8 kN ⋅ m at D
PROBLEM 7.40 CONTINUED Along DE:
ΣFy = 0: − ( 2 kN/m )( 2 m ) + 16 kN − V = 0
V = 12 kN
ΣM L = 0: M − x2 (16 kN ) + ( x2 + 2 m )( 2 kN/m )( 2 m ) = 0 M = −8 kN ⋅ m + (12 kN ) x2
M = 4 kN ⋅ m at E
Along EF:
ΣFy = 0: V − x4 ( 4 kN/m ) − 22 kN + 22 kN = 0 V = ( 4 kN/m ) x4 ΣM 0 = 0: M +
V = 12 kN at E
x4 ( 4 kN/m ) x4 − (1 m )( 22 kN ) = 0 2
M = 22 kN ⋅ m − ( 2 kN/m ) x42
M = 4 kN ⋅ m at E
Along FB:
ΣFy = 0: V + 22 kN = 0
V = 22 kN
ΣM N = 0: M − x3 ( 22 kN ) = 0 M = ( 22 kN ) x3 M = 22 kN ⋅ m at F
(b) From diagrams:
V M
max max
= 22.0 kN on FB W = 22.0 kN ⋅ m at F W
PROBLEM 7.41 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION ΣFy = 0: (12 m ) w − ( 6 m )( 3 kN/m ) = 0
(a)
w = 1.5 kN/m
Along AC:
ΣFy = 0: x (1.5 kN/m ) − V = 0
V = (1.5 kN/m ) x
V = 4.5 kN at C ΣM J = 0: M −
x (1.5 kN/m )( x ) = 0 2
M = ( 0.75 kN/m ) x 2
M = 6.75 N ⋅ m at C
Along CD:
ΣFy = 0: x (1.5 kN/m ) − ( x − 3 m )( 3 kN/m ) − V = 0 V = 9 kN − (1.5 kN/m ) x
V = 0 at x = 6 m
x x − 3m ΣM K = 0: M + ( 3 kN/m )( x − 3 m ) − (1.5 kN/m ) x = 0 2 2 M = −13.5 kN ⋅ m + ( 9 kN ) x − ( 0.75 kN/m ) x 2
( x = 6 m)
M = 13.5 kN ⋅ m at center
Finish by symmetry (b) From diagrams:
V M
max
max
= 4.50 kN at C and D
= 13.50 kN ⋅ m at center
PROBLEM 7.42 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam: ΣFy = 0: ( 4 m )( w ) − ( 2 m )(12 kN/m ) = 0 w = 6 kN/m
Along AC:
ΣFy = 0: − x ( 6 kN/m ) − V = 0
V = − ( 6 kN/m ) x
V = −6 kN at C ( x = 1 m ) ΣM J = 0: M +
x ( 6 kN/m )( x ) = 0 2
M = − ( 3 kN/m ) x 2
M = −3 kN ⋅ m at C
Along CD:
ΣFy = 0: − (1 m )( 6 kN/m ) + x1 ( 6 kN/m ) − v = 0 V = ( 6 kN/m )(1 m − x1 )
V = 0 at x1 = 1 m
ΣM K = 0: M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) −
x1 ( 6 kN/m ) x1 = 0 2
M = −3 kN ⋅ m − ( 6 kN ) x1 + ( 3 kN/m ) x12 M = −6 kN ⋅ m at center
( x1 = 1 m )
Finish by symmetry (b) From diagrams:
V
max
M
= 6.00 kN at C and D
max
= 6.00 kN at center
PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.9 ft, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION
(a) FBD Beam: ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0 w = 400 lb/ft
Along AC:
ΣFy = 0: x ( 400 lb ) − V = 0 V = 360 lb at C ΣM J = 0: M −
V = ( 400 lb ) x
( x = 0.9 ft )
x ( 400 lb/ft ) x = 0 2
M = ( 200 lb/ft ) x 2
M = 162 lb ⋅ ft at C
Along CD:
ΣFy = 0: ( 0.9 ft + x1 )( 400 lb/ft ) − 900 lb − V = 0 V = −540 lb + ( 400 lb/ft ) x1 ΣM K = 0: M + x1 ( 900 lb ) −
V = 0 at x1 = 1.35 ft
0.9 ft + x1 ( 400 lb/ft )( 0.9 ft + x1 ) = 0 2
M = 162 lb ⋅ ft − ( 540 lb ) x1 + ( 200 lb/ft ) x12 M = −202.5 lb ⋅ ft at center
( x1 = 1.35 ft )
Finish by symmetry (b) From diagrams:
V
max
M
= 540 lb at C + and D −
max
= 203 lb ⋅ ft at center
PROBLEM 7.44 Solve Prob. 7.43 assuming that a = 1.5 ft.
SOLUTION
(a) FBD Beam: ΣFy = 0: ( 4.5 ft ) w − 900 lb − 900 lb = 0 w = 400 lb/ft
Along AC:
x ( 400 lb/ft ) − V = 0
ΣFy = 0: V = ( 400 lb/ft ) x
V = 600 lb at C
ΣM J = 0: M −
( x = 1.5 ft )
x ( 400 lb/ft ) x = 0 2
M = ( 200 lb/ft ) x 2
M = 450 lb ⋅ ft at C
Along CD:
ΣFy = 0: x ( 400 lb/ft ) − 900 lb − V = 0 V = −900 lb + ( 400 lb/ft ) x
V = −300 at x = 1.5 ft
V = 0 at x = 2.25 ft ΣM K = 0: M + ( x − 1.5 ft )( 900 lb ) −
x ( 400 lb/ft ) x = 0 2
M = 1350 lb ⋅ ft − ( 900 lb ) x + ( 200 lb/ft ) x 2 M = 450 lb ⋅ ft at x = 1.5 ft M = 337.5 lb ⋅ ft at x = 2.25 ft ( center )
Finish by symmetry (b) From diagrams:
V
M
max max
= 600 lb at C − and D +
= 450 lb ⋅ ft at C and D
PROBLEM 7.45 Two short angle sections CE and DF are bolted to the uniform beam AB of weight 3.33 kN, and the assembly is temporarily supported by the vertical cables EG and FH as shown. A second beam resting on beam AB at I exerts a downward force of 3 kN on AB. Knowing that a = 0.3 m and neglecting the weight of the ngle sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.
SOLUTION FBD angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN = 3.165 kN 2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0 By symmetry:
M C = 0.3165 kN ⋅ m
PD = 3.165 kN; M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − x (1.11 kN/m ) − V = 0 V = − (1.11 kN/m ) x
V = −1.332 kN at C
ΣM J = 0: M + M = ( 0.555 kN/m ) x 2
( x = 1.2 m )
x (1.11 kN/m ) x = 0 2 M = − 0.7992 kN ⋅ m at C
Along CI:
ΣFy = 0: − (1.11 kN/m ) x + 3.165 kN − V = 0 V = 3.165 kN − (1.11 kN/m ) x
V = 1.5 kN at I
( x = 1.5 m )
ΣM k = 0: M + (1.11 kN/m ) x − ( x − 1.2 m )( 3.165 kN ) − ( 0.3165 kN ⋅ m ) = 0
PROBLEM 7.45 CONTINUED M = 3.4815 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2 M = − 0.4827 kN ⋅ m at C
M = 0.01725 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3 kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B, the diagram can be completed by symmetry. (b) From diagrams:
V M
max max
= 1.833 kN at C and D = 799 N ⋅ m at C and D
PROBLEM 7.46 Solve Prob. 7.45 when a = 0.6 m.
SOLUTION FBD angle CE:
T =
(a) By symmetry:
3.33 kN + 3 kN = 3.165 kN 2
ΣFy = 0: T − PC = 0
PC = T = 3.165 kN
ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
PD = 3.165 kN
M C = 0.3165 kN ⋅ m
M D = 0.3165 kN ⋅ m
Along AC:
ΣFy = 0: − (1.11 kN/m ) x − V = 0 V = − (1.11 kN/m ) x
V = − 0.999 kN at C
ΣM J = 0: M +
( x = 0.9 m )
x (1.11 kN/m ) x = 0 2
M = − ( 0.555 kN/m ) x 2
M = − 0.44955 kN ⋅ m at C
Along CI:
ΣFy = 0: − x (1.11 kN/m ) + 3.165 kN − V = 0 V = 3.165 kN − (1.11 kN/m ) x V = 1.5 kN at I
V = 2.166 kN at C
( x = 1.5 m )
ΣM K = 0: M − 0.3165 kN ⋅ m + ( x − 0.9 m )( 3.165 kN ) +
x (1.11 kN/m ) x = 0 2
PROBLEM 7.46 CONTINUED M = −2.532 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2 M = − 0.13305 kN ⋅ m at C
M = 0.96675 kN ⋅ m at I
Note: At I, the downward 3 kN force will reduce the shear V by 3 kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B, the diagram can be completed by symmetry. (b) From diagrams:
V
max
= 2.17 kN at C and D M
max
= 967 N ⋅ m at I
PROBLEM 7.47 Draw the shear and bending-moment diagrams for the beam AB, and determine the shear and bending moment (a) just to the left of C, (b) just to the right of C.
SOLUTION FBD CD: ΣFy = 0: −1.2 kN + C y = 0 ΣM C = 0: ( 0.4 m )(1.2 kN ) − M C = 0
C y = 1.2 kN
M C = 0.48 kN ⋅ m
FBD Beam: ΣM A = 0: (1.2 m ) B + 0.48 kN ⋅ m − ( 0.8 m )(1.2 kN ) = 0
B = 0.4 kN ΣFy = 0: Ay − 1.2 kN + 0.4 kN = 0
A y = 0.8 kN
Along AC:
ΣFy = 0: 0.8 kN − V = 0 ΣM J = 0: M − x ( 0.8 kN ) = 0
V = 0.8 kN M = ( 0.8 kN ) x
M = 0.64 kN ⋅ m at x = 0.8 m
Along CB:
ΣFy = 0: V + 0.4 kN = 0 ΣM K = 0: x1 ( 0.4 kN ) − M = 0
V = −0.4 kN M = ( 0.4 kN ) x1
M = 0.16 kN ⋅ m at x1 = 0.4 m
(a)
Just left of C:
V = 800 N M = 640 N ⋅ m
(b)
Just right of C:
V = −400 N M = 160.0 N ⋅ m
PROBLEM 7.48 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD angle:
ΣFy = 0: Fy − 600 N = 0
Fy = 600 N
ΣM Base = 0: M − ( 0.3 m )( 600 N ) = 0
M = 180 N ⋅ m
All three angles are the same. FBD Beam: ΣM A = 0: (1.8 m ) B − 3 (180 N ⋅ m ) − ( 0.3 m + 0.9 m + 1.5 m )( 600 N ) = 0
B = 1200 N ΣFy = 0: Ay − 3 ( 600 N ) + 1200 N = 0
A y = 600 N
Along AC:
ΣFy = 0: 600 N − V = 0
V = 600 N
ΣM J = 0: M − x ( 600 N ) = 0 M = ( 600 N ) x = 180 N ⋅ m at x = .3 m
Along CD:
ΣFy = 0: 600 N − 600 N − V = 0
V =0
ΣM K = 0: M + ( x − 0.3 m )( 600 N ) − 180 N ⋅ m − x ( 600 N ) = 0 M = 360 N ⋅ m
PROBLEM 7.48 CONTINUED Along DE:
ΣFy = 0: V − 600 N + 1200 N = 0
V = −600 N
ΣM N = 0: −M − 180 N ⋅ m − x2 ( 600 N ) + ( x2 + 0.3 m )(1200 N ) = 0 M = 180 N ⋅ m + ( 600 N ) x2 = 540 N ⋅ m at D, x2 = 0.6 m M = 180 N ⋅ m at E (x2 = 0)
Along EB:
ΣFy = 0: V + 1200 N = 0 ΣM L = 0: x1 (1200 N ) − M = 0
V = −1200 N M = (1200 N ) x1
M = 360 N ⋅ m at x1 = 0.3 m
From diagrams:
V M
max
max
= 1200 N on EB = 540 N ⋅ m at D +
PROBLEM 7.49 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Whole:
ΣM D = 0: (1.2 m )(1.5 kN ) − (1.2 m )( 6 kN ) − ( 3.6 m )(1.5 kN ) + (1.6 m ) G = 0 G = 6.75 kN
ΣFx = 0: − Dx + G = 0
Beam AB, with forces D and G replaced by equivalent force/couples at C and F
D x = 6.75 kN
ΣFy = 0: Dy − 1.5 kN − 6 kN − 1.5 kN = 0
D y = 9 kN
Along AD:
ΣFy = 0: −1.5 kN − V = 0 ΣM J = 0: x (1.5 kN ) + M = 0
V = −1.5 kN M = − (1.5 kN ) x
M = −1.8 kN at x = 1.2 m
Along DE:
ΣFy = 0: −1.5 kN + 9 kN − V = 0
V = 7.5 kN
ΣM K = 0: M + 5.4 kN ⋅ m − x1 ( 9 kN ) + (1.2 m + x1 )(1.5 kN ) = 0 M = 7.2 kN ⋅ m + ( 7.5 kN ) x1
M = 1.8 kN ⋅ m at x1 = 1.2 m
PROBLEM 7.49 CONTINUED Along EF:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM N = 0: − M + 5.4 kN ⋅ m − ( x4 + 1.2 m )(1.5 kN ) M = 3.6 kN ⋅ m − (1.5 kN ) x4 M = 1.8 kN ⋅ m at x4 = 1.2 m;
M = 3.6 kN ⋅ m at x4 = 0
Along FB:
ΣFy = 0: V − 1.5 kN = 0
V = 1.5 kN
ΣM L = 0: − M − x3 (1.5 kN ) = 0
M = ( −1.5 kN ) x3
M = −1.8 kN ⋅ m at x3 = 1.2 m
From diagrams:
V M
max
max
= 7.50 kN on DE
= 7.20 kN ⋅ m at D +
PROBLEM 7.50 Neglecting the size of the pulley at G, (a) draw the shear and bending-moment diagrams for the beam AB, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Whole:
(a)
ΣM A = 0: ( 0.5 m )
12 5 D + (1.2 m ) D − ( 2.5 m )( 480 N ) = 0 13 13 D = 1300 N
ΣFy = 0: Ay +
5 (1300 N ) − 480 N = 0 13
Ay = −20 N
A y = 20 N
Beam AB with pulley forces and Along AE: force at D replaced by equivalent force-couples at B, F, E ΣFy = 0: −20 N − V = 0 ΣM J = 0: M + x ( 20 N )
V = −20 N M = − ( 20 N ) x
M = −24 N ⋅ m at x = 1.2 m
Along EF:
ΣFy = 0: V − 288 N − 480 N = 0
V = 768 N
ΣM L = 0: −M − x2 ( 288 N ) − ( 28.8 N ⋅ m ) − ( x2 + 0.6 m )( 480 N ) = 0 M = −316.8 N ⋅ m − ( 768 N ) x2 M = −316.8 N ⋅ m at x2 = 0 ; M = −624 N ⋅ m at x2 = 0.4 m
Along FB:
ΣFy = 0: V − 480 N = 0 ΣM K = 0: − M − x1 ( 480 N ) = 0
V = 480 N M = − ( 480 N ) x1
M = −288 N ⋅ m at x1 = 0.6 m
PROBLEM 7.50 CONTINUED (b) From diagrams:
V
max
M
= 768 N along EF
max
= 624 N ⋅ m at E +
PROBLEM 7.51 For the beam of Prob. 7.43, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of M max . (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.)
