Ej Cargas Combinadas

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Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 0. 8–19.

100 kN 15 mm x

15 mm 200 mm 150 mm

a

a

Consider the equilibrium of the FBD of the top cut segment in Fig. a,

+ c © Fy = 0; a + © MC = 0;

N - 100 = 0

N = 100 kN

100(0.1) - M = 0

A = 0.2(0.03) = 0.006 m2

I =

#

M = 10 kN m

1 (0.03)(0.23) = 20.0(10 - 6) m4 12

The normal stress developed is the combination of axial and bending stress.Thus, s

=

My N ; A I

For the left edge fiber, y = C = 0.1 m . Then sL

= -

100(103) 10(103)(0.1) 0.006 20.0(10 - 6)

= - 66.67(106) Pa = 66.7 MPa (C)

Ans.

For the right edge fiber, y = 0.1 m . Th Then sR

= -

100 (103) 0.006

+

10(103)(0.1) 20.0(10 - 6)

= 33.3 MPa (T)

Ans.

Ans: sL

751

= 66.7 MPa (C), sR = 33.3 MPa (T)

© 2014 Pearson Education, Education, Inc., Upper Saddle River, River, NJ. All rights reserved. reserved. This material material is protected protected under all copyright copyright laws as they currently currently exist. No portion of this material may may be reproduced, in any form or by any means, means, without permission in writing writing from the publisher. publisher.

Determine the maximum and minimum normal stress in the bracket at section a–a when the load is applied at x = 300 mm. *8–20.

100 kN 15 mm x

15 mm 200 mm 150 mm

a

Consider the equilibrium of the FBD of the top cut segment in Fig. a,

+ c © Fy = 0; a + © MC = 0;

N - 100 = 0

N = 100 kN

#

M - 100(0.2) = 0

A = 0.2 (0.03) = 0.006 m2

I =

M = 20 kN m

1 (0.03)(0.23) = 20.0(10 - 6) m4 12


=

My N ; A I

For the left edge fiber, y = C = 0.1 m . Then sR

= -

100(103) 0.006

+

20.0(103)(0.1) 20.0(10 - 6)

= 83.33(106) Pa = 83.3 MPa (T)

Ans.

For the right edge fiber, y = C = 0.1 m . Thus sR

= -

100(103) 20.0(103)(0.1) 0.006 20.0(10 - 6)

= 117 MPa (C)

Ans.

752

a

© 2014 Pearson Education, Education, Inc., Upper Saddle River, River, NJ. All rights reserved. reserved. This material material is protected protected under all copyright copyright laws as they currently currently exist. No portion of this material may may be reproduced, in any form or by any means, means, without permission in writing writing from the publisher. publisher.

8–26. The column is built up by gluing the two identical boards together together.. Determin Determine e the maximum normal stress stress developed on the cross section when the eccentric force of P = 50 kN is applied.

P

150 mm 250 mm 75 mm 150 mm 50 mm

300 mm

Section Properties: The location of the centroid of the cross cross section, Fig. a, is y =

0.075(0.15)(0.3) + 0.3(0.3)(0.15) © yA = = 0.1875 m 0.15(0.3) + 0.3(0.15) ©A

The cross - sectional area and the moment of inertia about the section are

z

axis of the cross

A = 0.15(0.3) + 0.3(0.15) = 0.09 m2 Iz =

1 1 (0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 + (0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2 12 12

= 1.5609(10 - 3) m4 Equivalent Force System: Referring to Fig. b,

+ c © Fx = (FR)x;

- 50 = - F

© Mz = (MR)z;

- 50(0.2125) = - M

F = 50 kN

#

M = 10.625 kN m

Normal Stress: The normal stress is a combination of axial and bending stress. Thus Thus,,

s

=

N A

+

My I

By inspection, the maximum normal stress occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus Thus,,

smax

=

10.625(103)(0.2625) - 50(103) 0.09 1.5609(10 - 3) Ans.

