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7.1
Determine PROBLEM 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
Fx = 0:  F = 0
FBD JD:
Fy = 0: V  20 lb  20 lb = 0
F=0
V = 40.0 lb
M 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:
Fx = 0: 60 lb  V = 0 V = 60.0 lb Fy = 0:  F = 0 F=0 M 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:
Fy = 0: Ay  80 kN = 0 A y = 80 kN
M E = 0: (1.2 m ) Ax  (1.5 m )( 80 kN ) = 0
A x = 100 kN
= tan 1 = 21.801 0.75 m
FBD AJ:
Fx = 0: F  ( 80 kN ) sin 21.801  (100 kN ) cos 21.801 = 0
0.3 m
F = 122.6 kN
Fy = 0: V + ( 80 kN ) cos 21.801  (100 kN ) sin 21.801 = 0
V = 37.1 kN
M 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:
Fy = 0: Ay  100 N = 0
A y = 100 N
M F = 0: 2 ( 0.32 m ) cos 30 Ax  ( 0.48 m )(100 N ) = 0
A x = 86.603 N
FBD AJ:
Fx = 0: F  (100 N ) cos 30  ( 86.603 N ) sin 30 = 0
F = 129.9 N
Fy = 0: V + (100 N ) sin 30  ( 86.603 N ) cos 30 = 0
V = 25.0 N
M 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 twoforce member, so
Ay Ax = 0.36 m 0.15 m Ay = 5 Ax 12
M 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 Fx = 0: F  624 N = 0
FBD AJ:
Fy = 0: 260 N  V = 0
F = 624 N
V = 260 N
M 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:
M C = 0: ( 0.3 m ) Ax  ( 0.48 m )( 390 N ) = 0
A x = 624 N
AB is twoforce member, so
Ay Ax 5 = Ay = Ax 0.36 m 0.15 m 12
A y = 260 N
Fx = 0:  Ax + C x = 0
C x = A x = 624 N
Fy = 0: Ay + C y  390 N = 0 C y = 390 N  260 N = 130 N or C y = 130 N Fx = 0: F + 12 5 ( 624 N ) + (130 N ) = 0 13 13
FBD CK:
F = 626 N
Fy = 0:
F = 626 N
12 5 (130 N )  ( 624 N )  V = 0 13 13
V = 120 N
V = 120.0 N
M 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:
M B = 0: Ax ( 2r ) = 0
Ax = 0
Fx = 0: V  ( 30 lb ) cos 60 = 0
V = 15.00 lb
FBD AJ:
Fy = 0: F + ( 30 lb ) sin 60 = 0
F = 25.98 lb
F = 26.0 lb
M 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:
Fy = 0: By  30 lb = 0 M A = 0: 2rBx = 0
B y = 30 lb Bx = 0
Fx = 0: V  ( 30 lb ) cos 30 = 0
FBD BK:
V = 25.98 lb
V = 26.0 lb
Fy = 0: F + ( 30 lb ) sin 30 = 0
F = 15 lb
F = 15.00 lb
M 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 210N 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 Fx = 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
So, at J : yJ =
1 ( 0.32 m )2 = 0.04 m 2.56 m
Slope of parabola = tan =
dy = 2ax dx
2 At J : J = tan 1 ( 0.32 m ) = 14.036 2.56 m
So
= tan 1
4  14.036 = 39.094 3 Fx = 0: V  (175 N ) cos ( 39.094 ) = 0
V = 135.8 N
Fy = 0: F + (175 N ) sin ( 39.094 ) = 0 F = 110.35 N
F = 110.4 N
PROBLEM 7.9 CONTINUED
3 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 FBD Point A:
T1 = T2 = T 3 Fx = 0: 2T1  210 N = 0 5 Fy = 0: FC  Fx = 0: 4 (175 N ) = 0 5 T1 = 175 N FC = 140 N VC = 105 N
FBD BC:
3 (175 N )  VC = 0 5
3 4 M 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 Then FBD CK: And
a= 0.16 m
( 0.64 m )
2
=
1 ; 2.56 m
= tan 1
dy = tan 1 2ax dx
Fx = 0: (140 N ) cos  (105 N ) sin + F = 0
So
F = (105 N ) sin  (140 N ) cos dF = (105 N ) cos + (140 N ) sin d Fy = 0: V  (105 N ) cos  (140 N ) sin = 0
So
V = (105 N ) cos + (140 N ) sin
PROBLEM 7.10 CONTINUED
And
dV =  (105 N ) sin + (140 N ) cos d M 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 26.6 ) , the maxima are at the limits ( x = 0, or x = 0.64 m ) . Therefore, (a) (b) (c)
Fmax = 140.0 N Vmax = 156.5 N M max = 89.6 N m
at C at B at C
PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal forces at point J knowing that = 30o.
SOLUTION FBD AB:
4 3 M A = 0: r C + r C  2r ( 70 lb ) = 0 5 5
C = 100 lb
Fx = 0:  Ax + 4 (100 lb ) = 0 5
A x = 80 lb Fy = 0: Ay + 3 (100 lb )  70 lb = 0 5
A y = 10 lb
FBD AJ:
Fx = 0: F  ( 80 lb ) sin 30  (10 lb ) cos 30 = 0
F = 48.66 lb
F = 48.7 lb 60
Fy = 0: V  ( 80 lb ) cos 30 + (10 lb ) sin 30 = 0 V = 64.28 lb
V = 64.3 lb
30
M 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 FBD AB:
4 3 M A = 0: r C + r C  2r ( 70 lb ) = 0 5 5
C = 100 lb
Fx = 0:  Ax +
4 (100 lb ) = 0 5
A x = 80 lb
Fy = 0: Ay +
3 (100 lb )  70 lb = 0 5
A y = 10 lb
FBD AJ:
M J = 0: M  ( 8 in.)(1  cos )(10 lb )  ( 8 in.)( sin )( 80 lb ) = 0 M = ( 640 lb in.) sin + ( 80 lb in.) ( cos  1) dM = ( 640 lb in.) cos  ( 80 lb in.) sin = 0 d
for where So
FBD BK:
= tan 1 8 = 82.87 ,
d 2M =  ( 640 lb in.) sin  ( 80 lb in.) cos < 0 d 2
M = 565 lb in. at = 82.9 is a max for AC
M K = 0: M  ( 8 in.)(1  cos )( 70 lb ) = 0 M = ( 560 lb in.)(1  cos ) dM = ( 560 lb in.) sin = 0 d So, for = for = 0, where M = 0
2
, M = 560 lb in. is max for BC
M max = 565 lb in. at = 82.9
PROBLEM 7.13
Two members, each consisting of straight and 168mmradius quartercircle portions, are connected as shown and support a 480N load at D. Determine the internal forces at point J.
SOLUTION FBD Frame:
24 M A = 0: ( 0.336 m ) C  ( 0.252 m )( 480 N ) = 0 25
C = 375 N
Fy = 0: Ax 
24 C =0 25
Ax =
24 ( 375 N ) = 360 N 25
A x = 360 N
Fy = 0: Ay  480 N +
7 ( 375 N ) = 0 24
FBD CD:
A y = 375 N
M C = 0: ( 0.324 m )( 480 N )  ( 0.27 m ) B = 0
B = 576 N
Fx = 0: C x 
24 ( 375 N ) = 0 25
C x = 360 N
Fy = 0: 480 N +
7 ( 375 N ) + ( 576 N )  C y = 0 25
FBD CJ:
C y = 201 N
Fx = 0: V  ( 360 N ) cos 30  ( 201 N ) sin 30 = 0
V = 412 N
Fy = 0: F + ( 360 N ) sin 30  ( 201 N ) cos 30 = 0
F = 5.93 N
F = 5.93 N
M 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 168mmradius quartercircle portions, are connected as shown and support a 480N load at D. Determine the internal forces at point K.