SOLUTION FBD Beam:
ΣFy = 0: Lw − 2 P = 0
w=2
ΣM J = 0: M − M =
Along AC:
P L
x 2P x = 0 2 L
P 2 x L
P 2 a at x = a L
ΣM K = 0: M + ( x − a ) P − M = P (a − x) +
M =
x 2P x = 0 2 L
P 2 Pa 2 x = L L
at
x=a
L L M = P a − at x = 2 4
Along CD:
This is M min by symmetry–see moment diagram completed by symmetry. For minimum M
max
, set M max = −M min : P
a2 L = −P a − L 4
a 2 + La −
or
a=
Solving: Positive answer
L2 =0 4
−1 ± 2 L 2
a = 0.20711L = 0.932 ft
(a) (b)
M
max
= 0.04289PL = 173.7 lb ⋅ ft
PROBLEM 7.52 For the assembly of Prob. 7.45, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of M max . (See hint for Prob. 7.51.)
SOLUTION FBD Angle:
By symmetry of whole arrangement: T =
3.33 kN + 3 kN = 3.165 kN 2
ΣFy = 0: T − F = 0
F = 3.165 kN
ΣM 0 = 0: M − ( 0.1 m )( 3.165 kN ) = 0
ΣM J = 0: M +
M = 0.3165 kN ⋅ m
x (1.11 kN/m ) x = 0 2
M = − ( 0.555 kN/m ) x 2 = − ( 0.555 kN/m )(1.5 m − a )
Along AC:
at C
2
( this is M min )
x (1.11 kN/m ) x 2 − x − (1.5 m − a ) ( 3.165 kN ) = 0
ΣM K = 0: M − 0.3165 kN ⋅ m +
Along CI:
M = −4.431 kN ⋅ m + ( 3.165 kN )( x + a ) − ( 0.555 kN/m ) x 2 M max ( at x = 1.5 m ) = − 0.93225 kN ⋅ m + ( 3.165 kN ) a
For minimum M
max
, set M max = −M min :
− 0.93225 kN ⋅ m + ( 3.165 kN ) a = ( 0.555 kN/m )(1.5 m − a )
Yielding:
a 2 − ( 8.7027 m ) a + 3.92973 m 2 = 0
Solving:
a = 4.3514 ± 13.864 = 0.4778 m, 8.075 m
Second solution out of range, so
(a)
2
a = 0.478 m
M max = 0.5801 kN ⋅ m
(b)
M max = 580 N ⋅ m
PROBLEM 7.53 For the beam shown, determine (a) the magnitude P of the two upward forces for which the maximum value of the bending moment is as small as possible, (b) the corresponding value of M max . (See hint for Prob. 7.51.)
SOLUTION Ay = B = 60 kN − P
By symmetry: Along AC:
ΣM J = 0: M − x ( 60 kN − P ) = 0
M = ( 60 kN − P ) x
M = 120 kN ⋅ m − ( 2 m ) P at x = 2 m
Along CD:
ΣM K = 0: M + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0 M = 120 kN ⋅ m − Px M = 120 kN ⋅ m − ( 4 m ) P at x = 4 m
Along DE:
ΣM L = 0: M − ( x − 4 m ) P + ( x − 2 m )( 60 kN ) − x ( 60 kN − P ) = 0 M = 120 kN ⋅ m − ( 4 m ) P
(const)
Complete diagram by symmetry For minimum M
max
, set M max = −M min
120 kN ⋅ m − ( 2 m ) P = − 120 kN ⋅ m − ( 4 m ) P
P = 40.0 kN
(a)
M min = 120 kN ⋅ m − ( 4 m) P
(b)
M
max
= 40.0 kN ⋅ m
PROBLEM 7.54 For the beam and loading shown, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of M max . (See hint for Prob. 7.51.) SOLUTION ΣM A = 0: M A − (1.5 ft )(1 kip ) − ( 3.5 ft )( 4 kips )
FBD Beam:
+ ( 3.5 ft + a )( 2 kips ) = 0 M A = 8.5 kip ⋅ ft − ( 2 kips ) a
ΣFy = 0: Ay − 1 kip − 4 kips + 2 kips = 0 A y = 3 kips
Along AC:
ΣM J = 0: M − x ( 3 kips ) + 8.5 kip ⋅ ft − ( 2 kips ) a = 0 M = ( 3 kips ) x + ( 2 kips ) a − 8.5 kip ⋅ ft M = ( 2 kips ) a − 4 kip ⋅ ft at C ( x = 1.5 ft ) M = ( 2 kips ) a − 8.5 kip ⋅ ft at A ( M min )
Along DB:
ΣM K = 0: − M + x1 ( 2 kips ) = 0
M = ( 2 kips ) x1
M = ( 2 kips ) a at D ΣM L = 0: ( x2 + a )( 2 kips ) − x2 ( 4 kips ) − M = 0 M = ( 2 kips ) a − ( 2 kips ) x2
Along CD:
M = ( 2 kips ) a − 4 kip ⋅ ft at C
For minimum M
max
( see above )
, set M max ( at D ) = −M min ( at A)
( 2 kips ) a = − ( 2 kips ) a − 8.5 kip ⋅ ft 4a = 8.5 ft
a = 2.125 ft a = 2.13 ft
(a) So
M max = ( 2 kips ) a = 4.25 kip ⋅ ft
(b)
M
max
= 4.25 kip ⋅ ft
PROBLEM 7.55 Knowing that P = Q = 375 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of M max . (See hint for Prob. 7.51.)
SOLUTION ΣM A = 0: ( a ft ) D − ( 4 ft )( 375 lb ) − ( 8 ft )( 375 lb ) = 0
D=
FBD Beam:
4500 lb a
ΣFy = 0: Ay − 2 ( 375 lb ) +
4500 lb = 0 a
4500 A y = 750 − lb a
It is apparent that M = 0 at A and B, and that all segments of the M diagram are straight, so the max and min values of M must occur at C and D 4500 ΣM C = 0: M − ( 4 ft ) 750 − lb = 0 a
Segment AC:
18000 M = 3000 − lb ⋅ ft a ΣM D = 0: − ( 8 − a ) ft ( 375 lb ) − M = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M So
max
, set M max = −M min 3000 − a 2 = 48
18000 = 375 ( 8 − a ) a a = 6.9282 ft a = 6.93 ft
(a) M max = 375 ( 8 − a ) = 401.92 lb ⋅ ft
(b)
M
max
= 402 lb ⋅ ft
PROBLEM 7.56 Solve Prob. 7.55 assuming that P = 750 lb and Q = 375 lb.
SOLUTION ΣM D = 0: − ( a ft ) Ay + ( a − 4 ) ft ( 750 lb )
FBD Beam:
− ( 8 − a ) ft ( 375 lb ) = 0
6000 A y = 1125 − lb a
It is apparent that M = 0 at A and B, and that all segments of the M-diagram are straight, so M max and M min occur at C and D. 6000 ΣM C = 0: M − ( 4 ft ) 1125 − lb = 0 a
Segment AC:
24000 M = 4500 − lb ⋅ ft a ΣM D = 0: − M − ( 8 − a ) ft ( 375 lb ) = 0
Segment DB:
M = −375 ( 8 − a ) lb ⋅ ft
For minimum M max , set M max = −M min 4500 −
24000 = 375 ( 8 − a ) a
a 2 + 4a − 64 = 0 a = 6.2462 ft
Then
a = −2 ± 68 ( need + ) a = 6.25 ft
(a)
M max = 375 ( 8 − a ) = 657.7 lb ⋅ ft
(b)
M
max
= 658 lb ⋅ ft
PROBLEM 7.57 In order to reduce the bending moment in the cantilever beam AB, a cable and counterweight are permanently attached at end B. Determine the magnitude of the counterweight for which the maximum absolute value of the bending moment in the beam is as small as possible and the corresponding value of M max . Consider (a) the case when the distributed load is permanently applied to the beam, (b) the more general case when the distributed load may either be applied or removed.
SOLUTION x wx = 0 2
ΣM J = 0: −M −
M due to distributed load:
1 M = − wx 2 2 ΣM J = 0: −M + xw = 0
M due to counter weight:
(a) Both applied:
M = wx
M = Wx −
w 2 x 2
And here M =
dM W = W − wx = 0 at x = dx w
W2 > 0 so M max ; M min must be at x = L 2w
So M min = WL −
1 2 wL . For minimum M 2
max
set M max = −M min , so
W2 1 = −WL + wL2 or W 2 + 2wLW − w2 L2 = 0 2w 2 W = −wL ± 2w2 L2 (need +)
M max
(b) w may be removed:
W2 = = 2w
Without w, With w (see part a)
(
W =
)
2 −1 2
(
)
2 − 1 wL = 0.414wL
2
wL2
M max = 0.858wL2
M = Wx, M max = WL at A M = Wx − M min = WL −
w 2 x , 2
M max =
1 2 wL at x = L 2
W2 W at x = 2w w
PROBLEM 7.57 CONTINUED For minimum M max , set M max ( no w ) = −M min ( with w ) WL = −WL +
With
1 2 1 wL → W = wL → 2 4
M max =
1 2 wL 4
W =
1 wL 4
PROBLEM 7.58 Using the method of Sec. 7.6, solve Prob. 7.29.
SOLUTION
(a) and (b)
By symmetry: Ay = D = Shear Diag:
1 L wL w = 2 2 4
V jumps to Ay =
or A y = D =
wL 4
wL at A, 4
and stays constant (no load) to B. From B to C, V is linear wL L wL dV −w =− = −w , and it becomes at C. 4 2 4 dx (Note: V = 0 at center of beam. From C to D, V is again constant.) Moment Diag: M starts at zero at A wL dM = and increases linearly 4 to B. dV MB = 0 +
L wL wL2 . = 4 4 16
From B to C M is parabolic wL dM = V , which decreases to zero at center and − at C , 4 dx
M is maximum at center.
M max =
Then, M is linear with
wL2 1 L wL + 16 2 4 4
dM wL =− to D dy 4
MD = 0 V M
max
max
=
=
wL 4
3wL2 32
Notes: Symmetry could have been invoked to draw second half. Smooth transitions in M at B and C, as no discontinuities in V.
PROBLEM 7.59 Using the method of Sec. 7.6, solve Prob. 7.30.
SOLUTION
(a) and (b)
Shear Diag:
V = 0 at A and is linear
wL dV L dV = −w to − w = − at B. V is constant = 0 from 2 dx dx 2 B to C. V
max
=
wL 2
Moment Diag: M = 0 at A and is dM decreasing with V to B. parabolic dx MB =
1 L wL wL2 − = − 2 2 2 8
wL dM =− From B to C, M is linear dx 2 MC = −
wL2 L wL 3wL2 − = − 8 8 2 2
M
max
=
Notes: Smooth transition in M at B, as no discontinuity in V. It was not necessary to predetermine reactions at C. In fact they are given by −VC and − M C .
3wL2 8
PROBLEM 7.60 Using the method of Sec. 7.6, solve Prob. 7.31.
SOLUTION
(a) and (b)
Shear Diag: dV = 0 to B. V jumps down P V jumps to P at A, then is constant dx to zero at B, and is constant (zero) to C. V
max
= P
Moment Diag: dM M is linear = V = P to B. dy PL L M B = 0 + ( P) = . 2 2 PL dM = 0 at to C M is constant 2 dx M
max
=
PL 2
Note: It was not necessary to predetermine reactions at C. In fact they are given by −VC and − M C .
PROBLEM 7.61 Using the method of Sec. 7.6, solve Prob. 7.32.
SOLUTION
(a) and (b)
ΣM C = 0: LAy − M 0 = 0
Ay =
M0 L
Shear Diag: V jumps to −
M0 dV at A and is constant = 0 all the way to C L dx V
max
=
M0 W L
=
M0 W 2
Moment Diag: M dM M is zero at A and linear = V = − 0 throughout. L dx LM M M B− = − 0 = − 0 , 2 L 2 M but M jumps by + M 0 to + 0 at B. 2 M LM MC = 0 − 0 = 0 2 2 L M
max
PROBLEM 7.62 Using the method of Sec. 7.6, solve Prob. 7.33.
SOLUTION
(a) and (b)
ΣM B = 0: ( 0.6 ft )( 4 kips ) + ( 5.1 ft )( 8 kips ) + ( 7.8 ft )(10 kips ) − ( 9.6 ft ) Ay = 0
A y = 12.625 kips Shear Diag: dV V is piecewise constant, = 0 with discontinuities at each dx concentrated force. (equal to force)
V
= 12.63 kips W
max
Moment Diag:
dM = V throughout. M is zero at A, and piecewise linear dx M C = (1.8 ft )(12.625 kips ) = 22.725 kip ⋅ ft M D = 22.725 kip ⋅ ft + ( 2.7 ft )( 2.625 kips ) = 29.8125 kip ⋅ ft M E = 29.8125 kip ⋅ ft − ( 4.5 ft )( 5.375 kips ) = 5.625 kip ⋅ ft M B = 5.625 kip ⋅ ft − ( 0.6 ft )( 9.375 kips ) = 0
M
max
= 29.8 kip ⋅ ft W
PROBLEM 7.63 Using the method of Sec. 7.6, solve Prob. 7.36.
SOLUTION (a) and (b)
FBD Beam:
ΣM E = 0: (1.1 m )( 0.54 kN ) − ( 0.9 m ) C y + ( 0.4 m )(1.35 kN ) − ( 0.3 m )( 0.54 kN ) = 0 C y = 1.08 kN
ΣFy = 0: − 0.54 kN + 1.08 kN − 1.35 kN + E − 0.54 kN = 0 E = 1.35 kN Shear Diag:
dV V is piecewise constant, = 0 everywhere with discontinuities at dx each concentrated force. (equal to the force) V
max
= 810 N W
Moment Diag:
M is piecewise linear starting with M A = 0
M C = 0 − 0.2 m ( 0.54 kN ) = 0.108 kN ⋅ m
M D = 0.108 kN ⋅ m + ( 0.5 m )( 0.54 kN ) = 0.162 kN ⋅ m M E = 0.162 kN ⋅ m − ( 0.4 m )( 0.81 kN ) = − 0.162 kN ⋅ m M B = 0.162 kN ⋅ m + ( 0.3 m )( 0.54 kN ) = 0
M
max
= 0.162 kN ⋅ m = 162.0 N ⋅ m W
PROBLEM 7.64 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) and (b)
Shear Diag:
dV V = 0 at A and linear = −2 kN/m to C dx VC = −1.2 m ( 2 kN/m ) = −2.4 kN. At C, V jumps 6 kN to 3.6 kN, and is constant to D. From there, V is dV = +3 kN/m to B linear dx VB = 3.6 kN + (1 m )( 3 kN/m ) = 6.6 kN V
max
= 6.60 kN W
M A = 0.