= - 2.342 MPa = 2.34 MPa (C)

Ans: smax

7 58

= 2.34 MPa (C)

© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–27. The column is built up by gluing the two identical boards together. If the wood has an allowable normal stress of sallow = 6 MPa, determine the maximum allowable eccentric force P that can be applied to the column.

P

150 mm 250 mm 75 mm 150 mm 50 mm

300 mm

Section Properties: The location of the centroid c of the cross section, Fig. a, is y =

0.075(0.15)(0.3) + 0.3(0.3)(0.15) © yA = = 0.1875 m 0.15(0.3) + 0.3(0.15) ©A

The cross-sectional area and the moment of inertia about the z axis of the cross section are A = 0.15(0.3) + 0.3(0.15) = 0.09 m2 Iz =

1 1 (0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 + (0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2 12 12

= 1.5609(10 - 3) m4 Equivalent Force System: Referring to Fig. b,

+ c © Fx = (FR)x;

-P = -F

F = P

© Mz = (MR)z;

- P(0.2125) = - M

M = 0.2125P

Normal Stress: The normal stress is a combination of axial and bending stress. Thus, F =

N A

+

My I

By inspection, the maximum normal stress, Which is compression, occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A.Thus,

- 6(106) =

0.2125P(0.2625) -P 0.09 1.5609(10 - 3)

P = 128 076.92 N = 128 kN

Ans.

Ans: P = 128 kN 759


The frame supports the distributed load shown. Determine the state of stress acting at point D. Show the results on a differential element at this point. 8–38.

4 kN/ m

20 mm

D

60 mm B A

E

1.5 m

1.5 m

E

20 mm

D

5m

3m

50 mm

3m

sD

-

P A

sD

=

-

88.0 MPa

tD

=

8(103)

My

=

-

I

=

-

(0.1)(0.05)

12(103)(0.03) -

C

1 3 12 (0.05)(0.1)

Ans.

0

Ans.

Ans: s

D

773

=

-

88.0 MPa, tD

=

0


8–39. The frame supports the distributed load shown. Determine the state of stress acting at point E. Show the results on a differential element at this point.

4 kN/ m

20 mm

D

60 mm B A

E

1.5 m

D

1.5 m

20 mm

5m

3m

E

50 mm

3m

C

sE

tE

= -

=

P A

VQ It

-

=

My I

=

8(103) (0.1)(0.05)

+

8.25(103)(0.03) 1 3 12 (0.05)(0.1)

4.5(103)(0.04)(0.02)(0.05) 1 3 12 (0.05)(0.1) (0.05)

= 57.8 MPa

= 864 kPa

Ans.

Ans.

Ans: sE = 57.8 MPa, tE = 864 kPa 774


*8–40. The 500-kg engine is suspended from the jib crane at the position shown. Determine the state of stress at point A on the cross section of the boom at section a–a.

E

150 mm 20 mm 2m

2m

2m a 30



C

a

150 mm D

0.4 m

A

300 mm

20 mm B

20 mm Section a – a

Support Reactions: Referring to the free-body diagram of the entire boom, Fig. a,

a+© MC = 0;

FDE sin 30°(6) + FDE cos 30°(0.4) - 500(9.81)(2) = 0 FDE = 2931.50 N

Internal Loadings: Considering the equilibrium of the free-body diagram of the boom’s right cut segment,Fig. b ,

+ © F = 0; x

N - 2931.50 cos 30° = 0

N = 2538.75 N

+ c © Fy = 0;

2931.50 sin 30° - V = 0

V = 1465.75 N

:

a + © MO = 0;

2931.50 sin 30°(2) + 2931.50 cos 30°(0.4) - M = 0

#

M = 3947.00 N m

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the boom’s cross section are A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2

1 1 (0.15)(0.33) (0.13)(0.263) = 0.14709(10 - 3) m4 12 12

I =

Referring to Fig. c, Q A is QA = y1¿ A1¿ + y1¿ A2¿ = 0.065(0.13)(0.2) + 0.14(0.02)(0.15) = 0.589(10 - 3) m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

s

=

My N + A I

For point A. y = 0. Then

sA

=

- 2538.75 + 0 = - 0.2267 MPa = 0.227 MPa (C) 0.0112

Ans.

Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,

tA

=

VQA It

=

1465.75[0.589(10 - 3)] 0.14709(10 - 3)(0.02)

= 0.293 MPa

Ans.

The state of stress at point A is represented on the element shown in Fig. d .

775


8–40. Continued

776


8–41. The 500-kg engine is suspended from the jib crane at the position shown. Determine the state of stress at point B on the cross section of the boom at section a–a. Point B is just above the bottom flange.

E

150 mm 20 mm 2m

2m

2m a 30



C

a

150 mm D

0.4 m

A

300 mm

20 mm B

20 mm Section a – a

Support Reactions: Referring to the free-body diagram of the entire boom, Fig. a,

a +© MC = 0;

FDE sin 30°(6) + FDE cos 30°(0.4) - 500(9.81)(2) = 0 FDE = 2931.50 N

Internal Loadings: Considering the equilibrium of the free-body diagram of the boom’s right cut segment, Fig. b ,

+ © F = 0; x

N - 2931.50 cos 30° = 0

N = 2538.75 N

+ c © Fy = 0;

2931.50 sin 30° - V = 0

V = 1465.75 N

a+ © MO = 0;

2931.50 sin 30°(2) + 2931.50 cos 30°(0.4) - M = 0

:

#

M = 3947.00 N m

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the boom’s cross section are A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2 I =

1 1 (0.15)(0.33) (0.13)(0.263) = 0.14709(10 - 3) m4 12 12

Referring to Fig. c , Q B is QB = y2A2¿ = 0.14(0.02)(0.15) = 0.42(10 - 3) m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

s

=

N A

+

My I

For point B, y = 0.13 m. Then

sB

=

3947.00(0.13) - 2538.75 + = 3.26 MPa (T) 0.0112 0.14709(10 - 3)

Ans.

777


8–41. Continued

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,

tB

=

VQB It

=

1465.75[0.42(10 - 3)] 0.14709(10 - 3)(0.02)

= 0.209 MPa

Ans.

The state of stress at point B is represented on the element shown in Fig. d.

Ans: sB = 3.26 MPa (T), tB = 0.209 MPa 778


8–42. Determine the state of stress at point A on the cross section of the post at section a–a. Indicate the results on a differential element at the point.

1.5 ft

5 ft

400 lb

300 lb

Internal Loadings: Considering the equilibrium of the free-body diagram of the post’s upper cut segment, Fig. a ,

© Fy = 0;

Vy + 300 = 0

Vy = - 300 lb

© Fz = 0;

Vz + 400 = 0

Vz = - 400 lb

© Mx = 0;

T + 400(1.5) = 0

T = - 600 lb ft

© My = 0;

My + 400(5) = 0

My = - 2000 lb ft

© Mz = 0;

Mz - 300(5) = 0

Mz = 1500 lb ft

a

A

2.5 in. B

#

#

#

Section Properties: The moments of inertia about the y and z axes and the polar moment of inertia of the post’s cross section are Iy = Iz = J =

p

2

p

4

(2.54 - 24) = 5.765625p in4

(2.54 - 24) = 11.53125p in4

Referring to Fig. b , (Qz)A = 0 (Qy)A =

c

d

c

d

4(2.5) p 4(2) p 2 (2.52) (2 ) = 5.0833 in3 3p 2 3p 2

Normal Stress: The normal stress is contributed by bending stress only. Thus, s

= -

Mzy Iz

+

Myz Iy

For point A, y = 0 and z = - 2.5 in. Then sA

= -0 +

2 in.

a

- 2000(12)( - 2.5) = 3.31 ksi (T) 5.765625p

Ans.