SOLUTION FBD CD:
Fx = 0: C x = 0 M B = 0: ( 0.054 m )( 480 N )  ( 0.27 m ) C y = 0
C y = 96 N
Fy = 0: B  C y = 0
B = 96 N
FBD CK:
Fy = 0: V  ( 96 N ) cos 30 = 0
V = 83.1 N
Fx = 0: F  ( 96 N ) sin 30 = 0
F = 48.0 N
M 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
M A = 0: ( 5.4 ft ) Bx  ( 7.8 ft )( 90 lb )  ( 0.6 ft )( 90 lb ) = 0
B x = 140 lb
M 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
Fx = 0: Ex  140 lb = 0
E x = 140 lb
Fy = 0: 157.5 lb  90 lb  90 lb + E y = 0
E y = 22.5 lb
FBD EJ:
Fx = 0: V +
3 4 (140 lb ) + ( 22.5 lb  90 lb ) = 0 5 5
V = 30.0 lb
V = 30 lb
Fy = 0: F + 90 lb 
4 3 (140 lb )  ( 90 lb  22.5 lb ) = 0 5 5
F = 62.5 lb
F = 62.5lb
M 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
M 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
Fx = 0: 230 lb  90 lb  Ex = 0 E x = 140 lb
M A = 0: (1.8 ft )( 90 lb )  ( 7.2 ft ) E y = 0
E y = 22.5 lb
FBD KE:
Fx = 0: F  140 lb = 0
F = 140.0 lb
Fy = 0: V  22.5 lb = 0
V = 22.5 lb
M 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:
M A = 0: ( 5.4 ft ) Bx  ( 7.8 ft )( 90 lb ) = 0
B x = 130 lb
FBD BE with pulleys and cord:
M 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
Fx = 0: Ex  130 lb = 0
E x = 130 lb
Fy = 0: E y + 150 lb  90 lb  90 lb = 0
E y = 30 lb
FBD JE and pulley:
Fx = 0:  F  90 lb +
4 3 (130 lb ) + ( 90 lb  30 lb ) = 0 5 5
F = 50.0 lb
Fy = 0: V + 3 4 (130 lb ) + ( 30 lb  90 lb ) = 0 5 5
V = 30 lb
V = 30.0 lb
M 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:
M B = 0: ( 5.4 ft ) Ax  ( 7.8 ft )( 90 lb ) = 0
A x = 130 lb
FBD AE:
M E = 0:  ( 7.2 ft ) Ay  ( 4.8 ft )( 90 lb ) = 0 Ay = 60 lb
A y = 60 lb
Fx = 0:
F=0
FBD AK:
Fy = 0: 60 lb + 90 lb  V = 0
V = 30.0 lb
M K = 0: ( 4.8 ft )( 60 lb )  ( 2.4 ft )( 90 lb )  M = 0
M = 72.0 lb ft
PROBLEM 7.19
A 140mmdiameter 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 M A = ( 0.6 m ) Cx  ( 0.315 m )( 824.04 N ) = 0
(
)
C x = 432.62 N
FBD pipe:
By symmetry: N1 = N 2
Fy = 0: 2 N1 = 21 N1  W = 0 29
29 (824.04 N ) 42
= 568.98 N
Also note:
20 a = r tan = 70 mm 21 a = 66.67 mm
FBD BC:
M 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:
Fx = 0: F  21 20 ( 432.62 N )  ( 532.42 N ) = 0 29 29
F = 680 N
Fy = 0: 21 20 ( 532.42 N )  ( 432.62 N )  V = 0 29 29
V = 87.2 N
M 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 140mmdiameter 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 M C = 0: (.6 m ) Ax  (.315 m )( 824.04 N ) = 0
(
)
A x = 432.62 N
FBD pipe
By symmetry: N1 = N 2
Fy = 0: 2 N2 = 21 N1  W = 0 29
29 824.04 N 42 = 568.98 N
Also note: FBD AD:
a = r tan = ( 70 mm ) a = 66.67 mm
20 21
M 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:
Fx = 0: 21 20 ( 432.62 N ) + ( 291.6 N )  F = 0 29 29
F = 514 N
Fy = 0: 21 20 ( 291.6 N )  ( 432.62 N ) + V = 0 29 29
V = 87.2 N
M 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:
M D = 0: aP  2aA = 0
A= P 2 P =0 2 V = P 2 F=0
Fx = 0: V 
FBD AJ:
Fy = 0:
M J = 0: M  a
P =0 2 M = aP 2
(b) FBD Rod:
M D = 0: aP 
a4 A = 0 25
A=
5P 2
PROBLEM 7.21 CONTINUED
FBD AJ:
Fx = 0: 3 5P V = 0 5 2 3P 2
V =
Fy = 0: 4 5P F =0 5 2
F = 2P M =
3 aP 2
(c) FBD Rod:
3 4 M D = 0: aP  2a A  2a A = 0 5 5 A= 5P 14
3 5P Fx = 0: V  =0 5 14
V =
Fy = 0: 4 5P F =0 5 14
3P 14
F=
3 5P M J = 0: M  a =0 5 14
2P 7
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: Fx = 0: Ax = 0 M D = 0: aP  2aAy = 0 Ay = P 2
FBD AJ:
Fx = 0: V = 0
Fy = 0:
P F =0 2
F=
P 2
M J = 0: M = 0
(b) FBD Rod:
M A = 0 4 3 2a D + 2a D  aP = 0 5 5 Fx = 0: Ax  Fy = 0: Ay  P + 4 5 P=0 5 14 3 5 P=0 5 14
D= Ax = Ay =
5P 14 2P 7 11P 14
PROBLEM 7.22 CONTINUED
FBD AJ: Fx = 0: 2 P V = 0 7 V = Fy = 0: 11P F =0 14 F= M J = 0: a 2P M =0 7 M =
(c) FBD Rod:
2P 7
11P 14
2 aP 7
M A = 0:
a 4D  aP = 0 2 5
D=
5P 2
Fx = 0: Ax  Fy = 0: Ay  P 
4 5P =0 5 2 3 5P =0 5 2
Ax = 2P Ay =
5P 2
FBD AJ:
Fx = 0: 2 P  V = 0 V = 2P Fy = 0: 5P F =0 2 F= M J = 0: a ( 2 P )  M = 0 M = 2aP 5P 2
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 = 30.
SOLUTION
FBD CJ: Note = 180  60 = 60 = 2 3 r = r
sin =
3r
3 3 3 r = 2 2 120 4 = W 270 9 F = 2 3 W 9
Weight of section = W Fy = 0: F 
4 W cos 30 = 0 9
M 0 = 0: rF  ( r sin 60 )
2 3 3 3 3  M = r 2 2 9
4W M =0 9
2 3 1 4  Wr W = 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 = 120.
SOLUTION
(a) FBD Rod: Fx = 0: Ax = 0 M B = 0: rAy + 2r W 2r 2W  =0 3 3 2W 3
Ay =
FBD AJ: Note:
=
60 = 30 = 2 6 60 2W = 270 9
Weight of segment = W F = r sin =
r 3r sin 30 = /6 2W 2W + ( r  r sin 30 ) M =0 9 3
M J = 0: ( r cos  r sin 30 )
2W 9
M =
3r 3 r 3 1 3r 1  +  + = Wr 2 3 2 2 9 3
M = 0.1788Wr
PROBLEM 7.25
A quartercircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when
= 30o.
SOLUTION
FBD Rod:
Fx = 0: A x = 0 M B = 0: 2r
W  rAy = 0
Ay =
2W
= 15, weight of segment = W
FBD AJ: r = r sin = 2W
30 W = 90 3
r sin15 = 0.9886r /12 W cos 30  F = 0 3
Fy = 0:
F=
cos 30 
W 3 2 1  2 3
2W W M 0 = M + r F  + r cos15 =0 3
M = 0.0557 Wr
PROBLEM 7.26
A quartercircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when
= 30o.
SOLUTION
FBD Rod:
M A = 0: rB 
B=
2r
W =0
2W
FBD BJ:
= 15 =
r = r
12
/12
sin15 = 0.98862r 30 W = 90 3
Weight of segment = W Fy = 0: F 
W 2W cos 30  sin 30 = 0 3
3 1 + W F= 6
M 0 = 0: rF  ( r cos15 )
W M =0 3
3 1 cos15 M = rW +  0.98862 Wr 6 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:
M A = 0: rB 
2r
W =0
B=
2W
=
2
so r = r
0 sin ,
4
Weight of segment = W = Fx = 0: F  F = =
FBD BJ:
2 /2 W sin 2 = 0
4
4
W cos 2 
2W
2W
2W
( sin 2 ( sin
+ 2 cos 2 )
+ cos ) 4 W M =0
M 0 = 0: rF  ( r cos ) M = 2
r 4 Wr ( sin + cos )  sin cos W sin cos = M = 2Wr 1 1 sin 2 = sin 2 2 + cos  sin ) 2 Wr cos
But, so or
( sin
M =
dM 2 = Wr ( cos  sin ) = 0 at tan = 1 d
PROBLEM 7.27 CONTINUED
= 0.8603 rad
Solving numerically
and
M = 0.357Wr
at = 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:
Fx = 0: Ax = 0 M B = 0: 2r
W  rAy = 0
Ay =
2W
=
2
,
r =
r
sin 2 4 W = /2 2W
Weight of segment = W Fx = 0:  F  F = 2W 4
W cos 2 + = 2W
cos 2 = 0
(1  2 ) cos 2
(1  ) cos
FBD AJ:
2W M 0 = 0: M + F  M = But, so 2W
4 W =0 r + ( r cos ) 4W r
(1 + cos
 cos ) r 
sin cos
sin cos = M = 2r
1 1 sin 2 = sin 2 2
W (1  cos + cos  sin )
dM 2rW = ( sin  sin + cos  cos ) = 0 d for dM =0 d Only 0 and 1 in valid range At for
(1  ) sin
=0
= 0, 1, n ( n = 1, 2,
)
= 0 M = 0,
at = 57.3
at = 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 =
Along AB:
1 L ( w) 2 2
Ay = D =
wL 4
Fy = 0: M J = 0: M  x
Along BC:
wL V = 0 4 M =
V =
wL 4
wL =0 4
wL x (straight) 4
Fy = 0:
wL  wx1  V = 0 4 wL  wx1 4 V =0 at x1 = L 4
V = straight with M k = 0: M +
M =
x1 L wL wx1  + x1 =0 2 4 4
w L2 L 2 8 + 2 x1  x1 2
PROBLEM 7.29 CONTINUED
Parabola with Section CD by symmetry (b) From diagrams:
M =
3 L wL2 at x1 = 32 4
V
max
=
wL on AB and CD 4 = 3wL2 at center 32
M
max
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:
Fy = 0: wx  V = 0 straight with V = wL 2 at
V =  wx x= L 2
M J = 0: M +
x wx = 0 2
1 M =  wx 2 2
parabola with
Along BC:
M =
wL2 L at x = 8 2
Fy = 0:  w
L V = 0 2
1 V =  wL 2
L L M 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
= =
wL on BC 2 3wL2 at C 8
max
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:
Fy = 0: P  V = 0
V = P
M J = 0: M  Px = 0
M = Px
straight with M = Along BC:
PL at B 2
Fy = 0: P  P  V = 0
V =0
L M K = 0: M + Px1  P + x1 = 0 2 M = PL 2
(constant)
V M
max
(b) From diagrams:
= P along AB PL along BC 2
max
=
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 (a) FBD Beam:
M C = 0: LAy  M 0 = 0
Ay =
M0 L
C=
Fy = 0:  Ay + C = 0
M0 L
Along AB:
Fy = 0: 
M0 V = 0 L M0 +M =0 L
V =
M0 L M0 x L
M J = 0: x
M =
straight with M =  Along BC:
M0 at B 2
Fy = 0:  M K = 0: M + x
M0 V = 0 L
V =
M0 L
M0  M0 = 0 L