Moment Diag:
dM decreasing with V . From A to C, M is parabolic, dx MC = −
1 (1.2 m )( 2.4 kN ) = −1.44 kN ⋅ m 2
dM From C to D, M is linear dx
= 3.6 kN
M D = −1.44 kN ⋅ m + ( 0.6 m )( 3.6 kN ) = 0.72 kN ⋅ m.
dM From D to B, M is parabolic increasing with V dx M B = 0.72 kN ⋅ m +
1 (1 m )( 3.6 + 6.6 ) kN 2
= 5.82 kN ⋅ m M
max
= 5.82 kN ⋅ m W
Notes: Smooth transition in M at D. It was unnecessary to predetermine reactions at B, but they are given by −VB and −M B
PROBLEM 7.65 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION ΣM B = 0: ( 3 ft )(1 kip/ft )( 6 ft ) + ( 8 ft )( 6 kips )
(a) and (b)
+ (10 ft )( 6 kips ) − (12 ft ) Ay = 0 A y = 10.5 kips Shear Diag:
V is piecewise constant from A to E, with discontinuities at A, C, and E equal to the forces. VE = −1.5 kips. From E to B, V is linear dV = −1 kip/ft , dx so VB = −1.5 kips − ( 6 ft )(1 kip/ft ) = −7.5 kips V
max
= 10.50 kips W
Moment Diag: M A = 0, then M is piecewise linear to E M C = 0 + 2 ft (10.5 kips ) = 21 kip ⋅ ft M D = 21 kip ⋅ ft + ( 2 ft )( 4.5 kips ) = 30 kip ⋅ ft M E = 30 kip ⋅ ft − ( 2 ft )(1.5 kips ) = 27 kip ⋅ ft dM decreasing with V , and From E to B, M is parabolic dx M B = 27 kip ⋅ ft −
1 ( 6 ft )(1.5 kips + 7.5 kips ) = 0 2 M
max
= 30.0 kip ⋅ ft W
PROBLEM 7.66 Using the method of Sec. 7.6, solve Prob. 7.37.
SOLUTION (a) and (b)
FBD Beam: ΣFy = 0: Ay + ( 6 ft )( 2 kips/ft ) − 12 kips − 2 kips = 0
A y = 2 kips ΣM A = 0: M A + ( 3 ft )( 2 kips/ft )( 6 ft ) − (10.5 ft )(12 kips ) − (12 ft )( 2 kips ) = 0
M A = 114 kip ⋅ ft Shear Diag: dV VA = Ay = 2 kips. Then V is linear = 2 kips/ft to C, where dx VC = 2 kips + ( 6 ft )( 2 kips/ft ) = 14 kips. V is constant at 14 kips to D, then jumps down 12 kips to 2 kips and is constant to B V
Moment Diag:
max
= 14.00 kips W
M A = −114 kip ⋅ ft.
dM increasing with V and From A to C, M is parabolic dx M C = −114 kip ⋅ ft +
1 ( 2 kips + 14 kips )( 6 ft ) 2
M C = −66 kip ⋅ ft. Then M is piecewise linear. M D = −66 kip ⋅ ft + (14 kips )( 4.5 ft ) = −3 kip ⋅ ft M B = −3 kip ⋅ ft + ( 2 kips )(1.5 ft ) = 0 M
max
= 114.0 kip ⋅ ft W
PROBLEM 7.67 Using the method of Sec. 7.6, solve Prob. 7.38.
SOLUTION (a) and (b)
FBD Beam: kips ΣM B = 0: ( 3 ft ) 2 ( 6 ft ) + ( 9 ft )(12 kips ) − (15 ft ) Ay = 0 ft A y = 9.6 kips
Shear Diag: V jumps to Ay = 9.6 kips at A, is constant to C, jumps down 12 kips to −2.4 kips at C, is constant to D, and then is linear dV = −2 kips/ft to B dx VB = −2.4 kips − ( 2 kips/ft )( 6 ft ) = −14.4 kips
V
max
= 14.40 kips W
Moment Diag: dM = 9.6 kips/ft dx
M is linear from A to C
M C = 9.6 kips ( 6 ft ) = 57.6 kip ⋅ ft, M is linear from C to D
dM dx
= −2.4 kips/ft
M D = 57.6 kip ⋅ ft − 2.4 kips ( 3 ft ) M D = 50.4 kip ⋅ ft.
dM M is parabolic decreasing with V to B. dx M B = 50.4 kip ⋅ ft −
1 ( 2.4 kips + 14.4 kips )( 6 ft ) = 0 2
=0 M
max
= 57.6 kip ⋅ ft W
PROBLEM 7.68 Using the method of Sec. 7.6, solve Prob. 7.39.
SOLUTION (a) and (b)
FBD Beam: By symmetry:
1 ( 5 m )( 4 kN/m ) + 8 kN 2 or A y = B = 18 kN Ay = B =
Shear Diag: V jumps to 18 kN at A, and is constant to C, then drops 8 kN to 10 kN. dV = −4 kN/m , reaching −10 kN at After C, V is linear dx D VD = 10 kN − ( 4 kN/m )( 5 m ) passing through zero at the beam center. At D, V drops 8 kN to −18 kN and is then constant to B V
max
= 18.00 kN W
Moment Diag: dM M A = 0. Then M is linear = 18 kN/m to C dx M C = (18 kN )( 2 m ) = 36 kN ⋅ m, M is parabolic to D dM decreases with V to zero at center dx M center = 36 kN ⋅ m +
1 (10 kN )( 2.5 m ) = 48.5 kN ⋅ m = M max 2 M
Complete by invoking symmetry.
max
= 48.5 kN ⋅ m W
PROBLEM 7.69 Using the method of Sec. 7.6, solve Prob. 7.40.
SOLUTION (a) and (b)
FBD Beam: ΣM F = 0: (1 m )( 22 kN ) + (1.5 m )( 4 kN/m )( 3 m ) − ( 4 m ) Dy + ( 6 m )( 2 kN/m )( 2 m ) = 0
D y = 16 kN ΣFy = 0: 16 kN + 22 kN − Fy − ( 2 kN/m )( 2 m ) − ( 4 kN/m )( 3 m ) = 0
Fy = 22 kN Shear Diag: dV VA = 0, then V is linear = −2 kN/m to C; dx VC = −2 kN/m ( 4 m ) = −4 kN V is constant to D, jumps 16 kN to 12 kN and is constant to E. dV = −4 kN/m to F. Then V is linear dx VF = 12 kN − ( 4 kN/m )( 3 m ) = 0. V jumps to −22 kN at F, is constant to B, and returns to zero. V
Moment Diag: dM M A = 0, M is parabolic decreases with V to C. dx 1 M C = − ( 4 kN )( 2 m ) = −4 kN ⋅ m. 2
max
= 22.0 kN W
PROBLEM 7.69 CONTINUED dM Then M is linear = −4 kN to D. dx M D = −4 kN ⋅ m − ( 4 kN )(1 m ) = −8 kN ⋅ m dM = 12 kN , and From D to E, M is linear dx M E = −8 kN ⋅ m + (12 kN )(1m ) M E = 4 kN ⋅ m
dM M is parabolic decreasing with V to F. dx M F = 4 kN ⋅ m +
1 (12 kN )( 3 m ) = 22 kN ⋅ m. 2
dM Finally, M is linear = −22 kN , back to zero at B. dx M
max
= 22.0 kN ⋅ m W
PROBLEM 7.70 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) and (b)
FBD Beam: ΣM B = 0: (1.5 m )(16 kN ) + ( 3 m )( 8 kN ) + 6 kN ⋅ m − ( 4.5 m ) Ay = 0
A y = 12 kN Shear Diag: V is piecewise constant with discontinuities equal to the concentrated forces at A, C, D, B V max = 12.00 kN W
Moment Diag: dM =V After a jump of −6 kN ⋅ m at A, M is piecewise linear dx So
M C = −6 kN ⋅ m + (12 kN )(1.5 m ) = 12 kN ⋅ m M D = 12 kN ⋅ m + ( 4 kN )(1.5 m ) = 18 kN ⋅ m M B = 18 kN ⋅ m − (12 kN )(1.5 m ) = 0 M
max
= 18.00 kN ⋅ m W
PROBLEM 7.71 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a)
FBD Beam: ΣM A = 0: ( 8 m ) F + (11 m )( 2 kN ) + 10 kN ⋅ m − ( 6 m )( 8 kN ) − 12 kN ⋅ m − ( 2 m )( 6 kN ) = 0 F = 5 kN
ΣFy = 0: Ay − 6 kN − 8 kN + 5 kN + 2 kN = 0 A y = 7 kN Shear Diag: V is piecewise constant with discontinuities equal to the concentrated forces at A, C, E, F, G
Moment Diag: M is piecewise linear with a discontinuity equal to the couple at D. M C = ( 7 kN )( 2 m ) = 14 kN ⋅ m M D− = 14 kN ⋅ m + (1 kN )( 2 m ) = 16 kN ⋅ m
M D+ = 16 kN ⋅ m + 12 kN ⋅ m = 28 kN ⋅ m M E = 28 kN ⋅ m + (1 kN )( 2 m ) = 30 kN ⋅ m M F = 30 kN ⋅ m − ( 7 kN )( 2 m ) = 16 kN ⋅ m M G = 16 kN ⋅ m − ( 2 kN )( 3 m ) = 10 kN ⋅ m (b)
V M
max max
= 7.00 kN = 30.0 kN
PROBLEM 7.72 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum bending moment.
SOLUTION (a)
FBD Beam: ΣM B = 0: ( 3 ft )(1.2 kips/ft )( 6 ft ) − ( 8 ft ) Ay = 0
A y = 2.7 kips Shear Diag: V = Ay = 2.7 kips at A, is constant to C, then linear dV = −1.2 kips/ft to B. dx
VB = 2.7 kips − (1.2 kips/ft )( 6 ft )
VB = −4.5 kips V = 0 = 2.7 kips − (1.2 kips/ft ) x1 at
x1 = 2.25 ft
Moment Diag: dM M A = 0, M is linear = 2.7 kips to C. dx M C = ( 2.7 kips )( 2 ft ) = 5.4 kip ⋅ ft dM decreasing with V Then M is parabolic dx (b)
M max occurs where
dM =V =0 dx
M max = 5.4 kip ⋅ ft +
1 ( 2.7 kips ) x1; 2
x1 = 2.25 m
M max = 8.4375 kip ⋅ ft M max = 8.44 kip ⋅ ft, 2.25 m right of C Check:
M B = 8.4375 kip ⋅ ft −
1 ( 4.5 kips )( 3.75 ft ) = 0 2
PROBLEM 7.73 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum bending moment.
SOLUTION FBD Beam: (a)
ΣM B = 0: ( 6 ft )( 2 kips/ft )( 8 ft ) − (15 ft ) Ay = 0
A y = 6.4 kips Shear Diag: V = Ay = 6.4 kips at A, and is constant to C, then linear dV = −2 kips/ft to D, dx VD = 6.4 kips − ( 2 kips/ft )( 8 ft ) = −9.6 kips V = 0 = 6.4 kips − ( 2 kips/ft ) x1 at x1 = 3.2 ft
Moment Diag: dM M A = 0, then M is linear = 6.4 kips to M C = ( 6.4 kips )( 5 ft ) . dx dM M C = 32 kip ⋅ ft. M is then parabolic decreasing with V . dx (b)
M max occurs where M max = 32 kip ⋅ ft +
dM = V = 0. dx
1 ( 6.4 kips ) x1; 2
x1 = 3.2 ft
M max = 42.24 kip ⋅ ft M max = 42.2 kip ⋅ ft, 3.2 ft right of C M D = 42.24 kip ⋅ ft − M is linear from D to zero at B.
1 ( 9.6 kips )( 4.8 ft ) = 19.2 kip ⋅ ft 2
PROBLEM 7.74 For the beam shown, draw the shear and bending-moment diagrams and determine the maximum absolute value of the bending moment knowing that (a) P = 14 kN, (b) P = 20 kN.
SOLUTION (a)
FBD Beam: ΣFy = 0: Ay − (16 kN/m )( 2 m ) − 6 kN + P = 0 Ay = 38 kN − P (a)
A y = 24 kN
(b)
A y = 18 kN ΣM A = 0: ( 5 m ) P − ( 3.5 m )( 6 kN ) − 1 m (16 kN/m )( 2 m ) − M A = 0 M A = ( 5 m ) P − 53 kN ⋅ m
(b)
(a)
M A = 17 kN ⋅ m
(b)
M A = 47 kN ⋅ m
Shear Diags: dV VA = Ay . Then V is linear = −16 kN/m to C. dx VC = VA − (16 kN/m )( 2 m ) = VA − 32 kN (a)
VC = −8 kN
(b)
VC = −14 kN V = 0 = VA − (16 kN/m ) x1
(a)
x1 = 1.5 m
(b)
x1 = 1.125 m V is constant from C to D, decreases by 6 kN at D and is constant to B (at − P)
PROBLEM 7.74 CONTINUED Moment Diags: dM M A = M A reaction. Then M is parabolic decreasing with V . dx The maximum occurs where V = 0. M max = M A + (a)
M max = 17 kN ⋅ m +
1 VA x1. 2
1 ( 24 kN )(1.5 m ) = 35.0 kN ⋅ m 2 1.5 ft from A
(b)
M max = 47 kN ⋅ m +
1 (18 kN )(1.125 m ) = 57.125 kN ⋅ m 2
M max = 57.1 kN ⋅ m 1.125 ft from A M C = M max + (a) (b)
M C = 35 kN ⋅ m − M C = 57.125 kN ⋅ m −
1 VC ( 2 m − x1 ) 2
1 (8 kN )( 0.5 m ) = 33 kN ⋅ m 2
1 (14 kN )( 0.875 m ) = 51 kN ⋅ m 2
M is piecewise linear along C, D, B, with M B = 0 and M D = (1.5 m ) P
(a)
M D = 21 kN ⋅ m
(b)
M D = 30 kN ⋅ m
PROBLEM 7.75 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment knowing that (a) M = 0 , (b) M = 12 kN ⋅ m.
SOLUTION FBD Beam: ΣM A = 0: ( 4 m ) B − (1 m )( 20 kN/m )( 2 m ) − M = 0
B = 10 kN + (a)
M 4m
(a)
B = 10 kN
(b)
B = 13 kN ΣFy = 0: Ay − ( 20 kN/m )( 2 m ) + B = 0
Ay = 40 kN − B (a)
A y = 30 kN
(b)
A y = 27 kN
Shear Diags: (b)
dV = −20 kN/m to C. VA = Ay , then V is linear dx
VC = Ay − ( 20 kN/m )( 2 m ) = Ay − 40 kN (a)
VC = −10 kN
(b)
VC = −13 kN V = 0 = Ay − ( 20 kN/m ) x1 at x1 =
(a)
x1 = 1.5 m
(b)
x1 = 1.35 m V is constant from C to B.
Ay m 20 kN
PROBLEM 7.75 CONTINUED Moment Diags: dM M A = applied M . Then M is parabolic decreases with V dx
M is max where V = 0. M max = M + (a)
M
=
max
1 Ay x1. 2 1 ( 30 kN )(1.5 m ) = 22.5 kN ⋅ m 2 1.500 m from A
(b)
M max = 12 kN ⋅ m +
1 ( 27 kN )(1.35 m ) = 30.225 kN ⋅ m 2
M M C = M max −
max
= 30.2 kN 1.350 m from A
1 VC ( 2 m − x1 ) 2
(a)
M C = 20 kN ⋅ m
(b)
M C = 26 kN ⋅ m dM Finally, M is linear = VC to zero at B. dx
PROBLEM 7.76 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
SOLUTION FBD Beam: (a)
ΣM B = 0: ( 3 m )( 40 kN/m )( 6 m ) − ( 30 kN ⋅ m ) − ( 6 m ) Ay = 0
A y = 115 kN Shear Diag: dM = − 40 kN/m to B. VA = Ay = 115 kN, then V is linear dx
VB = 115 kN − ( 40 kN/m )( 6 m ) = −125 kN. V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 2.875 m Moment Diag: dM M A = 0. Then M is parabolic decreasing with V . Max M occurs dx where V = 0,
M max =
1 (115 kN/m )( 2.875 m ) = 165.3125 kN ⋅ m 2
1 (125 kN )( 6 m − x1 ) 2 1 = 165.3125 kN ⋅ m − (125 kN )( 6 − 2.875 ) m 2 = −30 kN ⋅ m as expected.