779


8–42. Continued

Shear Stress: The torsional shear stress at points A and B are

[(txy)T] A =

600(12)(2.5) Tc = = 0.4969 ksi 11.53125p J

The transverse shear stresses at points A and B are [(txz)V]A = [(txy)V] A =

Vz(Qz)A Iy t Vy(Qy)B Iz t

= 0 =

300(5.0833) = 0.08419 ksi 5.765625p(5 - 4)

Combining these two shear stress components, (txy)A = [(txy)T]A + [(txy)V]A = 0.4969 + 0.08419 = 0.581 ksi

Ans.

(txz)A = 0

Ans.

The state of stress at point A is represented on the element shown in Fig. c .

Ans: s A = 3.31 ksi (T), t A = 0.581 ksi 780


8–43. Determine the state of stress at point B on the cross section of the post at section a–a. Indicate the results on a differential element at the point.

1.5 ft

5 ft

400 lb

Internal Loadings: Considering the equilibrium of the free-body diagram of the post’s upper segment, Fig. a , Vy + 300 = 0

Vy = - 300 lb

© Fz = 0;

Vz + 400 = 0

Vz = - 400 lb

© Mx = 0;

T + 400(1.5) = 0

T = - 600 lb ft

© My = 0;

My + 400(5) = 0

My = - 2000 lb ft

© Mz = 0;

Mz - 300(5) = 0

Mz = 1500 lb ft

J =

p

2

4

#

#

#

(2.54 - 24) = 5.765625p in4

(2.54 - 24) = 11.53125p in4

Referring to Fig. b , (Qy)B = 0 (Qz)B =

4(2.5) 3p

c 2 (2.5 ) d - 4(2) c 2 (2 ) d = 5.0833 in 3 p

p

2

2

3

p

Normal Stress: The normal stress is contributed by bending stress only.Thus, s

= -

Mzy Iz

+

Myz Iy

For point B, y = 2 in. and z = 0. Then sB

=

1500(12)(2) 5.765625p

A B

Section Properties: The moments of inertia about the y and z axes and the polar moment of inertia of the post’s cross section are p

2 in.

a

2.5 in.

© Fy = 0;

Iy = Iz =

a

300 lb

+ 0 = - 1.987 ksi = 1.99 ksi (C)

Ans.

781


8–43. Continued

Shear Stress: The torsional shear stress at point B is

[(txz)T]B =

Tc J

=

600(12)(2) 11.53125p

= 0.3975 ksi

The transverse shear stress at point B is [(txy)V]B = [(txz)V]B =

Vy(Qy)A Iz t Vz(Qz)B Iy t

= 0 =

400(5.0833) = 0.1123 ksi 5.765625p(5 - 4)

Combining these two shear stress components, (txz)B = [(txz)T]B + [(txz)V]B = 0.3975 + 0.1123 = 0.510 ksi (txy)B = 0

Ans. Ans.

The state of stress at point B is represented on the element shown in Fig. c .

Ans: sB = 1.99 ksi (C), tB = 0.510 ksi 782


*8–44. Determine the normal stress developed at points A and B. Neglect the weight of the block.

6 kip 3 in.

Referring to Fig. a,

12 kip

6 in. a

© Fx = (FR)x;

- 6 - 12 = F

F = - 18.0 kip

© My = (MR)y;

6(1.5) - 12(1.5) = My

My = - 9.00 kip in

© Mz = (MR)z;

12(3) - 6(3) = Mz

Mz = 18.0 kip in

A

#

#

The cross-sectional area and moments of inertia about the y and z axes of the cross section are A = 6(3) = 18 in2 Iy =

1 (6)(3)3 = 13.5 in4 12

Iz =

1 (3)(63) = 54.0 in4 12


=

My z Mz y F + A Iz Iy

For point A, y = 3 in. and z = - 1.5 in.

sA

=

- 9.00( - 1.5) 18.0(3) - 18.0 + 18.0 54.0 13.5

= - 1.00 ksi = 1.00 ksi (C)

Ans.