x M = M 0 1  L M0 at B 2 V
max
straight with M = (b) From diagrams:
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:
M 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
=0
Fy = 0: 12.625 kips  10 kips  8 kips  4 kips + B = 0
B = 9.375 kips
Along AC:
Fy = 0: 12.625 kips  V = 0 V = 12.625 kips M J = 0: M  x (12.625 kips ) = 0 M = (12.625 kips ) x M = 22.725 kip ft at C
Along CD:
Fy = 0: 12.625 kips  10 kips  V = 0 V = 2.625 kips M 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 )
PROBLEM 7.33 CONTINUED
Along DE: Along DE:
Fy = 0: (12.625  10  8 ) kips  V = 0 V = 5.375 kips
M 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
( x1 = 4.5 ft )
Along EB:
Along EB:
Fy = 0: V + 9.375 kips = 0
V = 9.375 kips
M 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
= 12.63 kips on AC = 29.8 kip ft at D
M
max
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:
M C = 0:
(1.2 m )( 4 kN )  (1 m )(16 kN ) + ( 2 m )(8 kN ) + ( 3.2 m ) B = 0
B = 1.5 kN Fy = 0: 4 kN + C y  16 kN + 8 kN  1.5 kN = 0 C y = 13.5 kN
Along AC:
Fy = 0: 4 kN  V = 0 V = 4 kN M J = 0: M + x ( 4 kN ) = 0 M = 4.8 kN m at C M = 4 kN x
Along CD:
Fy = 0: 4 kN + 13.5 kN  V = 0 V = 9.5 kN M 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:
Fy = 0: V + 8 kN  1.5 kN = 0 V = 6.5 kN M 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:
Fy = 0: V  1.5 kN = 0 V = 1.5 kN M N = 0: x2 (1.5 kN ) + M = 0 M =  (1.5 kN ) x2 M = 1.8 kN m at E V M
max max
(b) From diagrams:
= 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:
Fy = 0: 3 kip  V = 0 M J = 0: M + x ( 3 kips ) = 0
V = 3 kips M = ( 3 kips ) x
M = 9 kip ft at C
Along CD:
Fy = 0: 3 kips  5 kips  V = 0
V = 8 kips
M 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:
Fy = 0: 3 kips  5 kips + 6 kips  V = 0 V = 2 kips M 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:
Fy = 0: 3 kips  5 kips + 6 kips  4 kips  V = 0 M N = 0: M + ( 4 kips ) x2 + ( 2.1 ft + x2 )( 5 kips )
V = 6 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
( x2
= 2.4 ft ) V
max max
(b) From diagrams:
= 8.00 kips on CD = 33.0 kip ft at B
M
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 (a) FBD Beam:
M E = 0:
(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 Fy = 0: 540 N + 1080 N  1350 N 540 N + E y = 0 E y = 1350 N
Along AC:
Fy = 0: 540 N  V = 0
V = 540 N M J = 0: x ( 540 N ) + M = 0 M =  ( 540 N ) x
Along CD:
Fy = 0: 540 N + 1080 N  V = 0
V = 540 N
M 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:
Fy = 0: V + 1350 N  540 N = 0
V = 810 N
M 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:
Fy = 0: V  540 N = 0
V = 540 N M = 540 N x2 = 0.3 m ) V M
max
M L = 0: M + x2 ( 540 N ) = 0 M = 162 N m at E
( x2
(b) From diagrams
= 810 N on DE
max
= 162.0 N m at D and E
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:
Fy = 0: Ay + ( 6 ft )( 2 kips/ft )  12 kips  2 kips = 0
A y = 2 kips
M 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:
Fy = 0: 2 kips + x ( 2 kips/ft )  V = 0 V = 2 kips + ( 2 kips/ft ) x V = 14 kips at C ( x = 6 ft ) M 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:
Fy = 0: V  12 kips  2 kips = 0
V = 14 kips
M 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
( x1 = 4.5 ft )
Fy = 0:
V  2 kips = 0
V = + 2 kips
M 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 max
= 14.00 kips on CD = 114.0 kip ft at A
M
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:
M A = (15 ft ) B  (12 ft )( 2 kips/ft )( 6 ft )  ( 6 ft )(12 kips ) = 0
B = 14.4 kips
Fy = 0: Ay  12 kips  ( 2 kips/ft )( 6 ft ) + 14.4 kips
A y = 9.6 kips
Along AC:
Fy = 0: 9.6 kips  V = 0 V = 9.6 kips M 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:
Fy = 0: 9.6 kips  12 kips  V = 0 V = 2.4 kips M 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:
Fy = 0: V  x3 ( 2 kips/ft ) + 14.4 kips = 0 V = 14.4 kips + ( 2 kips/ft ) x3 V = 2.4 kips at D M 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 = 57.6 kip ft at C
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:
Fy = 0: 18 kN  V = 0 M J = 0: M  x (18 kN )
V = 18 kN M = (18 kN ) x
M = 36 kN m at C ( x = 2 m )
Along CD:
Fy = 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 ) M K = 0: M + x1 ( 4 kN/m ) x1 + (8 kN ) x1  ( 2 m + x1 )(18 kN ) = 0 2
2 M = 36 kN m + (10 kN/m ) x1  ( 2 kN/m ) x1
M = 48.5 kN m at x1 = 2.5 m
Complete diagram by symmetry (b) From diagrams:
V
max
= 18.00 kN on AC and DB
max
M
= 48.5 kN m at center
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)
M 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
Fy = 0:  ( 2 m )( 2 kN/m ) + D y  ( 3 m )( 4 kN/m )  22 kN + 22 kN = 0
D y = 16 kN
Along AC:
Fy = 0:  x ( 2 kN/m )  V = 0 V =  ( 2 kN/m ) x M J = 0: M + M =  (1 kN/m ) x 2 V = 4 kN at C x ( 2 kN/m )( x ) 0 2 M = 4 kN m at C
Along CD:
Fy = 0:  ( 2 m )( 2 kN/m )  V = 0
V = 4 kN
M 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:
Fy = 0:  ( 2 kN/m )( 2 m ) + 16 kN  V = 0
V = 12 kN
M 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:
Fy = 0: V  x4 ( 4 kN/m )  22 kN + 22 kN = 0 V = ( 4 kN/m ) x4 M 0 = 0: M + V = 12 kN at E
x4 ( 4 kN/m ) x4  (1 m )( 22 kN ) = 0 2 M = 4 kN m at E
2 M = 22 kN m  ( 2 kN/m ) x4
Along FB:
Fy = 0: V + 22 kN = 0
V = 22 kN
M 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 = 22.0 kN m at F
PROBLEM 7.41
Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a)
Fy = 0: (12 m ) w  ( 6 m )( 3 kN/m ) = 0
w = 1.5 kN/m
Along AC:
Fy = 0: x (1.5 kN/m )  V = 0 V = 4.5 kN at C
M J = 0: M  M = ( 0.75 kN/m ) x 2
V = (1.5 kN/m ) x
x (1.5 kN/m )( x ) = 0 2 M = 6.75 N m at C
Along CD:
Fy = 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 M 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 M = 13.5 kN m at center
( x = 6 m)
= 4.50 kN at C and D
Finish by symmetry (b) From diagrams:
M V
max
max
= 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 bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION (a) FBD Beam:
Fy = 0: ( 4 m )( w )  ( 2 m )(12 kN/m ) = 0 w = 6 kN/m
Along AC:
Fy = 0:  x ( 6 kN/m )  V = 0
V =  ( 6 kN/m ) x
V = 6 kN at C ( x = 1 m ) M J = 0: M + M =  ( 3 kN/m ) x 2 x ( 6 kN/m )( x ) = 0 2 M = 3 kN m at C
Along CD:
Fy = 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 x1 ( 6 kN/m ) x1 = 0 2
M K = 0: M + ( 0.5 m + x1 )( 6 kN/m )(1 m ) 
2 M = 3 kN m  ( 6 kN ) x1 + ( 3 kN/m ) x1
M = 6 kN m at center
( x1 = 1 m )
V = 6.00 kN at C and D = 6.00 kN at center
Finish by symmetry (b) From diagrams:
max
M
max
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 bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION
(a) FBD Beam:
Fy = 0: ( 4.5 ft ) w  900 lb  900 lb = 0 w = 400 lb/ft
Along AC:
Fy = 0: x ( 400 lb )  V = 0 V = 360 lb at C M J = 0: M  M = ( 200 lb/ft ) x 2
V = ( 400 lb ) x
( x = 0.9 ft )
x ( 400 lb/ft ) x = 0 2 M = 162 lb ft at C
Along CD:
Fy = 0: ( 0.9 ft + x1 )( 400 lb/ft )  900 lb  V = 0 V = 540 lb + ( 400 lb/ft ) x1 M 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 )
= 540 lb at C + and D 
Finish by symmetry (b) From diagrams:
V
max
M
max
= 203 lb ft at center
PROBLEM 7.44
Solve Prob. 7.43 assuming that a = 1.5 ft.
SOLUTION
(a) FBD Beam:
Fy = 0: ( 4.5 ft ) w  900 lb  900 lb = 0 w = 400 lb/ft
Along AC:
Fy = 0: V = ( 400 lb/ft ) x
x ( 400 lb/ft )  V = 0 V = 600 lb at C
( x = 1.5 ft )
M J = 0: M  M = ( 200 lb/ft ) x 2
x ( 400 lb/ft ) x = 0 2 M = 450 lb ft at C
Along CD:
Fy = 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 M 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
max max
= 600 lb at C  and D +
M
= 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 bendingmoment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.