M B = M max −
(b)
M
max
= 165.3 kN ⋅ m ( 2.88 m from A )
PROBLEM 7.77 Solve Prob. 7.76 assuming that the 30 kN ⋅ m couple applied at B is counterclockwise
SOLUTION (a)
FBD Beam: ΣM B = 0: 30 kN ⋅ m + ( 3 m )( 40 kN/m )( 6 m ) − ( 6 m ) Ay = 0
A y = 125 kN Shear Diag: dV VA = Ay = 125 kN, V is linear = −40 kN/m to B. dx VB = 125 kN − ( 40 kN/m )( 6 m ) = −115 kN V = 0 = 115 kN − ( 40 kN/m ) x1 at x1 = 3.125 m
Moment Diag: dM M A = 0. Then M is parabolic decreases with V . Max M occurs dx where V = 0, M max =
1 (125 kN )( 3.125 m ) = 195.3125 kN ⋅ m 2
(b)
M
max
M B = 195.3125 kN ⋅ m −
= 195.3 kN ⋅ m ( 3.125 m from A ) 1 (115 kN )( 6 − 3.125) m 2
M B = 30 kN ⋅ m as expected.
PROBLEM 7.78 For beam AB, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
SOLUTION (a) Replacing the load at E with equivalent force-couple at C:
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) − ( 3 m )( 4 kN ) − (1.5 m )( 8 kN/m )( 3 m ) − 4 kN ⋅ m = 0
D = 10 kN ΣFy = 0: Ay + 10 kN − 2 kN − 4 kN − ( 8 kN/m )( 3 m ) = 0
A y = 20 kN Shear Diag: dV VA = Ay = 20 kN, then V is linear = −8 kN/m to C. dx VC = 20 kN − ( 8 kN/m )( 3 m ) = −4 kN V = 0 = 20 kN − ( 8 kN/m ) x1 at x1 = 2.5 m At C, V decreases by 4 kN to −8 kN. At D, V increases by 10 kN to 2 kN.
Moment Diag: dM M A = 0, then M is parabolic decreasing with V . Max M occurs dx where V = 0. M max =
1 ( 20 kN )( 2.5 m ) = 25 kN ⋅ m 2 (b)
M max = 25.0 kN ⋅ m, 2.50 m from A
PROBLEM 7.78 CONTINUED M C = 25 kN ⋅ m −
1 ( 4 kN )( 0.5 m ) = 24 kN ⋅ m. 2
At C, M decreases by 4 kN ⋅ m to 20 kN ⋅ m. From C to B, M is piecewise dM dM linear with = +2 kN to B. = −8 kN to D, then dx dx M D = 20 kN ⋅ m − ( 8 kN )( 3 m ) = −4 kN ⋅ m.
MB = 0
PROBLEM 7.79 Solve Prob. 7.78 assuming that the 4-kN force applied at E is directed upward.
SOLUTION (a) Replacing the load at E with equivalent force-couple at C.
ΣM A = 0: ( 6 m ) D − ( 8 m )( 2 kN ) + ( 3 m )( 4 kN ) − 4 kN ⋅ m − (1.5 m )( 8 kN/m )( 3 m ) = 0
D= ΣFy = 0: Ay +
22 kN 3
22 kN − ( 8 kN/m )( 3 m ) + 4 kN − 2 kN = 0 3
Ay =
44 kN 3
Shear Diag: VA = Ay =
44 dV kN, then V is linear = −8 kN/m to C. 3 dx VC =
44 28 kN − ( 8 kN/m )( 3 m ) = − kN 3 3
V =0=
44 11 kN − ( 8 kN/m ) x1 at x1 = m. 3 6
At C, V increases 4 kN to − At D, V increases
16 kN. 3
22 kN to 2 kN. 3
PROBLEM 7.79 CONTINUED Moment Diag: dM M A = 0. Then M is parabolic decreasing with V . Max M occurs dx where V = 0. M max =
1 44 11 121 kN m = kN ⋅ m 2 3 9 6
(b)
MC =
M max = 13.44 kN ⋅ m at 1.833 m from A
121 1 28 7 kN ⋅ m − kN m = 8 kN ⋅ m. 9 2 3 6
At C, M increases by 4 kN ⋅ m to 12 kN ⋅ m. Then M is linear 16 dM =− kN to D. 3 dx 16 kN ( 3 m ) = −4 kN ⋅ m. M is again linear M D = 12 kN ⋅ m − 3 dM = 2 kN to zero at B. dx
PROBLEM 7.80 For the beam and loading shown, (a) derive the equations of the shear and bending-moment curves, (b) draw the shear and bending-moment diagrams, (c) determine the magnitude and location of the maximum bending moment.
SOLUTION x Distributed load w = w0 1 − L ΣM A = 0:
1 total = w0 L 2
L 1 w0 L − LB = 0 3 2
B=
w0 L 6
1 wL w0 L + 0 = 0 2 6
Ay =
w0 L 3
ΣFy = 0: Ay −
Shear: VA = Ay =
w0 L , 3
x dV x = − w → V = VA − ∫ w0 1 − dx 0 L dx
Then
1 x 1 x 2 1 w0 2 w L V = 0 − w0 x + x = w0 L − + 2 L 3 3 L 2 L Note: At x = L, V = −
w0 L ; 6
2
x x x 2 V = 0 at − 2 + = 0 → = 1− L L L 3
1 3
Moment: M A = 0, Then
x x/ L x x dM = V → M = ∫0 Vdx = L ∫0 V d dx L L
x / L 1 x 1 x 2 x − + d M = w0 L ∫ 0 3 L 2 L L 2
1 x 1 x 2 1 x 3 M = w0 L2 − + 6 L 3 L 2 L
PROBLEM 7.80 CONTINUED x M max at =1− L
1 2 = 0.06415w0 L 3
1 x 1 x 2 V = w0 L − + 3 L 2 L
(a)
1 x 1 x 2 1 x 3 M = w0 L − + 6 L 3 L 2 L 2
(c)
M max = 0.0642 w0 L2 at x = 0.423L
PROBLEM 7.81 For the beam and loading shown, (a) derive the equations of the shear and bending-moment curves, (b) draw the shear and bending-moment diagrams, (c) determine the magnitude and location of the maximum bending moment.
SOLUTION x w = w0 4 − 1 L
Distributed load
dV = −w, and V ( 0 ) = 0, so dx
Shear: V =
x
∫0
− wdx = −
x/L
∫0
x Lwd L
2 x x/L x x x V =∫ wo L 1 − 4 d = w L − 2 0 L L 0 L L
Notes: At x = L, V = −w0 L x x = 2 L L
And V = 0 at
2
or
x 1 = L 2
M ( 0 ) = 0,
dM =V dx
1 x Also V is max where w = 0 = 4 L Vmax =
1 w0 L 8
Moment:
x x V d L L 2 x / L x − 2 x d x M = w0 L2 ∫ 0 L L L
M =
x
x/ L
∫0 vdx = L∫0
(a)
2 x x V = w0 L − 2 L L
1 x 2 2 x 3 M = w0 L2 − 3 L 2 L
PROBLEM 7.81 CONTINUED M max =
1 L w0 L2 at x = 24 2
1 M min = − w0 L2 at x = L 6 M max =
w0 L2 L at x = 24 2
(c)
M
max
= −M min =
w0 L2 at B 6
PROBLEM 7.82 For the beam shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum bending moment. (Hint: Derive the equations of the shear and bending-moment curves for portion CD of the beam.)
SOLUTION (a)
FBD Beam: ΣM B = 0:
( 3a )
1 w0 ( 3a ) − 5aAy = 0 A y = 0.9w0a 2 1 ΣFy = 0: 0.9w0a − w0 ( 3a ) + B = 0 2
B = 0.6w0a
Shear Diag: V = Ay = 0.9w0a from A to C and V = B = − 0.6w0a from B to D. x1 . If x1 is measured right to left, 3a dV x w dM = + w and = − V . So, from D, V = − 0.6w0a + ∫0 1 0 x1dx1, 3a dx1 dx1
Then from D to C, w = w0
2 1 x1 V = w0a − 0.6 + 6 a 2
x Note: V = 0 at 1 = 3.6, x1 = a
3.6 a
Moment Diag: dM M = 0 at A, increasing linearly = 0.9w0a to M C = 0.9w0a 2. dx1 dM = 0.6w0a to Similarly M = 0 at B increasing linearly dx M D = 0.6w0a 2. Between C and D 2 x1 1 x1 M = 0.6w0a + w0a ∫ 0.6 − dx1, 0 6 a 2
3 x 1 x M = w0a 2 0.6 + 0.6 1 − 1 a 18 a
PROBLEM 7.82 CONTINUED (b)
At
x1 = a
3.6, M = M max = 1.359w0a 2 x1 = 1.897a left of D
PROBLEM 7.83 Beam AB, which lies on the ground, supports the parabolic load shown. Assuming the upward reaction of the ground to be uniformly distributed, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.
SOLUTION ΣFy = 0: wg L −
(a) wg L =
Define ξ =
or
L 4w0
∫0
L2
( Lx − x ) dx = 0 2
4w0 1 2 1 3 2 LL − L = w0 L 3 3 L2 2
2w0 3
wg =
x x 2 2 − − w0 L L 3
x dx so dξ = → net load w = 4w0 L L
1 w = 4w0 − + ξ − ξ 2 6 V = V(0) −
ξ
1
1
1
∫0 4w0 L − 6 + ξ − ξ
2
dξ =
1
0 + 4w0 L ξ + ξ 2 − ξ 3 6 2 3
(
2 w0 L ξ − 3ξ 2 + 2ξ 3 3
V = M = M0 + =
(b)
x
2
ξ
∫0 Vdx = 0 + 3 w0 L ∫0 2
(ξ − 3ξ
2
)
)
+ 2ξ 3 d ξ
(
2 1 1 1 w0 L2 ξ 2 − ξ 3 + ξ 4 = w0 L2 ξ 2 − 2ξ 3 + ξ 4 3 2 3 2
Max M occurs where V = 0 → 1 − 3ξ + 2ξ 2 = 0 → ξ =
)
1 2
1 1 1 w L2 1 2 M ξ = = w0 L2 − + = 0 2 3 48 4 8 16
M max =
w0 L2 at center of beam 48
PROBLEM 7.84 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +325 lb ft at D and +800 lb ft at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.
SOLUTION FBD ACD:
(a)
ΣM D − = 0: 0.325 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0 C y = 3.325 kips
FBD EB:
ΣM E = 0: (1 ft ) B − 0.8 kip ⋅ ft = 0
B = 0.8 kip
FBD Beam: ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft ) − (1 ft )( 3.325 kips ) − (1 ft ) Q + 2 ft ( 0.8 kips ) = 0 Q = 1.275 kips ΣFy = 0: 3.325 kips + 0.8 kips − 1.275 kips − ( 2 kips/ft )(1ft ) − P = 0
(a)
P = 0.85 kip P = 850 lb Q = 1.275 kips
(b) Shear Diag: dV = −2 kips/ft from 0 at A to V is linear dx − ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise constant with discontinuities equal to forces at C, D, E, B
Moment Diag: dM decreasing with V from 0 at A to M is parabolic dx 1 − ( 2 kips )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with 2
PROBLEM 7.84 CONTINUED M D = −1 kip ⋅ ft + (1.325 kips )(1 ft ) = 0.325 kip ⋅ ft M E = 0.325 kip ⋅ ft + ( 0.475 kips )(1 ft ) = 0.800 kip ⋅ ft M B = 0.8 kip ⋅ ft − ( 0.8 kip )(1 ft ) = 0
PROBLEM 7.85 Solve Prob. 7.84 assuming that the bending moment was found to be +260 lb ft at D and +860 lb ft at E.
SOLUTION FBD ACD:
(a)
ΣM D = 0: 0.26 kip ⋅ ft − (1 ft ) C y + (1.5 ft )( 2 kips/ft )(1 ft ) = 0 C y = 3.26 kips
FBD DB: ΣM E = 0: (1 ft ) B − 0.86 kip ⋅ ft
B = 0.86 kip
FBD Beam: ΣM D = 0: (1.5 ft )( 2 kips/ft )(1 ft ) − (1 ft )( 3.26 kips ) + (1 ft ) Q + ( 2 ft )( 0.86 kips ) = 0 Q = 1.460 kips
Q = 1.460 kips
ΣFy = 0: 3.26 kips + 0.86 kips − 1.460 kips − P − ( 2 kips/ft )(1 ft ) = 0 P = 0.66 kips P = 660 lb
(b) Shear Diag: dV V is linear = −2 kips/ft from 0 at A to dx − ( 2 kips/ft )(1 ft ) = −2 kips at C. Then V is piecewise
constant with discontinuities equal to forces at C, D, E, B. Moment Diag: dM decreasing with V from 0 at A to M is parabolic dx 1 − ( 2 kips/ft )(1 ft ) = −1 kip ⋅ ft at C. Then M is piecewise linear with 2
PROBLEM 7.85 CONTINUED
M 0 = 0.26 kip ⋅ ft M E = 0.86 kip ⋅ ft, M B = 0
PROBLEM 7.86 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +7 kN · m at D and +5 kN · m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.
SOLUTION
FBD AD:
ΣM D = 0: 7 kN ⋅ m + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0 2 Ay − P = 8.2 kN
(1)
ΣM E = 0: ( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m ) − 5 kN ⋅ m = 0
FBD EB: 2B − Q = 6.2 kN
(2)
( 6 m ) B − (1 m ) P − ( 5 m ) Q − ( 3 m )( 0.6 kN/m )( 6 m ) = 0
ΣM A = 0:
6 B − P − 5Q = 10.8 kN
(3)
ΣM B = 0: (1 m ) Q + ( 5 m ) P + ( 3 m )( 0.6 kN/m )( 6 m ) − (6 m) A = 0 6 A − Q − 5P = 10.8 kN
P = 6.60 kN , Q = 600 N
Solving (1)–(4): A y = 7.4 kN ,
(4)
B = 3.4 kN
(b) Shear Diag: dV = − 0.6 kN/m throughout, and dx discontinuities equal to forces at A, C, F, B.
V is piecewise linear with
Note V = 0 = 0.2 kN − ( 0.6 kN/m ) x at x =
1 m 3
PROBLEM 7.86 CONTINUED Moment Diag: dM decreasing with V with “breaks” in M is piecewise parabolic dx slope at C and F. MC = M max = 7.1 kN ⋅ m +
1 ( 7.4 + 6.8) kN (1 m ) = 7.1 kN ⋅ m 2
1 1 ( 0.2 kN ) m = 7.133 kN ⋅ m 2 3
1 2 M F = 7.133 kN ⋅ m − ( 2.2 kN ) 3 m = 3.1 kN ⋅ m 2 3
PROBLEM 7.87 Solve Prob. 7.86 assuming that the bending moment was found to be +3.6 kN · m at D and +4.14 kN · m at E.
SOLUTION
FBD AD:
ΣM D = 0: 3.6 kN ⋅ m + (1 m ) P + (1 m )( 0.6 kN/m )( 2 m )
(a)
− ( 2 m ) Ay = 0 2 Ay − P = 4.8 kN
FBD EB:
ΣM E = 0:
(1)
( 2 m ) B − (1 m ) Q − (1 m )( 0.6 kN/m )( 2 m ) − 4.14 kN ⋅ m = 0
2B − Q = 5.34 kN
(2)
ΣM A = 0: ( 6 m ) B − ( 5 m ) Q − (1 m ) P − ( 3 m )( 0.6 kN/m )( 6 m ) = 0 6 B − P − 5Q = 10.8 kN
(3)
By symmetry: 6 A − Q − 5P = 10.8 kN
Solving (1)–(4)
(4)
P = 660 N , Q = 2.28 kN
Ay = 2.73 kN , B = 3.81 kN
(b) Shear Diag: dV = − 0.6 kN/m throughout, and V is piecewise linear with dx discontinuities equal to forces at A, C, F, B.