For point B, y = 3 in and z = 1.5 in.

sB

=

18.0(3) - 9.00(1.5) - 18.0 + 18.0 54 13.5

= - 3.00 ksi = 3.00 ksi (C)

Ans.

783

B a


8–45. Sketch the normal stress distribution acting over the cross section at section a–a. Neglect the weight of the block.

6 kip 3 in.

12 kip

6 in. a A

B a

Referring to Fig. a,

© Fx = (FR)x;

- 6 - 12 = F

F = - 18.0 kip

© My = (MR)y; 6(1.5) - 12(1.5) = My

#

My = - 9.00 kip in

#

© Mz = (MR)z; 12(3) - 6(3) = Mz

Mz = 18.0 kip in

The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 3 (6) = 18.0 in2 Iy =

1 (6)(33) = 13.5 in4 12

Iz =

1 (3)(63) = 54.0 in4 12


=

F A

-

Mzy Iz

+

Myz Iy

For point A, y = 3 in. and z = - 1.5 in. sA

=

18.0(3) - 9.00( - 1.5) - 18.0 + 18.0 54.0 13.5

= - 1.00 ksi = 1.00 ksi (C)

Ans.

For point B, y = 3 in. and z = 1.5 in. sB

=

- 9.00(1.5) 18.0(3) - 18.0 + 18.0 54.0 13.5

= - 3.00 ksi = 3.00 ksi (C)

Ans.

784


8–45. Continued

For point C , y = - 3 in. and z = 1.5 in. sC

=

- 9.00(1.5) 18.0( - 3) - 18.0 + 18.0 54.0 13.5

= - 1.00 ksi = 1.00 ksi (C)

Ans.

For point D, y = - 3 in. and z = - 1.5 in. sD

=

18.0( - 3) - 9.00( - 1.5) - 18.0 + 18.0 54.0 13.5

= 1.00 ksi (T)

Ans.

The normal stress distribution over the cross section is shown in Fig. b

Ans: s A = 1.00 ksi (C), sB = 3.00 ksi (C) 785


x

8–51. A post having the dimensions shown is subjected to the bearing load P. Specify the region to which this load can be applied without causing tensile stress to be developed at points A, B, C , and D.

z

a

a A

B

a

P

a

D

ez

e y C

a y

a

Equivalent Force System: As shown on FBD. Section Properties: =

2a(2a)

Iz

=

1 (2a)(2a)3 12

=

5a4

=

1 (2a)(2a)3 12

=

5 4 a 3

Iy

2

B

A

+

1 (2a)a 2

R

6a2

=

+

2

B

1 (2a) a3 36

+

1 (2a) a a 2

+

2

B

1 (2a) a3 36

+

a 1 (2a) a 2 3

a

+

bR 3 a

2

a bR 2

Normal Stress: s

N

=

A

At point B where y

P

0 0

30a4

= -a

7

P

30a4

7 -

6ey

+

-

6a2

A - 5a2

5a4 -

= -a

C - 5a2

-

18ez

When

ez

=

0,

ey

6

When

ey

=

0,

ez

6

6

+

Iy

Peyy

and z

5a + 6ey

My z

+

Iz

-P

=

=

Mzy

-

-

6eyy

Pez z 5 4 3a -

18ez z B

, we require sB

6( - a) ey

-

6

0.

18( - a) ez D

18ez

5a

Ans.

5 a 6 5 a 18

Repeat the same procedures for point A, C and D. The region where P can be applied without creating tensile stress at points A, B, C , and D is shown shaded in the diagram.

Ans: 6ey + 18ez 792

6

5a


8–55. Determine the state of stress at point A on the cross section of the post at section a–a. Indicate the results on a differential element at the point.