SOLUTION FBD angle CE:
(a) By symmetry:
T =
3.33 kN + 3 kN = 3.165 kN 2 PC = T = 3.165 kN M C = 0.3165 kN m
Fy = 0: T  PC = 0
M C = 0: M C  ( 0.1 m )( 3.165 kN ) = 0 By symmetry:
PD = 3.165 kN; M D = 0.3165 kN m
Along AC:
Fy = 0:  x (1.11 kN/m )  V = 0 V =  (1.11 kN/m ) x V = 1.332 kN at C
( x = 1.2 m )
M J = 0: M + M = ( 0.555 kN/m ) x 2
x (1.11 kN/m ) x = 0 2 M =  0.7992 kN m at C
Along CI:
Fy = 0:  (1.11 kN/m ) x + 3.165 kN  V = 0 V = 3.165 kN  (1.11 kN/m ) x M k = 0: M + (1.11 kN/m ) x  ( x  1.2 m )( 3.165 kN )  ( 0.3165 kN m ) = 0 V = 1.5 kN at I
( x = 1.5 m )
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:
(a) By symmetry:
T =
3.33 kN + 3 kN = 3.165 kN 2 PC = T = 3.165 kN M C = 0.3165 kN m
Fy = 0: T  PC = 0
M C = 0: M C  ( 0.1 m )( 3.165 kN ) = 0
By symmetry:
Along AC:
PD = 3.165 kN
M D = 0.3165 kN m
Fy = 0:  (1.11 kN/m ) x  V = 0 V =  (1.11 kN/m ) x V =  0.999 kN at C
( x = 0.9 m )
M J = 0: M + M =  ( 0.555 kN/m ) x 2
x (1.11 kN/m ) x = 0 2 M =  0.44955 kN m at C
Along CI:
Fy = 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 M K = 0: M  0.3165 kN m + ( x  0.9 m )( 3.165 kN ) + x (1.11 kN/m ) x = 0 2 V = 2.166 kN at C
( x = 1.5 m )
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 bendingmoment 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:
Fy = 0: 1.2 kN + C y = 0 M C = 0: ( 0.4 m )(1.2 kN )  M C = 0
C y = 1.2 kN
M C = 0.48 kN m
FBD Beam:
M A = 0: (1.2 m ) B + 0.48 kN m  ( 0.8 m )(1.2 kN ) = 0
B = 0.4 kN
Fy = 0: Ay  1.2 kN + 0.4 kN = 0
A y = 0.8 kN
Along AC:
Fy = 0: 0.8 kN  V = 0 M 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:
Fy = 0: V + 0.4 kN = 0 M 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 bendingmoment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD angle:
Fy = 0: Fy  600 N = 0 Fy = 600 N M = 180 N m
M Base = 0: M  ( 0.3 m )( 600 N ) = 0
All three angles are the same.
FBD Beam:
M 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
Fy = 0: Ay  3 ( 600 N ) + 1200 N = 0
A y = 600 N
Along AC:
Fy = 0: 600 N  V = 0
V = 600 N
M J = 0: M  x ( 600 N ) = 0 M = ( 600 N ) x = 180 N m at x = .3 m
Along CD:
Fy = 0: 600 N  600 N  V = 0
V =0
M 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:
Fy = 0: V  600 N + 1200 N = 0
V = 600 N
M 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:
Fy = 0: V + 1200 N = 0 M 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
= 1200 N on EB
= 540 N m at D +
max
PROBLEM 7.49
Draw the shear and bendingmoment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Whole:
M 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
Fx = 0:  Dx + G = 0
D x = 6.75 kN D y = 9 kN
Beam AB, with forces D and G replaced by equivalent force/couples at C and F
Fy = 0: Dy  1.5 kN  6 kN  1.5 kN = 0
Along AD:
Fy = 0: 1.5 kN  V = 0 M 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:
Fy = 0: 1.5 kN + 9 kN  V = 0
V = 7.5 kN
M 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:
Fy = 0: V  1.5 kN = 0
V = 1.5 kN
M 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:
Fy = 0: V  1.5 kN = 0 M L = 0:  M  x3 (1.5 kN ) = 0
V = 1.5 kN M = ( 1.5 kN ) x3
M = 1.8 kN m at x3 = 1.2 m
From diagrams:
M
V
max
= 7.50 kN on DE
max
= 7.20 kN m at D +
PROBLEM 7.50
Neglecting the size of the pulley at G, (a) draw the shear and bendingmoment diagrams for the beam AB, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION FBD Whole:
(a)
M A = 0: ( 0.5 m )
12 5 D + (1.2 m ) D  ( 2.5 m )( 480 N ) = 0 13 13 D = 1300 N
Fy = 0: Ay +
5 (1300 N )  480 N = 0 13 A y = 20 N
Ay = 20 N
Beam AB with pulley forces and Along AE: force at D replaced by equivalent forcecouples at B, F, E
Fy = 0: 20 N  V = 0 M J = 0: M + x ( 20 N ) V = 20 N M =  ( 20 N ) x
M = 24 N m at x = 1.2 m
Along EF:
Fy = 0: V  288 N  480 N = 0
V = 768 N
M 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:
Fy = 0: V  480 N = 0 M 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
= 768 N along EF
= 624 N m at E +
M
max
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 bendingmoment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.)
SOLUTION FBD Beam:
Fy = 0: Lw  2 P = 0
w=2
P L
M J = 0: M  M =
x 2P x = 0 2 L P 2 a at x = a L
M =
P 2 x L
Along AC:
M K = 0: M + ( x  a ) P  M = P (a  x) + P 2 Pa 2 x = L L
x 2P x = 0 2 L
at
x=a
L L M = P a  at x = 2 4
Along CD:
This is M min by symmetrysee moment diagram completed by symmetry. For minimum M
max
, set M max = M min :
P a2 L = P a  L 4
or Solving: Positive answer
a 2 + La  a=
L2 =0 4
1 2 L 2
(a) (b)
M
max
a = 0.20711L = 0.932 ft
= 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 F = 3.165 kN M = 0.3165 kN m
Fy = 0: T  F = 0
M 0 = 0: M  ( 0.1 m )( 3.165 kN ) = 0
M J = 0: M +
x (1.11 kN/m ) x = 0 2
2
Along AC:
M =  ( 0.555 kN/m ) x 2 =  ( 0.555 kN/m )(1.5 m  a )
at C
( this is M min )
M K = 0: M  0.3165 kN m +
Along CI:
x (1.11 kN/m ) x 2  x  (1.5 m  a ) ( 3.165 kN ) = 0
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 :
2
 0.93225 kN m + ( 3.165 kN ) a = ( 0.555 kN/m )(1.5 m  a )
Yielding: Solving:
a 2  ( 8.7027 m ) a + 3.92973 m 2 = 0 a = 4.3514 13.864 = 0.4778 m, 8.075 m
Second solution out of range, so
(a)
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 By symmetry: Along AC:
M J = 0: M  x ( 60 kN  P ) = 0 M = ( 60 kN  P ) x Ay = B = 60 kN  P
M = 120 kN m  ( 2 m ) P at x = 2 m
Along CD:
M 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:
M 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
(a)
P = 40.0 kN M
max
M min = 120 kN m  ( 4 m) P
(b)
= 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 FBD Beam:
M A = 0: M A  (1.5 ft )(1 kip )  ( 3.5 ft )( 4 kips ) + ( 3.5 ft + a )( 2 kips ) = 0
M A = 8.5 kip ft  ( 2 kips ) a
Fy = 0: Ay  1 kip  4 kips + 2 kips = 0
A y = 3 kips
Along AC:
M 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 ) M K = 0:  M + x1 ( 2 kips ) = 0 M = ( 2 kips ) a at D M L = 0: ( x2 + a )( 2 kips )  x2 ( 4 kips )  M = 0 M = ( 2 kips ) x1
Along DB:
Along CD:
M = ( 2 kips ) a  ( 2 kips ) x2 M = ( 2 kips ) a  4 kip ft at C
( see above )
For minimum M
max
, 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) So
M max = ( 2 kips ) a = 4.25 kip ft
a = 2.13 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
M A = 0: ( a ft ) D  ( 4 ft )( 375 lb )  ( 8 ft )( 375 lb ) = 0
FBD Beam:
D=
4500 lb a 4500 lb = 0 a
Fy = 0: Ay  2 ( 375 lb ) +
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 M C = 0: M  ( 4 ft ) 750  lb = 0 a 18000 M = 3000  lb ft a
M D = 0:  ( 8  a ) ft ( 375 lb )  M = 0
Segment AC:
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)
M max = 375 ( 8  a ) = 401.92 lb ft
a = 6.93 ft
(b)
M
max
= 402 lb ft
PROBLEM 7.56
Solve Prob. 7.55 assuming that P = 750 lb and Q = 375 lb.
SOLUTION FBD Beam:
M D = 0:  ( a ft ) Ay + ( a  4 ) ft ( 750 lb )  ( 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 Mdiagram are straight, so M max and M min occur at C and D.
Segment AC:
6000 M C = 0: M  ( 4 ft ) 1125  lb = 0 a 24000 M = 4500  lb ft a
M 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  a 2 + 4a  64 = 0 a = 6.2462 ft 24000 = 375 ( 8  a ) a a = 2 68 ( need + )
(a)
a = 6.25 ft
Then
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 M due to distributed load:
M J = 0: M  x wx = 0 2
1 M =  wx 2 2
M due to counter weight:
M J = 0: M + xw = 0 M = wx
(a) Both applied:
M = Wx 
w 2 x 2
dM W = W  wx = 0 at x = dx w
And here M =
W2 > 0 so M max ; M min must be at x = L 2w 1 2 wL . For minimum M 2
max
So M min = WL 
set M max = M min , so
W2 1 = WL + wL2 or W 2 + 2wLW  w2 L2 = 0 2w 2
W = wL 2w2 L2 (need +)
W =
(
2  1 wL = 0.414wL
)
(b) w may be removed: Without w,
M max
W2 = = 2w
(
2 1 2
)
2
wL2
M max = 0.858wL2
M = Wx, M max = WL at A M = Wx  M min = WL  w 2 x , 2 M max = W2 W at x = 2w w
With w (see part a)
1 2 wL at x = L 2
PROBLEM 7.57 CONTINUED
For minimum M max , set M max ( no w ) = M min ( with w )
WL = WL + 1 2 1 wL W = wL 2 4 M max = W = 1 2 wL 4 1 wL 4
With
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
or A y = D =
wL 4
V jumps to Ay =
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 =
wL2 1 L wL + 16 2 4 4
Then, M is linear with
dM wL = to D dy 4
MD = 0 V M
max
=
wL 4
max
=
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
=
3wL2 8
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 .
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)
M C = 0: LAy  M 0 = 0 Shear Diag: V jumps to 
Ay =
M0 L
M0 dV at A and is constant = 0 all the way to C L dx
V
max
=
M0 L
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
=
M0 2
PROBLEM 7.62
Using the method of Sec. 7.6, solve Prob. 7.33.
SOLUTION
(a) and (b)
M 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
Moment Diag:
max
= 12.63 kips
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
PROBLEM 7.63
Using the method of Sec. 7.6, solve Prob. 7.36.