Note that V = 0 = 1.47 kN − ( 0.6 kN/m ) x at x = 2.45 m Moment Diag: dM decreasing with V , with “breaks” in M is piecewise parabolic dx slope at C and F.
PROBLEM 7.87 CONTINUED MC =
1 ( 2.73 + 2.13) kN (1 m ) = 2.43 kN ⋅ m 2
M max = 2.43 kN ⋅ m + M F = 4.231 kN ⋅ m −
1 (1.47 kN )( 2.45 m ) = 4.231 kN ⋅ m 2 1 ( 0.93 kN )(1.55 m ) = 3.51 kN ⋅ m 2
PROBLEM 7.88 Two loads are suspended as shown from cable ABCD. Knowing that dC = 1.5 ft, determine (a) the distance dB, (b) the components of the reaction at A, (c) the maximum tension in the cable.
SOLUTION ΣM A = 0: (10 ft ) Dy − 8 ft ( 450 lb ) − 4 ft ( 600 lb ) = 0
FBD cable:
Dy = 600 1b
ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0 Ay = 450 lb
ΣFx = 0: Ax − Dx = 0
FBD pt D:
(1)
600 lb D T = x = CD : Dx = 800 lb 3 4 5
= Ax
So TCD = 1000 lb
And
800 lb 450 lb = dB 4 ft
FBD pt A:
d B = 2.25 ft
(a) (b)
A x = 800 lb A y = 450 lb
TAB =
So
(800 lb )2 + ( 450 lb )2 (c)
= 918 lb Tmax = TCD = 1000 lb
Note: TCD is Tmax as cable slope is largest in section CD.
PROBLEM 7.89 Two loads are suspended as shown from cable ABCD. Knowing that the maximum tension in the cable is 720 lb, determine (a) the distance dB, (b) the distance dC.
SOLUTION ΣM A = 0: (10 ft ) Dy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
FBD cable:
D y = 600 1b ΣFy = 0: Ay + 600 lb − 600 lb − 450 lb = 0 A y = 450 lb
FBD pt D:
ΣFx = 0: Ax − Bx = 0
Since Ax = Bx ; And Dy > Ay , Tension TCD > TAB So
TCD = Tmax = 720 lb Dx =
( 720 lb )2 − ( 600 lb )2
dC 2ft = 600 lb 398lb
FBD pt. A:
dB 4ft = 450 lb 398lb
= 398 lb = Ax
dC = 3.015 ft
(a)
d B = 4.52 ft
(b)
dC = 3.02 ft
PROBLEM 7.90 Knowing that dC = 4 m, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION (a) FBD cable: ΣM E = 0:
( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN ) − ( 3 m ) Ax − (16 m ) Ay = 0
3 Ax + 16 Ay = 25.6 kN
(1)
FBD ABC: ΣM C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax − ( 8 m ) Ay = 0 Ax − 8 Ay = −4.8 kN
(2)
Ax = 3.2 kN
Solving (1) and (2)
Ay = 1 kN
So A = 3.35 kN (b)
cable:
17.35°
ΣFx = 0: − Ax + Ex = 0 Ex = Ax = 3.2 kN
ΣFy = 0: Ay − (1.2 + 0.8 + 1.2 ) kN + E y = 0 E y = 3.2 kN − Ay = ( 3.2 − 1) kN = 2.2 kN
So E = 3.88 kN
34.5°
PROBLEM 7.91 Knowing that dC = 2.25 m, determine (a) the reaction at A, (b) the reaction at E.
SOLUTION FBD Cable:
(a)
ΣM E = 0: ( 4 m )(1.2 kN ) + ( 8 m )( 0.8 kN ) + (12 m )(1.2 kN ) − (3 m) Ax − (16 m ) Ay = 0 3 Ax + 16 Ay = 25.6 kN
(1)
ΣM C = 0: ( 4 m )(1.2 kN ) − ( 0.75 m ) Ax − ( 8 m ) Ay = 0
FBD ABC:
0.75 Ax + 8 Ay = 4.8 kN
Solving (1) and (2)
Ax =
(2)
32 kN, 3
Ay = − 0.4 kN
So
A = 10.67 kN
2.15°
Note: this implies d B < 3 m (in fact d B = 2.85 m) (b) FBD cable:
ΣFx = 0: −
32 kN + Ex = 0 3
Ex =
32 kN 3
ΣFy = 0: − 0.4 kN − 1.2 kN − 0.8 kN − 1.2 kN + E y = 0 E y = 3.6 kN
E = 11.26 kN
18.65°
PROBLEM 7.92 Cable ABCDE supports three loads as shown. Knowing that dC = 3.6 ft, determine (a) the reaction at E, (b) the distances dB and dD.
SOLUTION FBD Cable: ΣM A = 0: ( 2.4 ft ) Ex + ( 8 ft ) E y − ( 2 ft )( 360 )
(a)
− ( 4 ft )( 720 lb ) − ( 6 ft )( 240 lb ) = 0 0.3Ex + E y = 630 lb
FBD CDE:
(1)
ΣM C = 0: − (1.2 ft ) Ex + ( 4 ft ) E y − ( 2 ft )( 240 lb ) = 0 − 0.3Ex + E y = +120 lb
Solving (1) and (2)
Ex = 850 lb
(a) (b) cable:
ΣFx = 0: − Ax + Ex = 0
(2) E y = 375 lb
E = 929 lb
23.8°
Ax = Ex = 850 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + 375 lb = 0
Point A:
Ay = 945 lb dB 945 lb = 2 ft 850 lb
d B = 2.22 ft
PROBLEM 7.92 CONTINUED ΣM D = 0: ( 2 ft )( 375 lb ) − ( d D − 2.4 ft )( 850 lb ) = 0
Segment DE:
d D = 3.28 ft
PROBLEM 7.93 Cable ABCDE supports three loads as shown. Determine (a) the distance dC for which portion CD of the cable is horizontal, (b) the corresponding reactions at the supports.
SOLUTION Segment DE:
ΣFy = 0: E y − 240 lb = 0
E y = 240 lb
ΣM A = ( 2.4 ft ) Ex + ( 8 ft )( 240 lb ) − ( 6 ft )( 240 lb )
FBD Cable:
− ( 4 ft )( 720 lb ) − ( 2 ft )( 360 lb ) = 0 E x = 1300 lb
So From Segment DE: ΣM D = 0: ( 2 ft ) E y − ( dC − 2.4 ft ) Ex = 0 dC = 2.4 ft +
Ey Ex
( 2 ft ) = ( 2.4 ft ) +
240 lb ( 2 ft ) = 2.7692 ft 1300 lb
(a)
dC = 2.77 ft
From FBD Cable: ΣFx = 0: − Ax + Ex = 0
A x = 1300 lb
ΣFy = 0: Ay − 360 lb − 720 lb − 240 lb + E y = 0
A y = 1080 lb
(b)
A = 1.690 kips
39.7°
E = 1.322 kips
10.46°
PROBLEM 7.94 An oil pipeline is supported at 6-m intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents, the tension in each hanger is 4 kN. Knowing that dC = 12 m, determine (a) the maximum tension in the cable, (b) the distance d D .
SOLUTION FBD Cable:
Note: A y and Fy shown are forces on cable, assuming the 4 kN loads at A and F act on supports. ΣM F = 0: ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN ) − ( 30 m ) Ay − ( 5 m ) Ax = 0 Ax + 6 Ay = 48 kN
FBD ABC:
(1)
ΣM C = 0: ( 6 m )( 4 kN ) + ( 7 m ) Ax − (12 m ) Ay = 0 7 Ax − 12 Ay = −24 kN
Solving (1) and (2)
A x = 8 kN
(2) 20 kN 3
Ay =
From FBD Cable: ΣFx = 0: − Ax + Fx = 0
Fx = Ax = 8 kN
ΣFy = 0: Ay − 4 ( 4 kN ) + Fy = 0 20 28 Fy = 16 kN − Ay = 16 − kN > Ay kN = 3 3
So
TEF > TAB
Tmax = TEF =
Fx2 + Fy2
FBD DEF: (a)
Tmax =
(18 kN )
2
2
28 + kN = 12.29 kN 3
28 ΣM D = 0: (12 m ) kN − d D ( 8 kN ) − ( 6 m )( 4 kN ) = 0 3
(b)
d D = 11.00 m
PROBLEM 7.95 Solve Prob. 7.94 assuming that dC = 9 m.
SOLUTION FBD Cable:
Note: 4 kN loads at A and F act directly on supports, not on cable. ΣM A = 0: ( 30 m ) Fy − ( 5 m ) Fx − ( 6 m ) 1( 4 kN ) + 2 ( 4 kN ) + 3 ( 4 kN ) + 4 ( 4 kN ) = 0
Fx − 6 Fy = −48 kN
(1)
ΣM C = 0: (18 ) Fy − ( 9 m ) Fx − (12 m )( 4 kN ) − ( 6 m )( 4 kN ) = 0
FBD CDEF:
Fx − 2Fy = −8 kN
(2)
Fx = 12 kN
Solving (1) and (2) TEF =
Fy = 10 kN
(10 kN )2 + (12 kN )2
= 15.62 kN
Since slope EF > slope AB this is Tmax Tmax = 15.62 kN
(a) Also could note from FBD cable ΣFy = 0: Ay + Fy − 4 ( 4 kN ) = 0 Ay = 16 kN − 12 kN = 4 kN
Thus FBD DEF:
(b)
Ay < Fy
and
TAB < TEF
ΣM D = 0: (12 m )(10 kN ) − d D (12 kN ) − ( 6 m )( 4 kN ) = 0 d D = 8.00 m
PROBLEM 7.96 Cable ABC supports two boxes as shown. Knowing that b = 3.6 m, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a.
SOLUTION
(
)
W = ( 8 kg ) 9.81 m/s 2 = 78.48 N
FBD BC:
ΣM A = 0: ( 3.6 m ) P − ( 2.4 m )
3W − aW = 0 2
a P = W 1 + 3.6 m ΣFx = 0: −T1x + P = 0 ΣFy = 0: T1y − W − T1 y
But
T1x
=
3 W =0 2
2.8 m a P=
So
T1x = P 5W 2
T1 y = 5W 2.8 m = a 2P
so 5Wa 5.6 m
(2)
a = 1.6258 m,
Solving (1) and (2):
(1)
P = 1.4516W
(a)
P = 1.4516 ( 78.48 ) = 113.9 N
(b)
a = 1.626 m
PROBLEM 7.97 Cable ABC supports two boxes as shown. Determine the distances a and b when a horizontal force P of magnitude 100 N is applied at C.
SOLUTION FBD pt C:
Segment BC: 2.4 m − a 0.8 m = 100 N 117.72 N
a = 1.7204 m a = 1.720 m
ΣM A = 0: b (100 N ) − ( 2.4 m )(117.72 N ) 2 − (1.7204 m ) 117.72 N = 0 3
b = 4.1754 m b = 4.18 m
PROBLEM 7.98 Knowing that WB = 150 lb and WC = 50 lb, determine the magnitude of the force P required to maintain equilibrium.
SOLUTION FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0 3Dx = 4 Dy
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 50 lb ) = 0
FBD BCD:
2D y − Dx = 60 lb
Solving (1) and (2)
FBD Cable:
D x = 120 lb
(2) D y = 90 lb
ΣM A = 0: ( 42 ft )( 90 lb ) − ( 30 ft )( 50 lb ) − (12 ft )(150 lb ) − (15 ft ) P = 0
P = 32.0 lb
PROBLEM 7.99 Knowing that WB = 40 lb and WC = 22 lb, determine the magnitude of the force P required to maintain equilibrium.
SOLUTION FBD CD:
ΣM C = 0: (12 ft ) Dy − ( 9 ft ) Dx = 0 4D y = 3Dx
FBD BCD:
(1)
ΣM B = 0: ( 30 ft ) D y − (15 ft ) Dx − (18 ft )( 22 lb ) = 0 10 Dy − 5Dx = 132 lb
Solving (1) and (2)
D x = 52.8 lb
(2) D y = 39.6 lb
FBD Whole: ΣM A = 0: ( 42 ft )( 39.6 lb ) − ( 30 ft )( 22 lb ) − (12 ft )( 40 lb ) − (15 ft ) P = 0
P = 34.9 lb
PROBLEM 7.100 If a = 4.5 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.
SOLUTION FBD pt C:
By symmetry:
TBC = TCD = T
1 ΣFy = 0: 2 T − 180 kN = 0 5
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
Segment DE: ΣM E = 0: ( 7.5 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 108.0 kN
Segment AB:
ΣM A = 0: ( 4.5 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0
Q = 45.0 kN
PROBLEM 7.101 If a = 6 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.
SOLUTION FBD pt C:
TBC = TCD = T
By symmetry:
1 ΣFy = 0: 2 T − 180 kN = 0 5
Tx = 180 kN
T = 90 5 kN
Ty = 90 kN
FBD DE:
ΣM E = 0: ( 9 m )( P − 180 kN ) + ( 6 m )( 90 kN ) = 0
P = 120.0 kN
FBD AB:
ΣM A = 0: ( 6 m )(180 kN ) − ( 6 m )( Q + 90 kN ) = 0 Q = 90.0 kN
PROBLEM 7.102 A transmission cable having a mass per unit length of 1 kg/m is strung between two insulators at the same elevation that are 60 m apart. Knowing that the sag of the cable is 1.2 m, determine (a) the maximum tension in the cable, (b) the length of the cable.
SOLUTION (a) Since h = 1.2 m L = 30 m we can approximate the load as evenly distributed in the horizontal direction.
(
)
w = 1 kg/m 9.81 m/s 2 = 9.81 N/m. w = ( 60 m )( 9.81 N/m ) w = 588.6 N Also we can assume that the weight of half the cable acts at the FBD half-cable:
1 chord point. 4
ΣM B = 0: (15 m )( 294.3 N ) − (1.2 m ) Tmin = 0 Tmin = 3678.75 N = Tmax ΣFy = 0: Tmax Tmax
y
y
x
− 294.3 N = 0
= 294.3 N
Tmax = 3690.5 N Tmax = 3.69 kN
(b)
2 4 2 y 2 y sB = xB 1 + B − B + " 3 xB 5 xB 2 4 2 1.2 2 1.2 = ( 30 m ) 1 + − + " = 30.048 m 3 30 5 30
so
s = 2sB = 60.096 m s = 60.1 m
Note: The more accurate methods of section 7.10, which assume the load is evenly distributed along the length instead of horizontally, yield Tmax = 3690.5 N and s = 60.06 m. Answers agree to 3 digits at least.
PROBLEM 7.103 Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Knowing that the mass per unit length of the cables is 0.4 kg/m, determine (a) the required sag h, (b) the maximum tension in each cable.