100 mm

100 mm

3 kN

4 kN 400 mm 50 mm 50 mm A

50 mm

B

50 mm

z

Section a – a

Internal Loadings: Considering the equilibrium of the free-body diagram of the post’s upper segment, Fig. a, © Fy =

0;

Vy

+

3

=

0

Vy

= -3

kN

© Fz =

0;

Vz

+

4

=

0

Vz

= -4

kN

© Mx =

0; T

0

© My =

0; My

+

4(0.4)

=

0

My

= - 1.6

kN # m

© Mz =

0; Mz

-

3(0.4)

=

0

Mz

= - 1.2

kN # m

=

Section Properties: The moment of inertia about the y and z axes of the post’s cross section is Iy

=

1 (0.1)(0.13) 12

=

Iz

=

8.3333(10 - 6) m4

Referring to Fig. b , (Qy)A

=

0

(Qz)A

=

0.025(0.05)(0.1)

=

0.125(10 - 3) m3

Normal Stress: The normal stress is contributed by bending stress only.Thus,

s

= -

For point A, y

sA

=

Mzy Iz

+

= - 0.05

Myz Iy

m and z

1.2(103)( - 0.05) 8.3333(10 - 6)

0

+

=

=

0. Then 7.20 MPa (T)

Ans.

Shear Stress: Then transverse shear stress at point A is

[(txy)V]A

[(txz)V]A

=

=

Vy(Qy)A Izt Vz(Qz)A Iy t

=

=

0 4(103)[0.125(10 - 3)] 8.3333(10 - 6)(0.1)

Ans.

=

0.6 MPa

Ans.

The state of stress at point A is represented on the elements shown in Figs. c and d , respectively.

796

400 mm

a

a

x

y


8–55. Continued

Ans: s A

797

=

7.20 MPa (T), t A

=

0.6 MPa


*8–56. Determine the state of stress at point B on the cross section of the post at section a–a. Indicate the results on a differential element at the point.

100 mm

100 mm

3 kN

4 kN 400 mm 50 mm 50 mm A

50 mm

B

50 mm

z

Section a – a

Internal Loadings: Considering the equilibrium of the free-body diagram of the post’s upper segment, Fig. a, © Fy =

0;

Vy

+

3

=

0

Vy

= -3

kN

© Fz =

0;

Vz

+

4

=

0

Vz

= -4

kN

© Mx =

0; T

© My =

0; My

+

4(0.4)

=

0

My

= - 1.6

© Mz =

0; Mz

-

3(0.4)

=

0

Mz

=

=

0 kN # m

1.2 kN # m

Section Properties: The moment of inertia about the y and z axes of the post’s cross section is =

Iy

Iz

=

1 (0.1)(0.13) 12

=

8.3333(10 - 6) m4

Referring to Fig. b , (Qz)B

=

0

(Qy)B

=

0.025(0.05)(0.1)

=

0.125(10 - 3) m3

Normal Stress: The normal stress is contributed by bending stress only.Thus,

s

=

-

Mzy Iz

For point B, y

sB

=

= -0 +

+

Myz Iy

0 and z

= - 0.05

m. Then

3 - 1.6(10 )( - 0.05)

8.3333(10 - 6)

=

9.60 MPa (T)

Ans.

Shear Stress: Then transverse shear stress at point B is

c( ) d txy

c( ) d txy

=

V

B

=

V

B

Vz(Qz)A Iy t Vy(Qy)A Iz t

=

=

0 3(103)[0.125(10 - 3)] 8.3333(10 - 6)(0.1)

Ans.

=

0.45 MPa

Ans.

The state of stress at point B is represented on the elements shown in Figs. c and d , respectively.

798

400 mm

a

a

x

y


8–56. Continued

799


8–59. If P = 60 kN, determine the maximum normal stress developed on the cross section of the column.

