ej cargas combinadas

7/24/2019 Ej Cargas Combinadas

1/30

751

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

a

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

For the left edge fiber, . Then

Ans.

For the right edge fiber, . Then

Ans.sR = - 100 (103)

0.006 +

10(103)(0.1)

20.0(10- 6)= 33.3 MPa (T)

y = 0.1 m

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

sL = -100(103)

0.006 -

10(103)(0.1)

20.0(10- 6)

y = C = 0.1 m

s =N

A ;

My

I

A = 0.2(0.03) = 0.006 m2I =1

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

+ MC = 0;100(0.1) - M = 0M = 10 kN # m

+ c Fy = 0;N - 100 = 0N = 100 kN

819. Determine the maximum and minimum normalstress in the bracket at section aa when the load is appliedat .x = 0

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

100 kN

x

200 mm150 mm

15 mm

15 mm

aa

Ans:

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


2/30

752

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

a


For the left edge fiber, . Then

Ans.

For the right edge fiber, . Thus

Ans.= 117 MPa (C)

sR = -100(103)

0.006 -

20.0(103)(0.1)

20.0(10- 6)

y = C = 0.1 m

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

sR = -100(103)

0.006 +

20.0(103)(0.1)

20.0(10- 6)

y = C = 0.1 m

s =

N

A ;

My

I

A = 0.2 (0.03) = 0.006 m2I =1

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

+ MC = 0;M - 100(0.2) = 0M = 20 kN # m

+ c Fy = 0;N - 100 = 0N = 100 kN

*820. Determine the maximum and minimum normalstress in the bracket at section aa when the load is appliedat x = 300 mm.

100 kN

x

200 mm150 mm

15 mm

15 mm

aa



3/30

758

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

The cross - sectional area and the moment of inertia about the z axis of the crosssection are

= 1.5609(10- 3) m4

Iz =1

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

1

12(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2

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

y = yA

A =

0.075(0.15)(0.3) + 0.3(0.3)(0.15)

0.15(0.3) + 0.3(0.15) = 0.1875 m

826. The column is built up by gluing the two identicalboards together. Determine the maximum normal stressdeveloped on the cross section when the eccentric force of

is applied.P = 50 kN

150 mm

150 mm

250 mm

75 mm

300 mm

50 mm

P

Equivalent Force System: Referring to Fig. b,

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

By inspection, the maximum normal stress occurs at points along the edge wheresuch as point A. Thus,

Ans.= -2.342 MPa = 2.34 MPa (C)

smax = -50(103)

0.09 -

10.625(103)(0.2625)

1.5609(10 - 3)

y = 0.45 - 0.1875 = 0.2625 m

s =N

A +My

I

M = 10.625 kN # m-50(0.2125) = -MMz = (MR)z;

F = 50 kN-50 = -F+ c Fx = (FR)x;


Ans:

smax = 2.34 MPa (C)


4/30

759

827. The column is built up by gluing the two identicalboards together. If the wood has an allowable normal stressof , determine the maximum allowableeccentric force P that can be applied to the column.

sallow = 6 MPa

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

The cross-sectional area and the moment of inertia about the z axis of the crosssection are

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

y = yA

A =

0.075(0.15)(0.3) + 0.3(0.3)(0.15)

0.15(0.3) + 0.3(0.15) = 0.1875 m

150 mm

150 mm

250 mm

75 mm

300 mm

50 mm

P

= 1.5609(10 - 3) m4

Iz =1

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

1

12(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2

Equivalent Force System: Referring to Fig.b,

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

By inspection, the maximum normal stress, Which is compression, occurs at pointsalong the edge where such as point A.Thus,

Ans.P = 128 076.92 N = 128 kN

-6(106) = -P

0.09 -

0.2125P(0.2625)

1.5609(10- 3)

y = 0.45 - 0.1875 = 0.2625 m

F =N

A +My

I

M = 0.2125P-P(0.2125) = -MMz = (MR)z;

F = P-P = -F+c Fx = (FR)x;


Ans:

P = 128 kN


5/30

773

838. The frame supports the distributed load shown.Determine the state of stress acting at point D. Show theresults on a differential element at this point.

Ans.

Ans.tD = 0

sD = - 88.0 MPa

sD = -P

A -

My

I = -

8(103)

(0.1)(0.05) -

12(103)(0.03)112 (0.05)(0.1)

3

4 kN/m

D

B

A

C

E

1.5 m 1.5 m

20 mm

50 mm

20 mm

60 mm

3 m

3 m

5 m

D

E


Ans:

sD = - 88.0 MPa, tD = 0


6/30

774

Ans.