SOLUTION Half-cable FBDs:
T1x = T2 x to create zero horizontal force on tower → thus T01 = T02 ΣM B = 0: (15 m ) w ( 30 m ) − h1T0 = 0
FBD I:
( 450 m ) w = 2
h1 FBD II:
T0
ΣM B = 0: ( 2 m ) T0 − (10 m ) w ( 20 m ) = 0
T0 = (100 m ) w
( 450 m ) w = 4.50 m = 2
h1
(a) ΣFx = 0: T1x − T0 = 0 T1x = (100 m ) w
FBD I:
ΣFy = 0: T1y − ( 30 m ) w = 0 T1y = ( 30 m ) w T1 =
(100 m )2 + ( 30 m )2 w
(
= (104.4 m )( 0.4 kg/m ) 9.81 m/s 2 = 409.7 N
)
(100 m ) w
PROBLEM 7.103 CONTINUED ΣFy = 0: T2y − ( 20 m ) w = 0
FBD II:
T2y = ( 20 m ) w T2 x = T1x = (100 m ) w T2 =
(100 m )2 + ( 20 m )2 w = 400.17 N (b)
T1 = 410 N T2 = 400 N
*
Since h L it is reasonable to approximate the cable weight as being distributed uniformly along the horizontal. The methods of section 7.10 are more accurate for cables sagging under their own weight.
PROBLEM 7.104 The center span of the George Washington Bridge, as originally constructed, consisted of a uniform roadway suspended from four cables. The uniform load supported by each cable was w = 9.75 kips/ft along the horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the original configuration (a) the maximum tension in each cable, (b) the length of each cable.
SOLUTION FBD half-span:
W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips ΣM B = 0: ( 875 ft )(17, 065 kips ) − ( 316 ft ) T0 = 0 T0 = 47, 246 kips Tmax = T02 + W 2 =
( 47, 246 kips )2 + (17, 063 kips )2 (a)
Tmax = 50, 200 kips
2 4 2 y 2 y s = x 1 + − + " 3 x 5 x 2 4 2 316 ft 2 316 ft − sB = (1750 ft ) 1 + + " 3 1750 ft 5 1750 ft
= 1787.3 ft
(b) *
To get 3-digit accuracy, only two terms are needed.
l = 2sB = 3575 ft
PROBLEM 7.105 Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length of each cable.
SOLUTION FBD half-span:
(a)
2075 ft ΣM B = 0: ( 23032.5 kips ) − ( 464 ft ) T0 = 0 2 T0 = 47, 246 kips
Tmax = T02 + W 2 = (b)
( 47, 246 kips )2 + ( 23,033 kips )2
2 2 y 2 s = x 1 + − 3 5 x
= 56, 400 kips
4 y + " x
2 4 2 464 ft 2 464 ft sB = ( 2075 ft ) 1 + − + " 3 2075 ft 5 2075 ft
sB = 2142 ft
l = 2s B
l = 4284 ft
PROBLEM 7.106 To mark the positions of the rails on the posts of a fence, a homeowner ties a cord to the post at A, passes the cord over a short piece of pipe attached to the post at B, and ties the free end of the cord to a bucket filled with bricks having a total mass of 20 kg. Knowing that the mass per unit length of the rope is 0.02 kg/m and assuming that A and B are at the same elevation, determine (a) the sag h, (b) the slope of the cable at B. Neglect the effect of friction.
SOLUTION FBD pulley:
ΣM P = 0: (Tmax − WB ) r = 0 Tmax = WB = 196.2 N FBD half-span:*
2 T0 = Tmax −W2 =
(196.2 N )2 − ( 4.91 N )2
= 196.139 N
25 m ΣM B = 0: ( 4.905 N ) − h (196.139 N ) = 0 2 h = 0.3126 m = 313 mm
(a) (b) *See note Prob. 7.103
θ B = sin −1
W 4.905 N = sin −1 = 1.433° Tmax 196.2 N
PROBLEM 7.107 A small ship is tied to a pier with a 5-m length of rope as shown. Knowing that the current exerts on the hull of the ship a 300-N force directed from the bow to the stern and that the mass per unit length of the rope is 2.2 kg/m, determine (a) the maximum tension in the rope, (b) the sag h. [Hint: Use only the first two terms of Eq. (7.10).]
SOLUTION (a) FBD ship:
ΣFx = 0: T0 − 300 N = 0
T0 = 300 N
FBD half-span:*
Tmax = T02 + W 2 = (b)
ΣM A = 0: hTx −
L W =0 4
2 2 4 s = x 1 + + " 3 x
L ( 2.5 m ) = 2
but
= ( 54 N ) = 305 N 2
LW 4Tx
yA = h =
LW 4Tx
so
yA W = 2Tx xA
2 2 53.955 N 1 + − " → L = 4.9732 m 3 600 N
So h = *See note Prob. 7.103
h=
( 300 N )2
LW = 0.2236 m 4Tx
h = 224 mm
PROBLEM 7.108 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The design of the bridge allowed for the effect of extreme temperature changes which cause the sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft, determine the change in length of the cables due to extreme temperature changes.
SOLUTION 2 2 y 2 s = x 1 + − 3 x 5
Knowing
l = 2sTOT
4 y + " x
2 2 2 h 2 h = L 1 + − + " 3 L/2 5 L/2
Winter: 2 4 2 386 ft 2 386 ft lw = ( 4260 ft ) 1 + " − + = 4351.43 ft 3 2130 ft 5 2130 ft
Summer: 2 4 2 394 ft 2 394 ft ls = ( 4260 ft ) 1 + − + " = 4355.18 ft 3 2130 ft 5 2130 ft
∆l = ls − lw = 3.75 ft
PROBLEM 7.109 A cable of length L + ∆ is suspended between two points which are at the same elevation and a distance L apart. (a) Assuming that ∆ is small compared to L and that the cable is parabolic, determine the approximate sag in terms of L and ∆ . (b) If L = 30 m and ∆ = 1.2 m, determine the approximate sag. [Hint: Use only the first two terms of Eq. (7.10).]
SOLUTION (a)
2 2 y s = x 1 + − " 3 x
L + ∆ = 2sTOT
2 2 h = L 1 + − " 3 L/2 2
2
∆ 2 2h 8 h = = →h= L 3 L 3 L (b)
For
L = 30 m,
∆ = 1.2 m
3 L∆ 8
h = 3.67 m
PROBLEM 7.110 Each cable of the side spans of the Golden Gate Bridge supports a load w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the maximum vertical distance h from each cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum tension in each cable, (b) the slope at B.
SOLUTION FBD AB:
ΣM A = 0: (1100 ft ) TBy − ( 496 ft ) TBx − ( 550 ft )W = 0 11TBy − 4.96TBx = 5.5W
(1)
FBD CB:
ΣM C = 0: ( 550 ft ) TBy − ( 278 ft ) TBx − ( 275 ft )
W =0 2
11TBy − 5.56TBx = 2.75W Solving (1) and (2)
TBy = 28,798 kips
Solving (1) and (2)
TBx = 51, 425 kips
Tmax = TB = TB2x + TB2y
So that
(2)
tan θ B =
TBy TBx
(a)
Tmax = 58,900 kips
(b)
θ B = 29.2°
PROBLEM 7.111 A steam pipe weighting 50 lb/ft that passes between two buildings 60 ft apart is supported by a system of cables as shown. Assuming that the weight of the cable is equivalent to a uniformly distributed loading of 7.5 lb/ft, determine (a) the location of the lowest point C of the cable, (b) the maximum tension in the cable. SOLUTION FBD AC:
FBD CB:
ΣM A = 0: (13.5 ft ) T0 −
a ( 57.5 lb/ft ) a = 0 2
(
)
T0 = 2.12963 lb/ft 2 a 2
(1)
60 ft − a ( 57.5 lb/ft )( 60 ft − a ) − ( 6 ft ) T0 = 0 2
ΣM B = 0:
(
)
6T0 = 28.75 lb/ft 2 3600 ft 2 − (120 ft ) a + a 2
Using (1) in (2)
(2)
0.55a 2 − (120 ft ) a + 3600 ft 2 = 0
Solving: a = (108 ± 72 ) ft
a = 36 ft (180 ft out of range)
So (a)
C is 36 ft from A
C is 6 ft below and 24 ft left of B
T0 = 2.1296 lb/ft 2 ( 36 ft ) = 2760 lb 2
W1 = ( 57.5 lb/ft )( 36 ft ) = 2070 lb
(b)
Tmax = TA = T02 + W12 =
( 2760 lb )2 + ( 2070 lb )2
= 3450 lb
PROBLEM 7.112 Chain AB supports a horizontal, uniform steel beam having a mass per unit length of 85 kg/m. If the maximum tension in the cable is not to exceed 8 kN, determine (a) the horizontal distance a from A to the lowest point C of the chain, (b) the approximate length of the chain.
SOLUTION
ΣM A = 0: y AT0 − yA =
a wa = 0 2
ΣM B = 0: − yBT0 +
wa 2 2T0
yB =
d = ( y B − yB ) =
Using
L = 6 m,
yields
w 2 b − a2 2T0
)
2
2 2 − ( wb ) ( 2d )2 Tmax
or
wb 2 2T0
2 T0 = TB2 − ( wb ) = Tmax − ( wb )
But ∴
(
b wb = 0 2
(
= w2 b 2 − a 2
)
2
(
= L2 w2 4b 2 − 4 Lb + L2
T2 4 L2 + d 2 b 2 − 4L3b + L4 − 4d 2 max w2
(
d = 0.9 m,
)
Tmax = 8 kN,
b = ( 2.934 ± 1.353) m
2
)
= 0
(
)
w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m
b = 4.287 m
( since b > 3 m ) (a)
a = 6 m − b = 1.713 m
PROBLEM 7.112 CONTINUED
2 T0 = Tmax − ( wb ) = 7156.9 N 2
yA wa = = 0.09979 xA 2T0
yB wb = = 0.24974 xB 2T0
2 2 2 yA 2 yB l = s A + sB = a 1 + +" + b 1+ + " 3 xA 3 xB
2 2 2 2 = (1.713 m ) 1 + ( 0.09979 ) + ( 4.287 m ) 1 + ( 0.24974 ) = 6.19 m 3 3
(b)
l = 6.19 m
PROBLEM 7.113 Chain AB of length 6.4 m supports a horizontal, uniform steel beam having a mass per unit length of 85 kg/m. Determine (a) the horizontal distance a from A to the lowest point C of the chain, (b) the maximum tension in the chain.
SOLUTION ΣM P = 0:
Geometry: y =
wx 2 2T0
x wx − yT0 = 0 2 y wx = x 2T0
so
and d = yB − y A =
FBD Segment:
(
w 2 b − a2 2T0
)
2 2 2 y 2 y l = s A + sB = a 1 + A + b 1 + B 3 a 3 b
Also
2 y A yB + 3 a b 2
l−L=
1 4d 2 = 6 b2 − a 2
(
)
2
(a
3
2
w2 3 a + b3 = 2 6T0
(
+b
3
)
(
)
2 3 3 2d a +b = 3 b2 − a 2 2
(
)
)
Using l = 6.4 m, L = 6 m, d = 0.9 m, b = 6 m − a, and solving for a, knowing that a < 3 ft a = 2.2196 m
(a) T0 =
Then
(
w 2 b − a2 2d
(
a = 2.22 m
)
)
w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m
And with
b = 6 m − a = 3.7804 m
And
Tmax = TB = T02 + ( wb ) =
T0 = 4338 N
2
( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2
Tmax = 5362 N
(b)
Tmax = 5.36 kN
PROBLEM 7.114 A cable AB of span L and a simple beam A′B′ of the same span are subjected to identical vertical loadings as shown. Show that the magnitude of the bending moment at a point C ′ in the beam is equal to the product , T0h where T0 is the magnitude of the horizontal component of the tension force in the cable and h is the vertical distance between point C and the chord joining the points of support A and B.
SOLUTION
FBD Cable:
ΣM B = 0: LACy + aT0 − ΣM B loads = 0
(1)
(Where ΣM B loads includes all applied loads) x ΣM C = 0: xACy − h − a T0 − ΣM C left = 0 L
FBD AC:
(Where ΣM C left includes all loads left of C) x (1) − ( 2 ): L
FBD Beam:
FBD AC:
(2)
hT0 −
x ΣM B loads + ΣM C left = 0 L
(3)
ΣM B = 0: LABy − ΣM B loads = 0
(4)
ΣM C = 0: xABy − ΣM C left − M C = 0
(5)
x ( 4 ) − ( 5): L
Comparing (3) and (6)
−
x ΣM B loads + ΣM C left + M C = 0 L M C = hT0
Q.E.D.
(6)
PROBLEM 7.115 Making use of the property established in Prob. 7.114, solve the problem indicated by first solving the corresponding beam problem. Prob. 7.89a.
SOLUTION FBD Beam:
ΣM D = 0: ( 2 ft )( 450 lb ) + ( 6 ft )( 600 lb ) − (10 ft ) A By = 0 A By = 450 lb
ΣM B = 0: M B − ( 4 ft )( 450 lb ) = 0
Section AB:
M B = 1800 lb ⋅ ft
ΣM A = 0: (10 ft ) DCy − ( 8 ft )( 450 lb ) − ( 4 ft )( 600 lb ) = 0
Cable:
DCy = 600 lb
( Note: Dy > Ay so Tmax = TCD ) 2 2 T0 = Tmax − DCy
T0 =
( 720 lb )2 − ( 600 lb )2
T0 = 398 lb
dB =
M B 1800 lb ⋅ ft = = 4.523 ft T0 398 lb d B = 4.52 ft
PROBLEM 7.116 Making use of the property established in Prob. 7.114, solve the problem indicated by first solving the corresponding beam problem. Prob. 7.92b.
SOLUTION FBD Beam:
ΣM E = 0: ( 2 ft )( 240 lb ) + ( 4 ft )( 720 lb ) + ( 6 ft )( 360 lb ) − ( 8 ft ) ABy = 0 A By = 690 lb
ΣM B = 0: M B − ( 2 ft )( 690 lb ) = 0 M B = 1380 lb ⋅ ft
ΣM B = 0: M C + ( 2 ft )( 360 lb ) − ( 4 ft )( 690 lb ) = 0 M C = 2040 lb ⋅ ft
ΣM D = 0: M D + ( 2 ft )( 720 lb ) + ( 4 ft )( 360 lb ) − ( 6 ft )( 690 lb ) = 0 M D = 1260 lb ⋅ ft hC = dC − 1.2 ft = 3.6 ft − 1.2 ft = 2.4 ft T0 =
Cable:
MC 2040 lb ⋅ ft = = 850 lb hC 2.4 ft M B 1380 lb ⋅ ft = = 1.6235 ft T0 850 lb
hB =
d B = hB + 0.6 ft h0 =
d B = 2.22 ft
M D 1260 lb ⋅ ft = = 1.482 ft T0 850 lb
d B = h0 + 1.8 ft
d D = 3.28 ft
PROBLEM 7.117 Making use of the property established in Prob. 7.114, solve the problem indicated by first solving the corresponding beam problem. Prob. 7.94b.
SOLUTION FBD Beam:
A By = F = 8 kN
By symmetry: M B = M E;
MC = M D
AC:
ΣM C = 0: M C + ( 6 m )( 4 kN ) − (12 m )( 8 kN ) = 0 M C = 72 kN ⋅ m
so
M D = 72 kN ⋅ m
Cable:
Since
M D = MC
hD = hC = 12 m − 3 m = 9 m d D = hD + 2 m = 11 m d D = 11.00 m
PROBLEM 7.118 Making use of the property established in Prob. 7.114, solve the problem indicated by first solving the corresponding beam problem. Prob. 7.95b.
SOLUTION FBD Beam:
By symmetry: M B = M E
and
MC = M D
Cable:
Since
M D = M C , hD = hC
hD = hC = dC − 3 m = 9 m − 3 m = 6 m
Then
d D = hD + 2 m = 6 m + 2 m = 8 m d D = 8.00 m
PROBLEM 7.119 Show that the curve assumed by a cable that carries a distributed load w( x) is defined by the differential equation d 2 y / dx 2 = w( x ) / T0 , where T0 is the tension at the lowest point.