SOLUTION
(a) and (b)
FBD Beam:
M 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
Fy = 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
Moment Diag:
max
= 810 N
M is piecewise linear starting with M A = 0
M C = 0  0.2 m ( 0.54 kN ) = 0.108 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 D = 0.108 kN m + ( 0.5 m )( 0.54 kN ) = 0.162 kN m
M
max
= 0.162 kN m = 162.0 N m
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
Moment Diag:
max
= 6.60 kN
M A = 0.
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 = 3.6 kN
dM From C to D, M is linear dx
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 + = 5.82 kN m M
max
1 (1 m )( 3.6 + 6.6 ) kN 2 = 5.82 kN m
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 bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
SOLUTION
(a) and (b) M B = 0: ( 3 ft )(1 kip/ft )( 6 ft ) + ( 8 ft )( 6 kips ) + (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
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
PROBLEM 7.66
Using the method of Sec. 7.6, solve Prob. 7.37.
SOLUTION
(a) and (b)
FBD Beam:
Fy = 0: Ay + ( 6 ft )( 2 kips/ft )  12 kips  2 kips = 0
A y = 2 kips
M 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
max
= 14.00 kips
Moment Diag:
M A = 114 kip ft.
dM increasing with V and From A to C, M is parabolic dx M C = 114 kip ft + 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
1 ( 2 kips + 14 kips )( 6 ft ) 2
= 114.0 kip ft
PROBLEM 7.67
Using the method of Sec. 7.6, solve Prob. 7.38.
SOLUTION
(a) and (b)
FBD Beam:
kips M 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
Moment Diag:
M is linear from A to C dM = 9.6 kips/ft dx
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  =0 M
max
1 ( 2.4 kips + 14.4 kips )( 6 ft ) = 0 2 = 57.6 kip ft
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
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
PROBLEM 7.69
Using the method of Sec. 7.6, solve Prob. 7.40.
SOLUTION
(a) and (b)
FBD Beam: M 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
Fy = 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
max
= 22.0 kN
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
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
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: M 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
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
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: M 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
Fy = 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 bendingmoment diagrams, (b) determine the magnitude and location of the maximum bending moment.
SOLUTION
(a)
FBD Beam:
M 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 1 ( 2.7 kips ) x1; 2 x1 = 2.25 m
M max = 5.4 kip ft +
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 bendingmoment diagrams, (b) determine the magnitude and location of the maximum bending moment.
SOLUTION
FBD Beam:
(a) M 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 x1 = 3.2 ft
1 ( 6.4 kips ) x1; 2
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 bendingmoment 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:
Fy = 0: Ay  (16 kN/m )( 2 m )  6 kN + P = 0 Ay = 38 kN  P (a) (b)
A y = 24 kN A y = 18 kN
M 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
(a) (b) (b)
M A = 17 kN m 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) (b) VC = 8 kN VC = 14 kN V = 0 = VA  (16 kN/m ) x1 (a) (b) x1 = 1.5 m 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) 1 VA x1. 2
M max = 17 kN m +
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) 1 VC ( 2 m  x1 ) 2
M C = 35 kN m  M C = 57.125 kN m 
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 B M = 0 and M D = (1.5 m ) P
(a) (b)
M D = 21 kN m M D = 30 kN m
PROBLEM 7.75
For the beam shown, draw the shear and bendingmoment 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:
M A = 0: ( 4 m ) B  (1 m )( 20 kN/m )( 2 m )  M = 0
B = 10 kN +
(a) (a) (b)
M 4m
B = 10 kN B = 13 kN
Fy = 0: Ay  ( 20 kN/m )( 2 m ) + B = 0
Ay = 40 kN  B
(a) (b)
A y = 30 kN 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) (b)
VC = 10 kN VC = 13 kN
V = 0 = Ay  ( 20 kN/m ) x1 at x1 = Ay m 20 kN
(a) (b)
x1 = 1.5 m x1 = 1.35 m V is constant from C to B.
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)
1 Ay x1. 2 1 ( 30 kN )(1.5 m ) = 22.5 kN m 2 1.500 m from A
M
max
=
(b)
M max = 12 kN m +
1 ( 27 kN )(1.35 m ) = 30.225 kN m 2
M M C = M max 
(a) (b)
max
= 30.2 kN 1.350 m from A
1 VC ( 2 m  x1 ) 2
M C = 20 kN m 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 bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
SOLUTION
FBD Beam:
(a) M 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
M B = M max 
1 (125 kN )( 6 m  x1 ) 2 1 = 165.3125 kN m  (125 kN )( 6  2.875 ) m 2 = 30 kN m as expected. (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:
M 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 M
max
(b)
= 195.3 kN m ( 3.125 m from A ) 1 (115 kN )( 6  3.125) m 2
M B = 195.3125 kN m 
M B = 30 kN m as expected.
PROBLEM 7.78
For beam AB, (a) draw the shear and bendingmoment 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 forcecouple at C:
M 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
Fy = 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 4kN force applied at E is directed upward.
SOLUTION
(a) Replacing the load at E with equivalent forcecouple at C.
M 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=
Fy = 0: Ay +
22 kN 3
22 kN  ( 8 kN/m )( 3 m ) + 4 kN  2 kN = 0 3
Ay = Shear Diag:
VA = Ay =
44 kN 3
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 44 11 kN  ( 8 kN/m ) x1 at x1 = m. 3 6 16 kN. 3
V =0=
At C, V increases 4 kN to  At D, V increases
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 M max = 13.44 kN m at 1.833 m from A
(b)
MC =
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 bendingmoment curves, (b) draw the shear and bendingmoment diagrams, (c) determine the magnitude and location of the maximum bending moment.
SOLUTION
x Distributed load w = w0 1  L M A = 0: L 1 w0 L  LB = 0 3 2 1 wL w0 L + 0 = 0 2 6 w0 L , 3 1 total = w0 L 2
B= Ay =
w0 L 6 w0 L 3
Fy = 0: Ay 
Shear:
VA = Ay = Then
x dV x =  w V = VA  w0 1  dx 0 L dx
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 = 
2
w0 L ; 6
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
(a)
1 x 1 x 2 V = w0 L  + 3 L 2 L
2
1 x 1 x 2 1 x 3 M = w0 L  + 6 L 3 L 2 L (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 bendingmoment curves, (b) draw the shear and bendingmoment diagrams, (c) determine the magnitude and location of the maximum bending moment.
SOLUTION
Distributed load x w = w0 4  1 L
Shear:
V =
dV = w, and V ( 0 ) = 0, so dx
0
x
 wdx = 
0
x/L
x Lwd L
2 x x/L x x x  2 V = wo L 1  4 d = w L 0 L L 0 L L
Notes: At x = L, V = w0 L And V = 0 at
x x = 2 L L
2
or
x 1 = L 2
1 x Also V is max where w = 0 = 4 L Vmax = 1 w0 L 8 M ( 0 ) = 0,
M =
Moment:
dM =V dx
x x V d L L 2 x / L x  2 x d x M = w0 L2 0 L L L
0 vdx = L0
x
x/ L
(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 M
max
(c)
= M min =
w0 L2 at B 6
PROBLEM 7.82
For the beam shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum bending moment. (Hint: Derive the equations of the shear and bendingmoment curves for portion CD of the beam.)
SOLUTION (a) FBD Beam:
M B = 0: 1 w0 ( 3a )  5aAy = 0 A y = 0.9w0a 2 1 Fy = 0: 0.9w0a  w0 ( 3a ) + B = 0 2
( 3a )
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
2 1 x1 V = w0a  0.6 + 6 a
Then from D to C, w = w0
x Note: V = 0 at 1 = 3.6, x1 = a
2
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 0.6  1 x1 dx1, M = 0.6w0a + w0a 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 bendingmoment curves, (b) determine the maximum bending moment.
SOLUTION (a)
wg L = Fy = 0: wg L 
0
L 4w0
L2
( Lx  x ) dx = 0
2
4w0 1 2 1 3 2 LL  L = w0 L 3 3 L2 2
wg =
2w0 3
Define =
x dx so d = net load w = 4w0 L L
x x 2 2   w0 L L 3
or
1 w = 4w0  +  2 6 V = V(0) 
0 4w0 L  6 + 
1 1 1
1
2
d =
0 + 4w0 L + 2  3 6 2 3
V = M = M0 + = 2 w0 L  3 2 + 2 3 3
(
)
)
0 Vdx = 0 + 3 w0 L 0
2
x
2
(  3
(
2
+ 2 3 d
)
2 1 1 1 w0 L2 2  3 + 4 = w0 L2 2  2 3 + 4 3 2 3 2 1 2
(b)
Max M occurs where V = 0 1  3 + 2 2 = 0 =
1 1 1 w L2 1 2 M = = 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 bendingmoment diagrams for the beam.
SOLUTION FBD ACD: (a)
M 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:
M E = 0: (1 ft ) B  0.8 kip ft = 0
B = 0.8 kip
FBD Beam: M 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 Fy = 0: 3.325 kips + 0.8 kips  1.275 kips  ( 2 kips/ft )(1ft )  P = 0 P = 0.85 kip
P = 850 lb Q = 1.275 kips
(a)
(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)
M 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:
M E = 0: (1 ft ) B  0.86 kip ft B = 0.86 kip
FBD Beam:
M 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  P  ( 2 kips/ft )(1 ft ) = 0 P = 0.66 kips
P = 660 lb Q = 1.460 kips
Fy = 0: 3.26 kips + 0.86 kips  1.460 kips
(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 bendingmoment diagrams for the beam.