Ans.tE = VQ

It =

4.5(103)(0.04)(0.02)(0.05)1

12 (0.05)(0.1)3(0.05)

= 864 kPa

sE = -P

A -My

I =

8(103)

(0.1)(0.05) +

8.25(103)(0.03)1

12(0.05)(0.1)3

= 57.8 MPa

839. The frame supports the distributed load shown.Determine the state of stress acting at point E. Show theresults on a differential element at this point.


4 kN/m

D

B

A

C

E

1.5 m 1.5 m

20 mm

50 mm

20 mm

60 mm

3 m

3 m

5 m

D

E

Ans:

sE = 57.8 MPa, tE = 864 kPa


7/30

775

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

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

Section Properties: The cross-sectional area and the moment of inertia about thecentroidal axis of the booms cross section are

Referring to Fig. c, QA is

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

For point Then

Ans.

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

Ans.

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

tA = VQA

It =

1465.75[0.589(10- 3)]

0.14709(10 - 3)(0.02)= 0.293 MPa

sA = -2538.75

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

A. y = 0.

s =N

A +My

I

QA = y1A1 + y1A2 = 0.065(0.13)(0.2) + 0.14(0.02)(0.15) = 0.589(10- 3) m3

I =1

12(0.15)(0.33) -

1

12(0.13)(0.263) = 0.14709(10- 3) m4

A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2

M = 3947.00 N # m

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

V = 1465.75 N2931.50 sin 30 - V = 0+c Fy = 0;

N = 2538.75 NN - 2931.50 cos 30 = 0:+ Fx = 0;

FDE = 2931.50 N

FDE sin 30(6) + FDE cos 30(0.4) - 500(9.81)(2) = 0+MC = 0;

*840. The 500-kg engine is suspended from the jib crane atthe position shown. Determine the state of stress at pointAon the cross section of the boom at section aa.


Section a a

20 mm

20 mm

20 mm

150 mm

150 mm

300 mmA

B

E

C

D

2 m2 m 2 m

0.4 m

30a

a

a

a


8/30

776

840. Continued



9/30

777

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

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

Section Properties: The cross-sectional area and the moment of inertia about thecentroidal axis of the booms cross section are

Referring to Fig. c, QBis


For point Then

Ans.sB = -2538.75

0.0112 +

3947.00(0.13)

0.14709(10 - 3)= 3.26 MPa (T)

B, y = 0.13 m.

s =N

A +My

I

QB = y2A2 = 0.14(0.02)(0.15) = 0.42(10

- 3) m3

I =1

12(0.15)(0.33) -

1

12(0.13)(0.263) = 0.14709(10- 3) m4

A = 0.15(0.3) - 0.13(0.26) = 0.0112 m2

M = 3947.00 N # m

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

V = 1465.75 N2931.50 sin 30 - V = 0+ c Fy = 0;

N = 2538.75 NN - 2931.50 cos 30 = 0:+ Fx = 0;

FDE = 2931.50 N

FDE sin 30(6) + FDE cos 30(0.4) - 500(9.81)(2) = 0+MC = 0;

841. The 500-kg engine is suspended from the jib crane atthe position shown. Determine the state of stress at pointBon the cross section of the boom at section aa. Point B is

just above the bottom flange.


Section a a

20 mm

20 mm

20 mm

150 mm

150 mm

300 mmA

B

E

C

D

2 m2 m 2 m

0.4 m

30a

a

a

a


10/30

778

841. Continued



Ans.

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

tB = VQB

It =

1465.75[0.42(10- 3)]

0.14709(10 - 3)(0.02)= 0.209 MPa

Ans:

sB = 3.26 MPa (T), tB = 0.209 MPa


11/30

779

Internal Loadings: Considering the equilibrium of the free-body diagramof the posts upper cut segment, Fig.a,

Section Properties: The moments of inertia about theyandzaxes and thepolar moment of inertia of the posts cross section are

Referring to Fig. b,

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

For point and Then

Ans.sA = -0 + -2000(12)(-2.5)

5.765625p= 3.31 ksi (T)

z = -2.5 in.A, y = 0

s = -Mzy

Iz+

Myz

Iy

(Qy)A =4(2.5)

3p cp

2(2.52) d - 4(2)

3p cp

2(22) d = 5.0833 in3

(Qz)A = 0

J = p

2(2.54 - 24) = 11.53125pin4

Iy = Iz = p

4

(2.54 - 24) = 5.765625pin4

Mz = 1500 lb # ftMz = 0; Mz - 300(5) = 0My = -2000 lb # ftMy = 0; My + 400(5) = 0T = -600 lb # ftMx = 0; T + 400(1.5) = 0Vz = -400 lbFz = 0; Vz + 400 = 0

Vy = -300 lbFy = 0; Vy + 300 = 0

842. Determine the state of stress at point A on the crosssection of the post at section aa. Indicate the results on adifferential element at the point.


a a

5 ft

1.5 ft

300 lb400 lb

2 in.