SOLUTION
ΣFy = 0: Ty ( x + ∆x ) − Ty ( x ) − w ( x ) ∆ x = 0
FBD Elemental segment:
Ty ( x + ∆x )
So
T0
Ty ( x )
−
T0
Ty
But
T0 dy dx
So
− x + ∆x
∆x
In
lim :
∆x → 0
=
dy dx
x
=
w( x) ∆x T0
dy dx
=
w( x) T0
w( x) d2y = 2 T0 dx
Q.E.D.
PROBLEM 7.120 Using the property indicated in Prob. 7.119, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w0 cos(π x / L ) , where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable.
SOLUTION
w ( x ) = w0 cos
πx L
From Problem 7.119 w( x) πx d2y w = = 0 cos 2 T0 T0 L dx
So
dy using dx
dy W0 L πx = sin dx T0π L
y =
But
And
0
w0 L2 πx 1 − cos using y ( 0 ) = 0 2 L T0π
w L2 π L y = h = 0 2 1 − cos 2 T0π 2
TBy =
T0 =
so
T0 = Tmin Tmax = TA = TB :
= 0
w0 L2 π 2h Tmin =
so TBy T0
=
dy dx
= x = L2
w0 L2 π 2h
w0 L T0π
w0 L
π
2 TB = TBy + T02 =
w0 L
π
L 1+ πh
2
PROBLEM 7.121 If the weight per unit length of the cable AB is w0 / cos 2 θ , prove that the curve formed by the cable is a circular arc. (Hint: Use the property indicated in Prob. 7.119.)
SOLUTION Elemental Segment:
Load on segment* But
dx = cosθ ds,
w0 ds cos 2 θ w0 w( x) = cos3 θ
w ( x ) dx = so
From Problem 7.119
d2y w( x) w0 = = T0 dx 2 T0 cos3 θ
In general
d2y d dy d dθ = ( tan θ ) = sec2 θ = 2 dx dx dx dx dx
So
dθ w0 w0 = = 3 2 dx T0 cosθ T0 cos θ sec θ
or
T0 cosθ dθ = dx = rdθ cosθ w0
Giving r =
T0 = constant. So curve is circular arc w0
*For large sag, it is not appropriate to approximate ds by dx.
Q.E.D.
PROBLEM 7.122 Two hikers are standing 30-ft apart and are holding the ends of a 35-ft length of rope as shown. Knowing that the weight per unit length of the rope is 0.05 lb/ft, determine (a) the sag h, (b) the magnitude of the force exerted on the hand of a hiker.
SOLUTION Half-span:
w = 0.05 lb/ft,
L = 30 ft, sB = c sinh
sB =
35 ft 2
yB xB
15 ft 17.5 ft = c sinh c
c = 15.36 ft
Solving numerically, Then
yB = c cosh
xB 15 ft = (15.36 ft ) cosh = 23.28 ft c 15.36 ft
(a)
hB = yB − c = 23.28 ft − 15.36 ft = 7.92 ft
(b)
TB = wyB = ( 0.05 lb/ft )( 23.28 ft ) = 1.164 lb
PROBLEM 7.123 A 60-ft chain weighing 120 lb is suspended between two points at the same elevation. Knowing that the sag is 24 ft, determine (a) the distance between the supports, (b) the maximum tension in the chain.
SOLUTION
sB = 30 ft,
w=
hB = 24 ft,
120 lb = 2 lb/ft 60 ft xB =
yB2 = c 2 + sB2 = ( hB + c )
L 2
2
= hB2 + 2chB + c 2
( 30 ft ) − ( 24 ft ) sB2 − hB2 = 2hB 2 ( 24 ft ) 2
c=
2
c = 6.75 ft
Then
sB = c sinh
xB s → xB = c sinh −1 B c c
30 ft xB = ( 6.75 ft ) sinh −1 = 14.83 ft 6.75 ft
(a)
L = 2 xB = 29.7 ft
Tmax = TB = wyB = w ( c + hB ) = ( 2 lb/ft )( 6.75 ft + 24 ft ) = 61.5 lb
(b)
Tmax = 61.5 lb
PROBLEM 7.124 A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape. Neglect the elongation of the tape due to the tension.
SOLUTION
sB = 100 ft,
w=
4 lb = 0.02 lb/ft 200 ft
Tmax = 16 lb Tmax = TB = wyB yB =
TB 16 lb = = 800 ft w 0.02 lb/ft c 2 = yB2 − sB2
c=
But
(800 ft )2 − (100 ft )2
yB = xB cosh
= 793.73 ft
xB y → xB = c cosh −1 B c c
800 ft = ( 793.73 ft ) cosh −1 = 99.74 ft 793.73 ft L = 2 xB = 2 ( 99.74 ft ) = 199.5 ft
PROBLEM 7.125 An electric transmission cable of length 130 m and mass per unit length of 3.4 kg/m is suspended between two points at the same elevation. Knowing that the sag is 30 m, determine the horizontal distance between the supports and the maximum tension.
SOLUTION
sB = 65 m,
(
hB = 30 m
)
w = ( 3.4 kg/m ) 9.81 m/s 2 = 33.35 N/m yB2 = c 2 + s 2B
( c + hB )2
= c 2 + sB2
( 65 m ) − ( 30 m ) sB2 − hB2 = 2hB 2 ( 30 m ) 2
c=
2
= 55.417 m
Now
sB = c sinh
xB s 65 m → xB = c sinh −1 B = ( 55.417 m ) sinh −1 c c 55.417 m = 55.335 m L = 2 xB = 2 ( 55.335 m ) = 110.7 m
Tmax = wyB = w ( c + hB ) = ( 33.35 N/m )( 55.417 m + 30 m ) = 2846 N Tmax = 2.85 kN
PROBLEM 7.126 A 30-m length of wire having a mass per unit length of 0.3 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the force P for which h = 12 m, (b) the corresponding span L.
SOLUTION FBD Cable:
s = 30 m
30 m = 15 m so sB = 2
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m hB = 12 m yB2 = ( c + hB ) = c 2 + s 2B 2
c=
So
c=
Now
sB = c sinh
sB2 − hB2 2hB
(15 m )2 − (12 m )2 2 (12 m )
= 3.375 m
xB s 15 m → xB = c sinh −1 B = ( 3.375 m ) sinh −1 c c 3.375 m xB = 7.4156 m
P = T0 = wc = ( 2.943 N/m )( 3.375 m ) L = 2 xB = 2 ( 7.4156 m )
(a) (b)
P = 9.93 N L = 14.83 m
PROBLEM 7.127 A 30-m length of wire having a mass per unit length of 0.3 kg/m is attached to a fixed support at A and to a collar at B. Knowing that the magnitude of the horizontal force applied to the collar is P = 30 N, determine (a) the sag h, (b) the corresponding span L.
SOLUTION FBD Cable:
sT = 30 m,
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m P = T0 = wc c=
c=
P w
30 N = 10.1937 m 2.943 N/m
yB2 = ( hB + c ) = c 2 + sB2 2
h 2 + 2ch − sB2 = 0
sB =
30 m = 15 m 2
h 2 + 2 (10.1937 m ) h − 225 m 2 = 0 h = 7.9422 m sB = c sinh
(a)
h = 7.94 m
xA s 15 m → xB = c sinh −1 B = (10.1937 m ) sinh −1 c c 10.1937 m = 12.017 m L = 2 xB = 2 (12.017 m )
(b)
L = 24.0 m
PROBLEM 7.128 A 30-m length of wire having a mass per unit length of 0.3 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the sag h for which L = 22.5 m, (b) the corresponding force P.
SOLUTION FBD Cable:
sT = 30 m → sB =
30 m = 15 m 2
(
)
w = ( 0.3 kg/m ) 9.81 m/s 2 = 2.943 N/m L = 22.5 m sB = c sinh
xB L/2 = c sinh c c
15 m = c sinh
11.25 m c
Solving numerically: c = 8.328 m yB2 = c 2 + sB2 = ( 8.328 m ) + (15 m ) = 294.36 m 2 2
2
yB = 17.157 m
hB = yB − c = 17.157 m − 8.328 m
(a) P = wc = ( 2.943 N/m )( 8.328 m )
(b)
hB = 8.83 m
P = 24.5 N
PROBLEM 7.129 A 30-ft wire is suspended from two points at the same elevation that are 20 ft apart. Knowing that the maximum tension is 80 lb, determine (a) the sag of the wire, (b) the total weight of the wire.
SOLUTION
L = 20 ft sB =
xB =
30 ft = 15 ft 2
sB = c sinh
xB 10 ft = c sinh c c
Solving numerically: yB = c cosh
20 ft = 10 ft 2
c = 6.1647 ft
xB 10 ft = ( 6.1647 ft ) cosh c 1.1647 ft yB = 16.217 ft
hB = yB − c = 16.217 ft − 6.165 ft
Tmax = wyB
So
W =
and
(a)
hB = 10.05 ft
(b)
Wm = 148.0 lb
W = w ( 2 sB )
Tmax 80 lb ( 2sB ) = ( 30 ft ) yB 16.217 ft
PROBLEM 7.130 Determine the sag of a 45-ft chain which is attached to two points at the same elevation that are 20 ft apart.
SOLUTION
sB =
45 ft = 22.5 ft 2 xB =
L = 20 ft
L = 10 ft 2
sB = c sinh
xB c
22.5 ft = c sinh
Solving numerically: yB = c cosh
10 ft c
c = 4.2023 ft
xB c
= ( 4.2023 ft ) cosh
10 ft = 22.889 ft 4.2023 ft
hB = yB − c = 22.889 ft − 4.202 ft hB = 18.69 ft
PROBLEM 7.131 A 10-m rope is attached to two supports A and B as shown. Determine (a) the span of the rope for which the span is equal to the sag, (b) the corresponding angle θB.
SOLUTION
y = c cosh
We know
yB = c + h = c cosh
At B,
1 = cosh
or
sB = c sinh c=
h 2c
h h − 2c c
h = 4.933 c
Solving numerically
So
x c
xB s h → T = c sinh c 2 2c
sT 10 m = = 0.8550 m h 4.933 2sinh 2 sinh 2c 2
h = 4.933c = 4.933 ( 0.8550 ) m = 4.218 m
h = 4.22 m
(a) From
x y = c cosh , c
At B,
tan θ =
dy dx
θ = tan −1 5.848
dy x = sinh dx c = sinh
B
L = h = 4.22 m W
L 4.933 = sinh = 5.848 2c 2
(b)
θ = 80.3° W
PROBLEM 7.132 A cable having a mass per unit length of 3 kg/m is suspended between two points at the same elevation that are 48 m apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 1800 N.
SOLUTION
(
)
w = ( 3 kg/m ) 9.81 m/s 2 = 29.43 N/m
L = 48 m,
Tmax ≤ 1800 N
Tmax = wyB → yB = yB ≤ yB = c cosh
xB c
Tmax w
1800 N = 61.162 m 29.43 N/m 61.162 m = c cosh
Solving numerically
48m / 2 * c
c = 55.935 m
h = yB − c = 61.162 m − 55.935 m h = 5.23 m W
*Note: There is another value of c which will satisfy this equation. It is much smaller, thus corresponding to a much larger h.
PROBLEM 7.133 An 8-m length of chain having a mass per unit length of 3.72 kg/m is attached to a beam at A and passes over a small pulley at B as shown. Neglecting the effect of friction, determine the values of distance a for which the chain is in equilibrium.
SOLUTION
TB = wa
Neglect pulley size and friction But
TB = wyB
so
yB = c cosh c cosh
But sB = c sinh So
yB = a xB c
1m =a c
8m − a 1m = c sinh c 2
xB c
4 m = c sinh
(
1m c 1m + cosh c c 2
16 m = c 3e1/c − e−1/c
Solving numerically
)
c = 0.3773 m, 5.906 m
1m ( 0.3773 m ) cosh 0.3773 m = 2.68 m W 1m = a = c cosh c ( 5.906 m ) cosh 1 m = 5.99 m W 5.906 m
PROBLEM 7.134 A motor M is used to slowly reel in the cable shown. Knowing that the weight per unit length of the cable is 0.5 lb/ft, determine the maximum tension in the cable when h = 15 ft.
SOLUTION
w = 0.5 lb/ft
L = 30 ft yB = c cosh
hB = 15 ft
xB c
hB + c = c cosh
L 2c
15 ft − 1 15 ft = c cosh c
Solving numerically c = 9.281 ft yB = ( 9.281 ft ) cosh
15 ft = 24.281 ft 9.281 ft
Tmax = TB = wyB = ( 0.5 lb/ft )( 24.281 ft ) Tmax = 12.14 lb W
PROBLEM 7.135 A motor M is used to slowly reel in the cable shown. Knowing that the weight per unit length of the cable is 0.5 lb/ft, determine the maximum tension in the cable when h = 9 ft.
SOLUTION
w = 0.5 lb/ft,
L = 30 ft,
yB = hB + c = c cosh
hB = 9 ft
xB L = c cosh 2c c
15 ft 9 ft = c cosh − 1 c
Solving numerically c = 13.783 ft yB = hB + c = 9 ft + 13.783 ft = 21.783 ft Tmax = TB = wyB = ( 0.5 lb/ft )( 21.78 ft ) Tmax = 11.39 lb W
PROBLEM 7.136 To the left of point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the weight per unit length of the cable is 1.5 lb/ft, determine the force F when a = 10.8 ft.
SOLUTION
yD = c cosh
xD c
h + c = c cosh
a c
10.8 m − 1 12 m = c cosh c
Solving numerically Then
c = 6.2136 m
yB = ( 6.2136 m ) cosh
10.8 m = 18.2136 m 6.2136 m
F = Tmax = wyB = (1.5 lb/ft )(18.2136 m ) F = 27.3 lb
W
PROBLEM 7.137 To the left of point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the weight per unit length of the cable is 1.5 lb/ft, determine the force F when a = 18 ft.
SOLUTION
yD = c cosh
xD c
c + h = c cosh
a c
a h = c cosh − 1 c 18 ft 12 ft = c cosh − 1 c
Solving numerically
c = 15.162 ft
yB = h + c = 12 ft + 15.162 ft = 27.162 ft F = TD = wyD = (1.5 lb/ft )( 27.162 ft ) = 40.74 lb F = 40.7 lb
W
PROBLEM 7.138 A uniform cable has a mass per unit length of 4 kg/m and is held in the position shown by a horizontal force P applied at B. Knowing that P = 800 N and θA = 60°, determine (a) the location of point B, (b) the length of the cable.
SOLUTION
(
)
w = 4 kg/m 9.81 m/s 2 = 39.24 N/m P = T0 = wc
c=
P 800 N = w 39.24 N/m
c = 20.387 m y = c cosh
x c
dy x = sinh dx c tan θ = −
dy dx
= − sinh −a
−a a = sinh c c
a = c sinh −1 ( tan θ ) = ( 20.387 m ) sinh −1 ( tan 60° ) a = 26.849 m y A = c cosh
a 26.849 m = ( 20.387 m ) cosh = 40.774 m c 20.387 m
b = y A − c = 40.774 m − 20.387 m = 20.387 m
So s = c sinh
(a)
a 26.849 m = ( 20.387 m ) sinh = 35.31 m c 20.387 m
B is 26.8 m right and 20.4 m down from A W
(b)
s = 35.3 m W
PROBLEM 7.139 A uniform cable having a mass per unit length of 4 kg/m is held in the position shown by a horizontal force P applied at B. Knowing that P = 600 N and θA = 60°, determine (a) the location of point B, (b) the length of the cable.