SOLUTION (a)
2 Ay  P = 8.2 kN M D = 0: 7 kN m + (1 m )( 0.6 kN/m )( 2 m )  ( 2 m ) Ay = 0 (1)
FBD AD:
M E = 0: ( 2 m ) B  (1 m ) Q  (1 m )( 0.6 kN/m )( 2 m )
FBD EB:
2B  Q = 6.2 kN M A = 0:
 5 kN m = 0 (2)
( 6 m ) B  (1 m ) P  ( 5 m ) Q  ( 3 m )( 0.6 kN/m )( 6 m ) = 0
6 B  P  5Q = 10.8 kN  (6 m) A = 0 6 A  Q  5P = 10.8 kN (4) (3)
M B = 0: (1 m ) Q + ( 5 m ) P + ( 3 m )( 0.6 kN/m )( 6 m )
Solving (1)(4):
A y = 7.4 kN , B = 3.4 kN
P = 6.60 kN , Q = 600 N
(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 ( 2.2 kN ) 3 m = 3.1 kN m 2 3
M F = 7.133 kN m 
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
M D = 0: 3.6 kN m + (1 m ) P + (1 m )( 0.6 kN/m )( 2 m )  ( 2 m ) Ay = 0 2 Ay  P = 4.8 kN (1)
FBD AD:
(a)
FBD EB:
M E = 0:
( 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)
M 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 (4)
P = 660 N , Q = 2.28 kN
Solving (1)(4)
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 1 (1.47 kN )( 2.45 m ) = 4.231 kN m 2 1 ( 0.93 kN )(1.55 m ) = 3.51 kN m 2
M max = 2.43 kN m + M F = 4.231 kN m 
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 FBD cable:
M A = 0: (10 ft ) Dy  8 ft ( 450 lb )  4 ft ( 600 lb ) = 0
Dy = 600 1b
Fy = 0: Ay + 600 lb  600 lb  450 lb = 0
Ay = 450 lb
Fx = 0: Ax  Dx = 0
(1) = Ax
FBD pt D: So And
600 lb D T = x = CD : Dx = 800 lb 3 4 5
TCD = 1000 lb
FBD pt A:
800 lb 450 lb = dB 4 ft
(a) (b)
d B = 2.25 ft
A x = 800 lb A y = 450 lb
TAB =
(800 lb )2 + ( 450 lb )2
(c)
= 918 lb
Tmax = TCD = 1000 lb
So
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 FBD cable:
M A = 0: (10 ft ) Dy  ( 8 ft )( 450 lb )  ( 4 ft )( 600 lb ) = 0 D y = 600 1b Fy = 0: Ay + 600 lb  600 lb  450 lb = 0 A y = 450 lb
FBD pt D:
Fx = 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
= 398 lb = Ax
dC 2ft = 600 lb 398lb
dC = 3.015 ft
FBD pt. A:
dB 4ft = 450 lb 398lb
(a) (b)
d B = 4.52 ft 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: M E = 0:
( 4 m )(1.2 kN ) + (8 m )( 0.8 kN ) + (12 m )(1.2 kN )  ( 3 m ) Ax  (16 m ) Ay = 0
(1)
3 Ax + 16 Ay = 25.6 kN
FBD ABC:
M C = 0: ( 4 m )(1.2 kN ) + (1 m ) Ax  ( 8 m ) Ay = 0
Ax  8 Ay = 4.8 kN
(2)
Ay = 1 kN
Solving (1) and (2)
Ax = 3.2 kN
So A = 3.35 kN (b) cable:
Fx = 0:  Ax + Ex = 0
Ex = Ax = 3.2 kN
17.35
Fy = 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)
M 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)
M C = 0: ( 4 m )(1.2 kN )  ( 0.75 m ) Ax  ( 8 m ) Ay = 0
FBD ABC: Solving (1) and (2)
0.75 Ax + 8 Ay = 4.8 kN
Ax = 32 kN, 3 Ay =  0.4 kN
(2)
So
A = 10.67 kN
2.15
Note: this implies d B < 3 m (in fact d B = 2.85 m) (b) FBD cable:
Fx = 0:  32 kN + Ex = 0 3 Ex = 32 kN 3
Fy = 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: (a)
M A = 0: ( 2.4 ft ) Ex + ( 8 ft ) E y  ( 2 ft )( 360 )  ( 4 ft )( 720 lb )  ( 6 ft )( 240 lb ) = 0 0.3Ex + E y = 630 lb
(1)
FBD CDE:
M C = 0:  (1.2 ft ) Ex + ( 4 ft ) E y  ( 2 ft )( 240 lb ) = 0  0.3Ex + E y = +120 lb
(2)
E y = 375 lb
Solving (1) and (2)
Ex = 850 lb
(a) (b) cable:
Fx = 0:  Ax + Ex = 0
E = 929 lb
23.8
Ax = Ex = 850 lb
Fy = 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
M 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:
Fy = 0: E y  240 lb = 0
E y = 240 lb
FBD Cable:
M A = ( 2.4 ft ) Ex + ( 8 ft )( 240 lb )  ( 6 ft )( 240 lb )  ( 4 ft )( 720 lb )  ( 2 ft )( 360 lb ) = 0 E x = 1300 lb
So From Segment DE:
M D = 0: ( 2 ft ) E y  ( dC  2.4 ft ) Ex = 0
dC = 2.4 ft + Ey Ex
( 2 ft ) = ( 2.4 ft ) +
(a)
240 lb ( 2 ft ) = 2.7692 ft 1300 lb dC = 2.77 ft
From FBD Cable:
Fx = 0:  Ax + Ex = 0
A x = 1300 lb
Fy = 0: Ay  360 lb  720 lb  240 lb + E y = 0
A y = 1080 lb
(b)
A = 1.690 kips E = 1.322 kips
39.7 10.46
PROBLEM 7.94
An oil pipeline is supported at 6m 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.
M 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
(1)
FBD ABC:
M C = 0: ( 6 m )( 4 kN ) + ( 7 m ) Ax  (12 m ) Ay = 0 7 Ax  12 Ay = 24 kN
(2)
Ay =
20 kN 3
Solving (1) and (2) From FBD Cable:
A x = 8 kN
Fx = 0:  Ax + Fx = 0
Fx = Ax = 8 kN
Fy = 0: Ay  4 ( 4 kN ) + Fy = 0 20 28 Fy = 16 kN  Ay = 16  kN > Ay kN = 3 3
So FBD DEF: (a)
TEF > TAB
Tmax = TEF =
2
Fx2 + Fy2
2
Tmax =
(18 kN )
28 + kN = 12.29 kN 3
28 M 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.
M 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 M C = 0: (18 ) Fy  ( 9 m ) Fx  (12 m )( 4 kN )  ( 6 m )( 4 kN ) = 0
(1)
FBD CDEF: Solving (1) and (2)
TEF =
Fx  2Fy = 8 kN
(2)
Fy = 10 kN
Fx = 12 kN
(10 kN )2 + (12 kN )2
= 15.62 kN
Since slope EF > slope AB this is Tmax (a) Also could note from FBD cable
Fy = 0: Ay + Fy  4 ( 4 kN ) = 0 Ay = 16 kN  12 kN = 4 kN Tmax = 15.62 kN
Thus FBD DEF: (b)
Ay < Fy
and
TAB < TEF
M 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 FBD BC:
W = ( 8 kg ) 9.81 m/s 2 = 78.48 N M A = 0: ( 3.6 m ) P  ( 2.4 m ) 3W  aW = 0 2
(
)
a P = W 1 + 3.6 m Fx = 0: T1x + P = 0 Fy = 0: T1y  W  3 W =0 2 so 5Wa 5.6 m
P = 1.4516W
(1)
T1x = P T1 y = 5W 2
But
T1 y T1x
=
2.8 m a P=
5W 2.8 m = a 2P
(2)
Solving (1) and (2): So (a) (b)
a = 1.6258 m,
P = 1.4516 ( 78.48 ) = 113.9 N
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
M 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:
M C = 0: (12 ft ) Dy  ( 9 ft ) Dx = 0 3Dx = 4 Dy
(1)
M B = 0: ( 30 ft ) D y  (15 ft ) Dx  (18 ft )( 50 lb ) = 0
FBD BCD: Solving (1) and (2)
2D y  Dx = 60 lb
D x = 120 lb D y = 90 lb
(2)
FBD Cable:
M 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:
M C = 0: (12 ft ) Dy  ( 9 ft ) Dx = 0 4D y = 3Dx
(1)
FBD BCD:
M B = 0: ( 30 ft ) D y  (15 ft ) Dx  (18 ft )( 22 lb ) = 0 10 Dy  5Dx = 132 lb
(2)
D y = 39.6 lb
Solving (1) and (2)
D x = 52.8 lb
FBD Whole:
M 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
T = 90 5 kN
1 Fy = 0: 2 T  180 kN = 0 5
Tx = 180 kN
Ty = 90 kN
Segment DE:
M E = 0: ( 7.5 m )( P  180 kN ) + ( 6 m )( 90 kN ) = 0
P = 108.0 kN
Segment AB:
M 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: By symmetry:
TBC = TCD = T
T = 90 5 kN
1 Fy = 0: 2 T  180 kN = 0 5
Tx = 180 kN FBD DE:
Ty = 90 kN
M E = 0: ( 9 m )( P  180 kN ) + ( 6 m )( 90 kN ) = 0
P = 120.0 kN
FBD AB:
M 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 direction. L = 30 m we can approximate the load as evenly distributed in the horizontal
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 halfcable:
(
)
1 chord point. 4
M B = 0: (15 m )( 294.3 N )  (1.2 m ) Tmin = 0 Tmin = 3678.75 N = Tmax Fy = 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 +  + 3 30 5 30
= 30.048 m
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
Halfcable FBDs:
T1x = T2 x to create zero horizontal force on tower thus T01 = T02
FBD I:
M B = 0: (15 m ) w ( 30 m )  h1T0 = 0
h1 FBD II:
( 450 m ) w =
2
T0
M B = 0: ( 2 m ) T0  (10 m ) w ( 20 m ) = 0
T0 = (100 m ) w (a) FBD I: Fx = 0: T1x  T0 = 0 T1x = (100 m ) w h1
( 450 m ) w = 4.50 m =
2
(100 m ) w
Fy = 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
(
)
PROBLEM 7.103 CONTINUED
FBD II: Fy = 0: T2y  ( 20 m ) w = 0 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 halfspan:
W = ( 9.75 kips/ft )(1750 ft ) = 17, 062.5 kips M 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)
*
l = 2sB = 3575 ft
To get 3digit accuracy, only two terms are needed.