2.5 in.

A

B


12/30

780

842. Continued


Shear Stress: The torsional shear stress at pointsA and Bare

The transverse shear stresses at pointsA and Bare

Combining these two shear stress components,

Ans.

Ans.

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

(txz)A = 0

(t

xy)A = [(t

xy)T]A + [(t

xy)V]A = 0.4969 + 0.08419 = 0.581 ksi

[(txy)V]A =Vy(Qy)B

Iz t =

300(5.0833)

5.765625p(5 - 4) = 0.08419 ksi

[(txz)V]A =Vz(Qz)A

Iy t = 0

[(txy)T]A = Tc

J =

600(12)(2.5)

11.53125p= 0.4969 ksi

Ans:

sA = 3.31 ksi (T), tA = 0.581 ksi


13/30

781


a a

5 ft

1.5 ft

300 lb400 lb

2 in.

2.5 in.

A

B

Internal Loadings: Considering the equilibrium of the free-body diagram of theposts upper segment, Fig.a,

Section Properties: The moments of inertia about theyand zaxes and the polarmoment of inertia of the posts cross section are

Referring to Fig. b,


For point in. and . Then

Ans.sB =1500(12)(2)

5.765625p+ 0 = -1.987 ksi = 1.99 ksi (C)

z = 0B, y = 2

s = -Mzy

Iz+Myz

Iy

(Qz)B =4(2.5)

3p cp

2(2.52) d - 4(2)

3p cp

2(22) d = 5.0833 in3

(Qy)B = 0

J = p

2(2.54 - 24) = 11.53125pin4

Iy = Iz = p

4(2.54 - 24) = 5.765625pin4

Mz = 1500 lb # ftMz = 0; Mz - 300(5) = 0My = -2000 lb # ftMy = 0; My + 400(5) = 0T = -600 lb # ftMx = 0; T + 400(1.5) = 0Vz = -400 lb

Fz = 0; Vz + 400 = 0

Vy = -300 lbFy = 0; Vy + 300 = 0

843. Determine the state of stress at point B on the crosssection of the post at section aa. Indicate the results on adifferential element at the point.


14/30

782

843. Continued


Shear Stress: The torsional shear stress at point Bis

The transverse shear stress at point Bis

Combining these two shear stress components,

Ans.

Ans.

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

(txy)B = 0

(txz)B = [(txz)T]B + [(txz)V]B = 0.3975 + 0.1123 = 0.510 ksi

[(txz)V]B =Vz(Qz)B

Iy t =

400(5.0833)

5.765625p(5 - 4) = 0.1123 ksi

[(txy)V]B =Vy(Qy)A

Iz t = 0

[(txz)T]B = Tc

J =

600(12)(2)

11.53125p= 0.3975 ksi

Ans:

sB = 1.99 ksi (C), tB = 0.510 ksi


15/30

783

Referring to Fig. a,

The cross-sectional area and moments of inertia about they and z axes of the crosssection are


For pointA, . and .

Ans.

For point B, and .

Ans.= -3.00 ksi = 3.00 ksi (C)

sB = -18.018.0

- 18.0(3)54

+ -9.00(1.5)13.5

z = 1.5 iny = 3 in

= -1.00 ksi = 1.00 ksi (C)

sA = -18.0

18.0 -

18.0(3)

54.0 +

-9.00(-1.5)

13.5

z = -1.5 iny = 3 in

s = F

A -

Mzy

Iz+Myz

Iy

Iz =1

12(3)(63) = 54.0 in4

Iy =1

12(6)(3)3 = 13.5 in4

A = 6(3) = 18 in2

Mz = (MR)z;

12(3) - 6(3) = Mz

Mz = 18.0 kip # in

My = (MR)y;6(1.5) - 12(1.5) = MyMy = -9.00 kip # in

Fx = (FR)x;-6 - 12 = F F = -18.0 kip

*844. Determine the normal stress developed at pointsAand B. Neglect the weight of the block.


a

a

6 in.