SOLUTION
(
)
w = ( 4 kg/m ) 9.81 m/s 2 = 39.24 N/m P = T0 = wc
c=
P 600 N = w 39.24 N/m
c = 15.2905 m
y = c cosh
At A: So
tan θ = −
dy dx
x c
dy x = sinh dx c = − sinh
−a
−a a = sinh c c
a = c sinh −1 ( tan θ ) = (15.2905 m ) sinh −1 ( tan 60° ) = 20.137 m a c
yB = h + c = c cosh a h = c cosh − 1 c
20.137 m = (15.2905 m ) cosh − 1 15.2905 m = 15.291 m
So s = c sinh
(a)
B is 20.1 m right and 15.29 m down from A
a 20.137 m = (15.291 m ) sinh = 26.49 m c 15.291 m
(b)
s = 26.5 m
PROBLEM 7.140 The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of the cable is located at a distance a = 1.8 ft below the support A, determine (a) the location of the lowest point C, (b) the maximum tension in the cable.
SOLUTION
y A = c cosh
−a = c + 1.8 ft c
1.8 ft a = c cosh −1 1 + c yB = c cosh
b = c + 7.2 ft c
7.2 ft b = c cosh −1 1 + c But
1.8 ft 7.2 ft −1 a + b = 36 ft = c cosh −1 1 + + cosh 1 + c c
Solving numerically Then
c = 40.864 ft
7.2 ft b = ( 40.864 ft ) cosh −1 1 + = 23.92 ft 40.864 ft
(a) Tmax = wyB = ( 0.3 lb/ft )( 40.864 ft + 7.2 ft )
C is 23.9 ft left of and 7.20 ft below B W
(b)
Tmax = 14.42 lb W
PROBLEM 7.141 The cable ACB weighs 0.3 lb/ft. Knowing that the lowest point of the cable is located at a distance a = 6 ft below the support A, determine (a) the location of the lowest point C, (b) the maximum tension in the cable.
SOLUTION
y A = c cosh
−a = c + 6 ft c
6 ft a = c cosh −1 1 + c yB = c cosh
b = c + 11.4 ft c
11.4 ft b = c cosh −1 1 + c So Solving numerically
6 ft 11.4 ft −1 a + b = c cosh −1 1 + + cosh 1 + = 36 ft c c
c = 20.446 ft 11.4 ft b = ( 20.446 ft ) cosh −1 1 + = 20.696 ft 20.446 ft (a) C is 20.7 ft left of and 11.4 ft below B 20.696 ft Tmax = wyB = ( 0.3 lb/ft )( 20.446 ft ) cosh = 9.554 lb 20.446 ft (b)
Tmax = 9.55 lb
PROBLEM 7.142 Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ , (b) y = c sec θ .
SOLUTION
tan θ =
(a)
s = c sinh (b)
Also
dy x = sinh dx c x = c tan θ Q.E.D. c
(
)
y 2 = s 2 + c 2 cosh 2 x = sinh 2 x + 1
(
)
So
y 2 = c 2 tan 2 θ + 1 = c 2 sec2 θ
And
y = c sec θ Q.E.D.
PROBLEM 7.143 (a) Determine the maximum allowable horizontal span for a uniform cable of mass per unit length m′ if the tension in the cable is not to exceed a given value Tm . (b) Using the result of part a, determine the maximum span of a steel wire for which m′ = 0.34 kg/m and Tm = 32 kN.
SOLUTION
TB = Tmax = wyB = wc cosh Let
ξ =
L 2c
so
xB L 2c L = w cosh c 2 L 2c
Tmax =
wL cosh ξ 2ξ
dTmax wL 1 = sinh ξ − cosh ξ dξ ξ 2ξ
For
min Tmax ,
tanh ξ −
1
ξ
=0
Solving numerically ξ = 1.1997
(Tmax )min
=
wL cosh (1.1997 ) = 0.75444wL 2 (1.9997 )
(a)
If
Lmax =
(
Tmax T = 1.3255 max 0.75444w w
)
Tmax = 32 kN and w = ( 0.34 kg/m ) 9.81 m/s 2 = 3.3354 N/m Lmax = 1.3255
32.000 N = 12 717 m 3.3354 N/m (b)
Lmax = 12.72 km
PROBLEM 7.144 A cable has a weight per unit length of 2 lb/ft and is supported as shown. Knowing that the span L is 18 ft, determine the two values of the sag h for which the maximum tension is 80 lb.
SOLUTION
ymax = c cosh Tmax = wymax ymax =
ymax =
Tmax w
80 lb = 40 ft 2 lb/ft
c cosh
Solving numerically
L =h+c 2c
9 ft = 40 ft c
c1 = 2.6388 ft c2 = 38.958 ft h = ymax − c h1 = 40 ft − 2.6388 ft
h1 = 37.4 ft
h2 = 40 ft − 38.958 ft
h2 = 1.042 ft
PROBLEM 7.145 Determine the sag-to-span ratio for which the maximum tension in the cable is equal to the total weight of the entire cable AB.
SOLUTION
Tmax = wyB = 2wsB y B = 2sB c cosh
L L = 2c sinh 2c 2c
tanh
L 1 = 2c 2
L 1 = tanh −1 = 0.549306 2c 2 hB y −c L = B = cosh −1 c c 2c
= 0.154701 hB hB / c = L 2( L / 2c) =
0.5 ( 0.154701) = 0.14081 0.549306 hB = 0.1408 L
PROBLEM 7.146 A cable of weight w per unit length is suspended between two points at the same elevation that are a distance L apart. Determine (a) the sagto-span ratio for which the maximum tension is as small as possible, (b) the corresponding values of θ B and Tm .
SOLUTION
Tmax = wyB = wc cosh
(a)
L 2c
dTmax L L L sinh = w cosh − dc 2c 2c 2c min Tmax ,
For
tanh
dTmax =0 dc
L L 2c = 1.1997 = → 2c 2c L
yB L = cosh = 1.8102 c 2c h y = B − 1 = 0.8102 c c h 1 h 2c 0.8102 = = 0.3375 = L 2 c L 2 (1.1997 ) T0 = wc
(b) But
So
Tmax = wc cosh
T0 = Tmax cosθ B
L 2c
h = 0.338 L
Tmax L y = cosh = B 2c T0 c
Tmax = secθ B T0
y
θ B = sec−1 B = sec−1 (1.8102 ) c = 56.46°
Tmax = wyB = w
θ B = 56.5°
yB 2c L L = w (1.8102 ) c L 2 2 (1.1997 )
Tmax = 0.755wL
PROBLEM 7.147 For the beam and loading shown, (a) draw the shear and bending moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Beam:
(a)
ΣM A = 0: ( 6 ft ) E − ( 8 ft )( 4.5 kips ) − ( 4 ft )(12 kips ) − ( 2 ft )( 6 kips ) = 0 E = 16 kips ΣM E = 0: − ( 6 ft ) Ay + ( 4 ft )( 6 kips ) + ( 2 ft )(12 kips ) − ( 2 ft )( 4.5 kips ) = 0 A y = 6.5 kips
Shear Diag: V is piece wise constant with discontinuities equal to the forces at A, C, D, E, B Moment Diag: M is piecewise linear with slope changes at C, D, E MA = 0 M C = ( 6.5 kips )( 2 ft ) = 13 kip ⋅ ft M C = 13 kip ⋅ ft + ( 0.5 kips )( 2 ft ) = 14 kip ⋅ ft M D = 14 kip ⋅ ft − (11.5 kips )( 2 ft ) = −9 kip ⋅ ft M B = − 9 kip ⋅ ft + ( 4.5 kips )( 2 ft ) = 0
(b)
V M
max max
= 11.50 kips on DE = 14.00 kip ⋅ ft at D
PROBLEM 7.148 For the beam and loading shown, (a) draw the shear and bending moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Beam:
(a)
ΣM B = 0: ( 4 ft )(12 kips ) + ( 7 ft )( 2.5 kips/ft )( 6 ft ) − (10 ft ) Ay = 0 A y = 15.3 kips
Shear Diag: VA = Ay = 15.3 kips, then V is linear dV = −2.5 kips/ft to C. dx VC = 15.3 kips − ( 2.5 kips/ft )( 6 ft ) = 0.3 kips At C , V decreases by 12 kips to − 11.7 kips and is constant to B.
Moment Diag: M A = 0 and M is parabolic dM decreasing with V to C dx
MC =
1 (15.3 kips + 0.3 kip )( 6 ft ) = 46.8 kip ⋅ ft 2
M B = 46.8 kip ⋅ ft − (11.7 kips )( 4 ft ) = 0
(b)
V max = 15.3 kips M
max
= 46.8 kip ⋅ ft
PROBLEM 7.149 Two loads are suspended as shown from the cable ABCD. Knowing that hB = 1.8 m, determine (a) the distance hC, (b) the components of the reaction at D, (c) the maximum tension in the cable.
SOLUTION FBD Cable:
ΣFx = 0: − Ax + Dx = 0
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0
Dy = 7.8 kN ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kN
FBD AB:
ΣM B = 0: (1.8 m ) Ax − ( 3 m )( 8.2 kN ) = 0 Ax =
From above
FBD CD:
41 kN 3
Dx = Ax =
41 kN 3
41 kN = 0 ΣM C = 0: ( 4 m )( 7.8 kN ) − hC 3 hC = 2.283 m hC = 2.28 m
(a)
Dx = 13.67 kN
(b)
Dy = 7.80 kN
Since Ax = Bx and Ay > By , max T is TAB 2
TAB =
Ax2 + Ay2 =
2 41 kN + ( 8.2 kN ) 3
(c)
Tmax = 15.94 kN
PROBLEM 7.150 Knowing that the maximum tension in cable ABCD is 15 kN, determine (a) the distance hB, (b) the distance hC.
SOLUTION FBD Cable:
ΣFx = 0: − Ax + Dx = 0
Ax = Dx
ΣM A = 0: (10 m ) D y − ( 6 m )(10 kN ) − ( 3 m )( 6 kN ) = 0 D y = 7.8 kN ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kN
Since
Ax = Dx
and
Ay > Dy ,
Tmax = TAB
ΣFy = 0: 8.2 kN − (15 kN ) sin θ A = 0
FBD pt A:
θ A = sin −1
8.2 kN = 33.139° 15 kN
ΣFx = 0: − Ax + (15 kN ) cos θ A = 0 Ax = (15 kN ) cos ( 33.139° ) = 12.56 kN
FBD CD: From FBD cable:
hB = ( 3 m ) tan θ A = ( 3 m ) tan ( 33.139° )
(a)
hB = 1.959 m
ΣM C = 0: ( 4 m )( 7.8 kN ) − hC (12.56 kN ) = 0
(b)
hC = 2.48 m
PROBLEM 7.151 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when θ = 60°.
SOLUTION FBD Rod:
ΣM A = 0:
2r
π
B=
ΣFy′ = 0: F +
FBD BJ:
W − 2rB = 0
W
π
W W sin 60° − cos 60° = 0 π 3
F = − 0.12952W
W 3r W ΣM 0 = 0: r F − + +M =0 π 2π 3 1 1 M = Wr 0.12952 + − = 0.28868Wr π 2π
On BJ
M J = 0.289Wr
PROBLEM 7.152 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when θ = 150°.
SOLUTION FBD rod: ΣFy = 0: Ay − W = 0 Ay = W 2r ΣM B = 0: W − 2rAx = 0 W π Ax =
π
FBD AJ:
ΣFx′ = 0:
W
π
cos 30° +
5W sin 30° − F = 0 F = 0.69233W 6
W W ΣM 0 = 0: 0.25587r + r F − − M = 0 π 6 1 0.25587 M = Wr + 0.69233 − π 6 M = Wr ( 0.4166 )
On AJ
M = 0.417Wr
PROBLEM 7.153 Determine the internal forces at point J of the structure shown.
SOLUTION FBD ABC:
ΣM D = 0: ( 0.375 m )( 400 N ) − ( 0.24 m ) C y = 0
C y = 625 N ΣM B = 0: − ( 0.45 m ) C x + ( 0.135 m )( 400 N ) = 0
C x = 120 N
FBD CJ:
ΣFy = 0: 625 N − F = 0
F = 625 N ΣFx = 0: 120 N − V = 0
V = 120.0 N ΣM J = 0: M − ( 0.225 m )(120 N ) = 0
M = 27.0 N ⋅ m
PROBLEM 7.154 Determine the internal forces at point K of the structure shown.
SOLUTION FBD AK: ΣFx = 0: V = 0
V =0 ΣFy = 0: F − 400 N = 0
F = 400 N ΣM K = 0: ( 0.135 m )( 400 N ) − M = 0
M = 54.0 N ⋅ m
PROBLEM 7.155 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing the θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.
SOLUTION FBD Beam + channels:
(a)
T1 = T2 = T
By symmetry:
ΣFy = 0: 2T sin 60° − 3 kN = 0 T =
FBD Beam: With cable force replaced by equivalent force-couple system at F and G
3 kN 3 M = ( 0.5 m )
T1x = 3 2 3
3
T1y =
2 3
3 kN 2
kN = 0.433 kN ⋅ m
Shear Diagram: V is piecewise linear dV = − 0.6 kN/m with 1.5 kN dx discontinuities at F and H.
V F − = − ( 0.6 kN/m )(1.5 m ) = 0.9 kN V increases by 1.5 kN to + 0.6 kN at F + VG = 0.6 kN − ( 0.6 kN/m )(1 m ) = 0 Finish by invoking symmetry Moment Diagram: M is piecewise parabolic
dM decreasing with V with discontinuities of .433 kN at F and H. dx 1 M F − = − ( 0.9 kN )(1.5 m ) = − 0.675 kN ⋅ m 2 M increases by 0 .433 kN ⋅ m to − 0.242 kN ⋅ m at F + M G = − 0.242 kN ⋅ m +
1 ( 0.6 kN )(1 m ) = 0.058 kN ⋅ m 2
Finish by invoking symmetry
V
(b)
max
= 900 N
at F − and G + M
max
= 675 N ⋅ m
at F and G
PROBLEM 7.156 (a) Draw the shear and bending moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
PROBLEM 7.157 Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a.
SOLUTION FBD ABC:
ΣFy = 0: −40 lb − 80 lb + C y = 0 C y = 120 lb
FBD BC:
ΣM B = 0: ( 4 ft )(120 lb ) − (10 ft ) Cx = 0
C x = 48 lb From ABC:
ΣFx = 0: − P + Cx = 0 P = C x = 48 lb
(a)
P = 48.0 lb
ΣM C = 0: ( 4 ft )( 80 lb ) + a ( 40 lb ) − (15 ft )( 48 lb ) = 0 (b)
a = 10.00 ft
PROBLEM 7.158 Cable ABC supports two loads as shown. Determine the distances a and b when a horizontal force P of magnitude 60 lb is applied at A.
SOLUTION FBD ABC:
ΣFx = 0: C x − P = 0
Cx = 60 lb
ΣFy = 0: C y − 40 lb − 80 lb = 0 C y = 120 lb
FBD BC:
ΣM B = 0: b (120 lb ) − (10 ft )( 60 lb ) = 0 b = 5.00 ft
FBD AB:
ΣM B = 0: ( a − b )( 40 lb ) − ( 5 ft ) 60 lb = 0 a − b = 7.5 ft a = b + 7.5 ft = 5 ft + 7.5 ft
a = 12.50 ft
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