2P 15 mm 150 mm

P

15 mm

150 mm

15 mm

75 mm

100 mm

100 mm 100 mm

Equivalent Force System: Referring to Fig. a,

+ c © Fx = (FR)x ;

- 60 - 120 = - F

F = 180 kN

#

© My = (MR)y ;

- 60(0.075) = - My

My = 4.5 kN m

© Mz = (MR)z ;

- 120(0.25) = - Mz

Mz = 30 kN m

#

Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2

Iz =

1 1 (0.2)(0.33) (0.185)(0.273) = 0.14655(10 - 3) m4 12 12

Iy = 2

c 121 (0.015)(0.2 ) d + 121 (0.27)(0.015 ) = 20.0759(10 3

3

-6

) m4

Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, M y and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress.Thus,

s

=

smax

N A

=

-

s A

Mzy Iz

=

+

My z Iy

[ - 30(103)]( - 0.15) [ - 4.5(103)](0.1) - 180(103) + 0.01005 0.14655(10 - 3) 20.0759(10 - 6)

= - 71.0 MPa = 71.0 MPa (C)

Ans.

Ans: smax

802

= 71.0 MPa (C)


*8–60. Determine the maximum allowable force P, if the column is made from material having an allowable normal stress of sallow = 100 MPa.

2P 15 mm 150 mm

P

15 mm

150 mm

15 mm 100 mm 100 mm

Equivalent Force System: Referring to Fig. a,

+ c © Fx = (FR)x ;

- P - 2P = - F F = 3P

© My = (MR)y ;

- P(0.075) = - My My = 0.075P

© Mz = (MR)z ;

- 2P(0.25) = - Mz Mz = 0.5P

Section Properties: The cross-sectional area and the moment of inertia about the y and z axes of the cross section are A = 0.2(0.3) - 0.185(0.27) = 0.01005m2

Iz =

1 1 (0.2)(0.33) (0.185)(0.273) = 0.14655(10 - 3) m4 12 12

Iy = 2

c 121 (0.15)(0.2 ) d + 121 (0.27)(0.015 ) = 20.0759(10 3

3

-6

) m4

Normal Stress: The normal stress is the combination of axial and bending stress. Here, F is negative since it is a compressive force. Also, M y and Mz are negative since they are directed towards the negative sense of their respective axes. By inspection, point A is subjected to a maximum normal stress, which is in compression. Thus,

s

=

N A

-

Mzy Iz

- 100(106) = -

+

My z Iy

( - 0.5P)( - 0.15) - 0.075P(0.1) 3P + -3 0.01005 0.14655(10 ) 20.0759(10 - 6)

P = 84470.40 N = 84.5k N

Ans.

803

75 mm

100 mm


50 mm

8–75. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E on the cross section of the frame at section a–a. Indicate the results on an element.

25 mm

E

75 mm

Section a – a

0.5 m 0.5 m

1m

a B C

a

1m 30



Support Reactions: Referring to the free-body diagram of member BC shown in Fig. a,

1m

a + © MB = 0;

1m

F sin 45°(1) - 20(9.81)(2) = 0

+ © F = 0; x

:

+ c © Fy = 0;

b

F = 554.94 N

554.94 cos 45° - Bx = 0

Bx = 392.4 N

554.94 sin 45° - 20(9.81) - By = 0

By = 196.2 N

D

b

75 mm

F A

25 mm Section b – b

Internal Loadings: Consider the equilibrium of the free-body diagram of the right segment shown in Fig. b. + N - 392.4 = 0 N = 392.4 N : © Fx = 0;

+ c © Fy = 0;

V - 196.2 = 0

V = 196.2 N

a + © MC = 0;

196.2(0.5) - M = 0

M = 98.1 N # m

Section Properties: The cross -sectional area and the moment of inertia of the cross section are A = 0.05(0.075) = 3.75 A 10 - 3 B m2 I =

1 (0.05) A 0.0753 B = 1.7578 A 10 - 6 B m4 12

Referring to Fig. c, QE is QE = y ¿ A ¿ = 0.025(0.025)(0.05) = 31.25 A 10 - 6 B m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s

=

My N ; A I

For point E, y = 0.0375 - 0.025 = 0.0125 m. Then sE

=

392.4 3.75 A 10 - 3 B

+

98.1(0.0125) 1.7578 A 10 - 6 B

= 802 kPa

Ans.