6 kip

12 kip3 in.

A B


16/30

784

Referring to Fig. a,

The cross-sectional area and the moment of inertia about the y and z axes of thecross section are


For pointA, . and .

Ans.

For point B, . and .

Ans.= -3.00 ksi = 3.00 ksi (C)

sB = -18.0

18.0 -

18.0(3)

54.0 +

-9.00(1.5)

13.5

z = 1.5 iny = 3 in

= -1.00 ksi = 1.00 ksi (C)

sA = -18.0

18.0 - 18.0(3)

54.0 + -9.00(-1.5)

13.5

z = -1.5 iny = 3 in

s = F

A -

Mzy

Iz+Myz

Iy

Iz =1

12(3)(63) = 54.0 in4

Iy =1

12(6)(33) = 13.5 in4

A = 3 (6) = 18.0 in2

Mz = (MR)z; 12(3) - 6(3) = MzMz = 18.0 kip # in

My = (MR)y; 6(1.5) - 12(1.5) = MyMy = -9.00 kip # in

Fx = (FR)x; -6 - 12 = FF = -18.0 kip

845. Sketch the normal stress distribution acting over thecross section at section aa. Neglect the weight of the block.


a

a

6 in.

6 kip

12 kip3 in.

A B


17/30

785

For point C, and .

Ans.

For point D, . and .

Ans.

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

= 1.00 ksi (T)

sD = -18.0

18.0

-18.0(-3)

54.0

+ -9.00(-1.5)

13.5

z = -1.5 iny = -3 in

= -1.00 ksi = 1.00 ksi (C)

sC = -18.0

18.0 -

18.0(-3)

54.0 +

-9.00(1.5)

13.5

z = 1.5 iny = -3 in.

845. Continued


Ans:

sA = 1.00 ksi (C), sB = 3.00 ksi (C)


18/30

792


Equivalent Force System: As shown on FBD.

Section Properties:

Normal Stress:

At point B where and , we require .

Ans.

When

When

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

ey = 0,

ez 65

18a

ez = 0,

ey 6 56

a

6ey + 18ez 6 5a

0 7 -5a+ 6ey + 18ez

0 7P

30a4 C -5a2 - 6(-a) ey - 18( -a)ez D

sB 6 0z = -ay = -a

=P

30a4 A -5a2 - 6eyy - 18ezz B

=-P

6a2 -

Peyy

5a4 -

Pezz

53a

4

s =N

A -

Mzy

Iz+

Myz

Iy

=5

3a4

Iy =1

12(2a)(2a)3 + 2B 1

36(2a)a3 +

1

2(2a)aa a

3b2R

= 5a4

Iz =1

12(2a)(2a)3 + 2B 1

36(2a)a3 +

1

2(2a)aaa + a

3b2R

A = 2a(2a) + 2B12

(2a)aR = 6a2

851. A post having the dimensions shown is subjected tothe bearing load P. Specify the region to which this load canbe applied without causing tensile stress to be developed atpoints A, B, C, and D.

x

y

z

A

a a a a

a

a

D

ez

eyBC

P

Ans:

6ey + 18ez 6 5a


19/30

796



Section Properties: The moment of inertia about they and z axes of the posts crosssection is

Referring to Fig.b,


For point and . Then

Ans.

Shear Stress: Then transverse shear stress at point A is

Ans.

Ans.

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

[(txz)V]A =Vz(Qz)A

Iyt =

4(103)[0.125(10- 3)]

8.3333(10 - 6)(0.1)= 0.6 MPa

[(txy)V]A =Vy(Qy)A

Izt = 0

sA =1.2(103)(-0.05)

8.3333(10 - 6)+ 0 = 7.20 MPa (T)

z = 0A, y = -0.05 m

s = -Mzy

Iz+

Myz

Iy

(Qz)A = 0.025(0.05)(0.1) = 0.125(10- 3) m3

(Qy)A = 0

Iy = Iz =1

12(0.1)(0.13) = 8.3333(10 - 6) m4

Mz = -1.2 kN # mMz = 0; Mz - 3(0.4) = 0

My = -1.6 kN # mMy = 0; My + 4(0.4) = 0

Mx = 0; T = 0

Vz = -4 kNFz = 0; Vz + 4 = 0

Vy = -3 kNFy = 0; Vy + 3 = 0

855. Determine the state of stress at pointA on the crosssection of the post at section aa. Indicate the results on adifferential element at the point.