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 halfspan:
(a)
2075 ft M 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
y + x
4
= 56, 400 kips
2 2 y 2 s = x 1 +  3 x 5
l = 4284 ft
2 4 2 464 ft 2 464 ft sB = ( 2075 ft ) 1 +  + 3 2075 ft 5 2075 ft
sB = 2142 ft
l = 2s B
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:
M P = 0: (Tmax  WB ) r = 0 Tmax = WB = 196.2 N
FBD halfspan:*
2 T0 = Tmax  W 2 =
(196.2 N )2  ( 4.91 N )2
= 196.139 N
25 m M B = 0: ( 4.905 N )  h (196.139 N ) = 0 2 (a) (b) *See note Prob. 7.103 h = 0.3126 m = 313 mm
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 5m length of rope as shown. Knowing that the current exerts on the hull of the ship a 300N 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:
Fx = 0: T0  300 N = 0
FBD halfspan:*
T0 = 300 N
Tmax = T02 + W 2 = (b) M A = 0: hTx  L W =0 4 h= LW 4Tx
( 300 N )2
= ( 54 N ) = 305 N
2
2 2 4 s = x 1 + + 3 x
but
yA = h =
LW 4Tx
so
yA W = 2Tx xA
L ( 2.5 m ) = 2
2 2 53.955 N 1 +  3 600 N
L = 4.9732 m h = 224 mm
So h = *See note Prob. 7.103
LW = 0.2236 m 4Tx
PROBLEM 7.108
The center span of the VerrazanoNarrows 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
y + x
4
Knowing Winter:
l = 2sTOT
2 2 2 h 2 h = L 1 +  + 3 L/2 5 L/2
2 4 2 386 ft 2 386 ft lw = ( 4260 ft ) 1 +  + 3 2130 ft 5 2130 ft
= 4351.43 ft
Summer:
2 4 2 394 ft 2 394 ft ls = ( 4260 ft ) 1 +  + 3 2130 ft 5 2130 ft
= 4355.18 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
2 2
L + = 2sTOT
2 2 h = L 1 +  3 L/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:
M A = 0: (1100 ft ) TBy  ( 496 ft ) TBx  ( 550 ft )W = 0 11TBy  4.96TBx = 5.5W (1)
FBD CB:
M C = 0: ( 550 ft ) TBy  ( 278 ft ) TBx  ( 275 ft )
11TBy  5.56TBx = 2.75W Solving (1) and (2) Solving (1) and (2)
W =0 2
(2)
TBy = 28,798 kips TBx = 51, 425 kips
tan B = TBy TBx
2 2 Tmax = TB = TBx + TBy
So that
(a)
Tmax = 58,900 kips
(b)
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:
M A = 0: (13.5 ft ) T0 
a ( 57.5 lb/ft ) a = 0 2
T0 = 2.12963 lb/ft 2 a 2 M B = 0: 60 ft  a ( 57.5 lb/ft )( 60 ft  a )  ( 6 ft ) T0 = 0 2
(
)
(1)
6T0 = 28.75 lb/ft 2 3600 ft 2  (120 ft ) a + a 2
(
)
(2)
Using (1) in (2)
0.55a 2  (120 ft ) a + 3600 ft 2 = 0
a = 36 ft (180 ft out of range)
Solving: a = (108 72 ) ft
So (a)
2
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
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
M A = 0: y AT0  yA = wa 2 2T0
a wa = 0 2
M B = 0:  yBT0 + yB = wb 2 2T0
b wb = 0 2
d = ( y B  yB ) =
w 2 b  a2 2T0
2
(
)
)
2
But
2 2 T0 = TB  ( wb ) = Tmax  ( wb )
2 ( 2d )2 Tmax  ( wb )2
= w2 b 2  a 2
(
2
= L2 w2 4b 2  4 Lb + L2
(
)
or Using
L = 6 m,
T2 4 L2 + d 2 b 2  4L3b + L4  4d 2 max w2
(
)
=0
d = 0.9 m,
Tmax = 8 kN,
w = ( 85 kg/m ) 9.81 m/s2 = 833.85 N/m
(
)
yields
b = ( 2.934 1.353) 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 y + b 1 + B + 3 xB
2 2 y 1 + A + l = s A + sB = a 3 xA
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
M P = 0: x wx  yT0 = 0 2 so y wx = x 2T0 w 2 b  a2 2T0
Geometry:
y =
wx 2 2T0
and d = yB  y A =
(
)
FBD Segment:
Also
2 2 2 y 2 y l = s A + sB = a 1 + A + b 1 + B 3 a 3 b
lL=
2 y A yB + 3 a b
2
2
w2 3 a + b3 = 2 6T0
(
)
) )
1 4d 2 = 6 b2  a 2
(
)
2
(a
3
+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 = w 2 b  a2 2d
a = 2.22 m
Then And with And
(
)
w = ( 85 kg/m ) 9.81 m/s 2 = 833.85 N/m
b = 6 m  a = 3.7804 m
Tmax = TB = T02 + ( wb ) =
2
(
)
T0 = 4338 N
( 4338 N )2 + (833.85 N/m )2 ( 3.7804 m )2
(b)
Tmax = 5.36 kN
Tmax = 5362 N
PROBLEM 7.114
A cable AB of span L and a simple beam AB 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:
M B = 0: LACy + aT0  M B loads = 0
(1)
(Where M B loads includes all applied loads)
x M C = 0: xACy  h  a T0  M C left = 0 L
(2)
FBD AC:
(Where M C left includes all loads left of C)
x (1)  ( 2 ): L hT0  x M B loads + M C left = 0 L
(3)
FBD Beam:
M B = 0: LABy  M B loads = 0
(4)
M C = 0: xABy  M C left  M C = 0
(5) (6)
FBD AC:
x ( 4 )  ( 5): L

x M B loads + M C left + M C = 0 L M C = hT0
Comparing (3) and (6)
Q.E.D.
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:
M D = 0: ( 2 ft )( 450 lb ) + ( 6 ft )( 600 lb )  (10 ft ) A By = 0
A By = 450 lb
M B = 0: M B  ( 4 ft )( 450 lb ) = 0
Section AB:
M B = 1800 lb ft
M 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:
M E = 0: ( 2 ft )( 240 lb ) + ( 4 ft )( 720 lb ) + ( 6 ft )( 360 lb )  ( 8 ft ) ABy = 0
A By = 690 lb
M B = 0: M B  ( 2 ft )( 690 lb ) = 0 M B = 1380 lb ft
M B = 0: M C + ( 2 ft )( 360 lb )  ( 4 ft )( 690 lb ) = 0 M C = 2040 lb ft
M 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 = MC 2040 lb ft = = 850 lb hC 2.4 ft M B 1380 lb ft = = 1.6235 ft T0 850 lb d B = hB + 0.6 ft h0 = M D 1260 lb ft = = 1.482 ft T0 850 lb d D = 3.28 ft d B = 2.22 ft
Cable:
hB =
d B = h0 + 1.8 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:
By symmetry:
M B = M E;
A By = F = 8 kN
MC = M D
AC:
M 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 Cable:
and
MC = M D
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
FBD Elemental segment:
Fy = 0: Ty ( x + x )  Ty ( x )  w ( x ) x = 0
So
Ty ( x + x ) T0

Ty ( x ) T0
=
w( x) x T0
But
dy dx
Ty T0

x + x
=
dy dx
w( x) T0
So
dy dx
x
x
=
In
x 0
lim :
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( x / L ) , where x is measured from midspan. Also determine the maximum and minimum values of the tension in the cable.
SOLUTION
w ( x ) = w0 cos
x
L
From Problem 7.119
w( x) x d2y w = = 0 cos 2 T0 T0 L dx
So
dy W0 L x = sin dx T0 L
dy using dx
0
= 0
y = w L2 L y = h = 0 2 1  cos 2 T0 2 T0 = Tmin
Tmax = TA = TB : TBy T0 = dy dx
w0 L2 x 1  cos using y ( 0 ) = 0 2 L T0
But
so
T0 =
w0 L2 2h Tmin = w0 L2 2h
And
so
=
x= L 2
w0 L T0
TBy =
w0 L
2 TB = TBy + T02 =
w0 L
L 1+ h
2
PROBLEM 7.121
If the weight per unit length of the cable AB is w0 / cos 2 , 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 = cos ds,
w ( x ) dx = so
w0 ds cos 2 w0 w( x) = cos3
From Problem 7.119
d2y w( x) w0 = = T0 dx 2 T0 cos3
d2y d dy d d = ( tan ) = sec2 = 2 dx dx dx dx dx
In general
So
d w0 w0 = = 3 2 dx T0 cos T0 cos sec T0 cos d = dx = rd cos w0 T0 = constant. So curve is circular arc w0
or
Giving r =
Q.E.D.
*For large sag, it is not appropriate to approximate ds by dx.