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus, tE

=

VQA It

=

196.2 C 31.25 A 10 - 6 B D 1.7578 A 10 - 6 B (0.05)

= 69.8 kPa

Ans.

The state of stress at point E is represented on the element shown in Fig. d.

822

75 mm


8–75. Continued

Ans: sE = 802 kPa, tE = 69.8 kPa 823


*8–76. The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point F on the cross section of the frame at section b - b. Indicate the results on an element.

50 mm 25 mm

E

75 mm

Section a – a

0.5 m 0.5 m

1m

a B

Support Reactions: Referring to the free-body diagram of the entire frame shown in Fig. a,

C

a

1m 30



a + © MA = 0;

FBD sin 30°(3) - 20(9.81)(2) = 0

+ c © Fy = 0;

Ay - 261.6 cos 30° - 20(9.81) = 0

Ay = 422.75 N

Ax - 261.6 sin 30° = 0

Ax = 130.8 N

+ © F = 0; x

:

FBD = 261.6 N 1m b

Internal Loadings: Consider the equilibrium of the free-body diagram of the lower cut segment, Fig. b,

+ © F = 0; x

V = 130.8 N

+ c © Fy = 0;

422.75 - N = 0

N = 422.75 N

a + © MC = 0;

130.8(1) - M = 0

M = 130.8 N # m

Section Properties: The cross -sectional area and the moment of inertia about the centroidal axis of the cross section are A = 0.075(0.075) = 5.625 A 10 - 3 B m2

1 (0.075) A 0.0753 B = 2.6367 A 10 - 6 B m4 12

Referring to Fig. c, QE is QF = y ¿ A ¿ = 0.025(0.025)(0.075) = 46.875 A 10 - 6 B m3

Normal Stress: The normal stress is the combination of axial and bending stress. Thus, s

=

My N ; A I

For point F , y = 0.0375 - 0.025 = 0.0125 m. Then sF

=

- 422.75 5.625 A 10 - 3 B

-

130.8(0.0125) 2.6367 A 10 - 6 B

= - 695.24 kPa = 695 kPa (C)

Ans.

824

F A

25 mm Section b – b

130.8 - V = 0

I =

75 mm

1m D

:

b

75 mm


8–76. Continued

Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,

tA

=

VQA It

c

130.8 46.875 A 10 - 6 B

=

d

2.6367 A 10 - 6 B (0.075)

= 31.0 kPa

Ans.

The state of stress at point A is represented on the element shown in Fig. d.

825


The wall hanger has a thickness of 0.25 in. and is used to support the vertical reactions of the beam that is loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the state of stress at points C and D on the strap at A. Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown. 8–85.

10 kip

2 kip/ ft

A

B

2 ft

2 ft

6 ft

2 in. 2 in. 2 in. 3.75 in.

F

2.75 in. 3 in.

a + ©MB

=

0;

12(3)

+ 10(8) - FA(10) =

FA =

2

I =

A =

c 121 (0.25)(2) d 3

2(0.25)(2)

=

=

D C

1 in.

0

1 in.

11.60 kip

0.333 in4

1 in2

At point C ,

sC

tC

=

=

P A

=

2(5.80) 1

=

11.6 ksi

Ans.

0

Ans.

At point D ,

sD

tD

=

=

P A

-

Mc I

=

2(5.80) 1

-

[2(5.80)](1) 0.333

= - 23.2

ksi

Ans.

0

Ans.

Ans:

834

11.6 ksi, tC = 0, ksi, tD = 0

sC

=

sD

= - 23.2

Ej Cargas Combinadas

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