PROBLEM 7.122
Two hikers are standing 30ft apart and are holding the ends of a 35ft 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 Halfspan:
w = 0.05 lb/ft,
L = 30 ft, sB = c sinh yB xB
sB =
35 ft 2
15 ft 17.5 ft = c sinh c
Solving numerically, Then
yB = c cosh
c = 15.36 ft xB 15 ft = (15.36 ft ) cosh = 23.28 ft c 15.36 ft
(a) (b)
hB = yB  c = 23.28 ft  15.36 ft = 7.92 ft TB = wyB = ( 0.05 lb/ft )( 23.28 ft ) = 1.164 lb
PROBLEM 7.123
A 60ft 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=
120 lb = 2 lb/ft 60 ft xB =
2
hB = 24 ft,
L 2
2 2 yB = c 2 + sB = ( hB + c )
2 = hB + 2chB + c 2 2 2 ( 30 ft )  ( 24 ft ) sB  hB = 2hB 2 ( 24 ft ) 2 2
c=
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 200ft 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
2 2 c 2 = y B  sB
c=
(800 ft )2  (100 ft )2
= 793.73 ft
But
yB = xB cosh
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
2 yB = c 2 + s 2 B
(
)
( c + hB )2
c=
2 = c 2 + sB 2 2
2 2 ( 65 m )  ( 30 m ) sB  hB = 2hB 2 ( 30 m )
= 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 30m 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
2 yB = ( c + hB ) = c 2 + s 2 B 2 2 2 sB  hB 2hB
(
)
So
c=
c=
(15 m )2  (12 m )2 2 (12 m )
= 3.375 m
Now
sB = c sinh
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 30m 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
2
2 2 yB = ( hB + c ) = c 2 + sB 2 h 2 + 2ch  sB = 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 30m 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 11.25 m c
(
)
15 m = c sinh
Solving numerically: c = 8.328 m
2 2 yB = c 2 + sB = ( 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 )
hB = 8.83 m
(b)
P = 24.5 N
PROBLEM 7.129
A 30ft 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 =
20 ft = 10 ft 2
30 ft = 15 ft 2 xB 10 ft = c sinh c c c = 6.1647 ft
sB = c sinh
Solving numerically:
yB = c cosh
xB 10 ft = ( 6.1647 ft ) cosh c 1.1647 ft yB = 16.217 ft
hB = yB  c = 16.217 ft  6.165 ft
(a)
Tmax = wyB and W = w ( 2 sB )
hB = 10.05 ft
So
W =
Tmax 80 lb ( 2sB ) = ( 30 ft ) yB 16.217 ft
(b)
Wm = 148.0 lb
PROBLEM 7.130
Determine the sag of a 45ft 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 = 10 ft 2 xB c
L = 20 ft
sB = c sinh
22.5 ft = c sinh
10 ft c
Solving numerically:
yB = c cosh xB c
c = 4.2023 ft
= ( 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 10m 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 B.
SOLUTION
We know At B, or Solving numerically
sB = c sinh
y = c cosh
x c h 2c
yB = c + h = c cosh 1 = cosh
h h  2c c
h = 4.933 c
xB s h T = c sinh c 2 2c
So
c=
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 L = h = 4.22 m
(a) From At B,
x y = c cosh , c
tan = dy dx
dy x = sinh dx c
= sinh L 4.933 = sinh = 5.848 2c 2
B
= tan 1 5.848
(b)
= 80.3
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 w
Tmax = wyB yB = yB yB = c cosh xB c
1800 N = 61.162 m 29.43 N/m 61.162 m = c cosh 48m / 2 * c
Solving numerically
c = 55.935 m
h = yB  c = 61.162 m  55.935 m h = 5.23 m
*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 8m 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
Neglect pulley size and friction But
TB = wa TB = wyB
so
yB = a xB c
yB = c cosh c cosh
1m =a c
But sB = c sinh So
xB c
8m  a 1m = c sinh c 2 4 m = c sinh 1m c 1m + cosh c c 2
16 m = c 3e1/c  e1/c
(
)
Solving numerically
c = 0.3773 m, 5.906 m
1m ( 0.3773 m ) cosh 0.3773 m = 2.68 m 1m = a = c cosh c ( 5.906 m ) cosh 1 m = 5.99 m 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 xB c L 2c
hB = 15 ft
hB + c = c cosh
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
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,
hB = 9 ft
yB = hB + c = c cosh
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
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 a c
h + c = c cosh
10.8 m  1 12 m = c cosh c
Solving numerically Then
c = 6.2136 m 10.8 m = 18.2136 m 6.2136 m
yB = ( 6.2136 m ) cosh
F = Tmax = wyB = (1.5 lb/ft )(18.2136 m )
F = 27.3 lb
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 a c
c + h = c cosh
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
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 A = 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 =  dy dx =  sinh
a
a a = sinh c c
a = c sinh 1 ( tan ) = ( 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)
B is 26.8 m right and 20.4 m down from A
a 26.849 m = ( 20.387 m ) sinh = 35.31 m c 20.387 m
(b)
s = 35.3 m
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 A = 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
x c
dy x = sinh dx c
=  sinh a a = sinh c c
At A: So
tan = 
dy dx
a
a = c sinh 1 ( tan ) = (15.2905 m ) sinh 1 ( tan 60 ) = 20.137 m 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
a c
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
(b)
Tmax = 14.42 lb
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 the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan , (b) y = c sec .
SOLUTION
(a)
tan = s = c sinh
dy x = sinh dx c x = c tan Q.E.D. c
(b)
So And
Also
y 2 = s 2 + c 2 cosh 2 x = sinh 2 x + 1 y 2 = c 2 tan 2 + 1 = c 2 sec2 y = c sec 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 xB L 2c L = w cosh c 2 L 2c
wL cosh 2
=
L 2c
so
Tmax =
dTmax wL 1 = sinh  cosh d 2
For
min Tmax ,
tanh 
1
=0
Solving numerically = 1.1997
(Tmax )min
=
wL cosh (1.1997 ) = 0.75444wL 2 (1.9997 )
(a)
Lmax =
Tmax T = 1.3255 max 0.75444w w
If
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 =
L =h+c 2c ymax = Tmax w
80 lb = 40 ft 2 lb/ft 9 ft = 40 ft c
c cosh
Solving numerically
c1 = 2.6388 ft c2 = 38.958 ft h = ymax  c h1 = 40 ft  2.6388 ft h2 = 40 ft  38.958 ft h1 = 37.4 ft h2 = 1.042 ft
PROBLEM 7.145
Determine the sagtospan 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 L 1 = 2c 2
tanh
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 sagtospan ratio for which the maximum tension is as small as possible, (b) the corresponding values of B and Tm .
SOLUTION
(a)
Tmax = wyB = wc cosh
L 2c
dTmax L L L sinh = w cosh  dc 2c 2c 2c For min Tmax , 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 ) (b) But T0 = wc Tmax = wc cosh L 2c Tmax L y = cosh = B 2c T0 c h = 0.338 L
T0 = Tmax cos B y
Tmax = sec B T0
So
B = sec1 B = sec1 (1.8102 ) c
= 56.46
B = 56.5
Tmax = wyB = w
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)
M A = 0: ( 6 ft ) E  ( 8 ft )( 4.5 kips )  ( 4 ft )(12 kips )  ( 2 ft )( 6 kips ) = 0
E = 16 kips M 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)
M 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:
Fx = 0:  Ax + Dx = 0 Ax = Dx
M A = 0: (10 m ) D y  ( 6 m )(10 kN )  ( 3 m )( 6 kN ) = 0
Dy = 7.8 kN
Fy = 0: Ay  6 kN  10 kN + 7.8 kN = 0
A y = 8.2 kN
FBD AB:
M B = 0: (1.8 m ) Ax  ( 3 m )( 8.2 kN ) = 0 Ax = 41 kN 3 41 kN 3
From above
Dx = Ax =
FBD CD:
41 kN = 0 M C = 0: ( 4 m )( 7.8 kN )  hC 3 hC = 2.283 m
(a) (b)
hC = 2.28 m Dx = 13.67 kN Dy = 7.80 kN
Since Ax = Bx and Ay > By , max T is TAB
TAB =
2 2 Ax + Ay = 2 41 kN + ( 8.2 kN ) 3 2
(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:
Fx = 0:  Ax + Dx = 0 Ax = Dx
M A = 0: (10 m ) D y  ( 6 m )(10 kN )  ( 3 m )( 6 kN ) = 0 D y = 7.8 kN Fy = 0: Ay  6 kN  10 kN + 7.8 kN = 0 A y = 8.2 kN
Since
Ax = Dx
and
Ay > Dy ,
Tmax = TAB
Fy = 0: 8.2 kN  (15 kN ) sin A = 0
FBD pt A:
A = sin 1
8.2 kN = 33.139 15 kN
Fx = 0:  Ax + (15 kN ) cos A = 0 Ax = (15 kN ) cos ( 33.139 ) = 12.56 kN
FBD CD: From FBD cable:
hB = ( 3 m ) tan A = ( 3 m ) tan ( 33.139 )
(a)
hB = 1.959 m
M 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 = 60.
SOLUTION FBD Rod:
M A = 0: 2r
W  2rB = 0
B=
W
Fy = 0: F +
W W sin 60  cos 60 = 0 3
FBD BJ:
F =  0.12952W
W 3r W M 0 = 0: r F  + +M =0 2 3 1 1 M = Wr 0.12952 +  = 0.28868Wr 2
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 = 150.
SOLUTION FBD rod:
Fy = 0: Ay  W = 0 Ay = W 2r M B = 0: W  2rAx = 0 W Ax =
FBD AJ:
Fx = 0:
W
cos 30 +
5W sin 30  F = 0 F = 0.69233W 6
W W M 0 = 0: 0.25587r + r F   M = 0 6 1 0.25587 M = Wr + 0.69233  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:
M D = 0: ( 0.375 m )( 400 N )  ( 0.24 m ) C y = 0
C y = 625 N
M B = 0:  ( 0.45 m ) C x + ( 0.135 m )( 400 N ) = 0
C x = 120 N
FBD CJ:
Fy = 0: 625 N  F = 0
F = 625 N
Fx = 0: 120 N  V = 0
V = 120.0 N
M 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:
Fx = 0: V = 0
V =0
Fy = 0: F  400 N = 0
F = 400 N
M 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 = 30 and neglecting the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.
SOLUTION
FBD Beam + channels: (a) By symmetry:
T1 = T2 = T Fy = 0: 2T sin 60  3 kN = 0
T =
3 kN 3 M = ( 0.5 m )
T1x = 3 2 3
3 2 3
T1y =
3 kN 2
FBD Beam: With cable force replaced by equivalent forcecouple system at F and G
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 + Finish by invoking symmetry (b) 1 ( 0.6 kN )(1 m ) = 0.058 kN m 2
V
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:
Fy = 0: 40 lb  80 lb + C y = 0
C y = 120 lb
FBD BC:
M B = 0: ( 4 ft )(120 lb )  (10 ft ) Cx = 0
C x = 48 lb From ABC:
Fx = 0:  P + Cx = 0 P = C x = 48 lb
(a)
P = 48.0 lb
M 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:
Fx = 0: C x  P = 0
Cx = 60 lb
Fy = 0: C y  40 lb  80 lb = 0
C y = 120 lb
FBD BC:
M B = 0: b (120 lb )  (10 ft )( 60 lb ) = 0 b = 5.00 ft
FBD AB:
M 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