400 mm

3 kN4 kN

100 mm100 mm

400 mm

a

yx

a

z

50 mm

50 mm

50 mm

50 mm BA

Section a a


20/30

797


855. Continued

Ans:

sA = 7.20 MPa (T), tA = 0.6 MPa


21/30

798



Section Properties: The moment of inertia about theyand zaxes of the posts crosssection is

Referring to Fig.b,


For point and . Then

Ans.

Shear Stress: Then transverse shear stress at point Bis

Ans.

Ans.

The state of stress at point Bis represented on the elements shown in Figs.cand d,respectively.

c(txy)V dB

=

Vy(Qy)A

Izt =

3(103)[0.125(10- 3)]

8.3333(10- 6)(0.1)= 0.45 MPa

c(txy)V dB

=Vz(Qz)A

Iyt = 0

sB = -0 +-1.6(103)(-0.05)

8.3333(10 - 6)= 9.60 MPa (T)

z = -0.05 mB, y = 0

s = -Mzy

Iz+

Myz

Iy

(Qy)B = 0.025(0.05)(0.1) = 0.125(10- 3) m3

(Qz)B = 0

Iy = Iz =1

12(0.1)(0.13) = 8.3333(10- 6) m4

Mz = 1.2 kN # mMz = 0; Mz - 3(0.4) = 0

My = -1.6 kN # mMy = 0; My + 4(0.4) = 0

Mx = 0; T = 0

Vz = -4 kNFz = 0; Vz + 4 = 0

Vy = -3 kNFy = 0; Vy + 3 = 0

*856. Determine the state of stress at point B on the crosssection of the post at section aa. Indicate the results on adifferential element at the point.

400 mm

3 kN4 kN

100 mm100 mm

400 mm

a

yx

a

z

50 mm

50 mm

50 mm

50 mm BA

Section a a


22/30

799


856. Continued


23/30

802


859. If determine the maximum normal stressdeveloped on the cross section of the column.

P = 60 kN,

100 mm

15 mm

15 mm

15 mm

75 mm

150 mm

150 mm

100 mm

100 mm

P

2P

Equivalent Force System: Referring to Fig. a,

Section Properties: The cross-sectional area and the moment of inertia about the yand zaxes of the cross section are

Normal Stress: The normal stress is the combination of axial and bending stress.Here, F is negative since it is a compressive force. Also, and are negativesince they are directed towards the negative sense of their respective axes. Byinspection, pointA is subjected to a maximum normal stress.Thus,

Ans.= -71.0 MPa = 71.0 MPa (C)

smax = sA = -180(103)

0.01005 -

[-30(103)](-0.15)

0.14655(10 - 3)+

[-4.5(103)](0.1)

20.0759(10 - 6)

s =N

A -

Mzy

Iz+Myz

Iy

MzMy

Iy = 2 c 112

(0.015)(0.23) d + 112

(0.27)(0.0153) = 20.0759(10 - 6) m4

Iz =1

12(0.2)(0.33) -

1

12(0.185)(0.273) = 0.14655(10 - 3) m4

A = 0.2(0.3) - 0.185(0.27) = 0.01005 m2

Mz = (MR)z; -120(0.25) = -Mz Mz = 30 kN # m

My = (MR)y; -60(0.075) = -My My = 4.5 kN # m

+ c Fx = (FR)x; -60 - 120 = -F F = 180 kN

Ans:

smax = 71.0 MPa (C)


24/30

803


Equivalent Force System: Referring to Fig. a,

Section Properties: The cross-sectional area and the moment of inertia about the yand zaxes of the cross section are

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

Ans.P = 84470.40 N = 84.5k N

-100(106) = - 3P

0.01005 -

(-0.5P)(-0.15)

0.14655(10 - 3)+

-0.075P(0.1)

20.0759(10 - 6)

s =N

A -

Mzy

Iz+Myz

Iy

MzMy

Iy = 2 c1

12(0.15)(0.23) d +1

12(0.27)(0.0153) = 20.0759(10 - 6) m4

Iz =1

12(0.2)(0.33) -

1

12(0.185)(0.273) = 0.14655(10- 3)m4

A = 0.2(0.3) - 0.185(0.27) = 0.01005m2

Mz = 0.5P

Mz = (MR)z; -2P(0.25) = -Mz

My = 0.075P

My = (MR)y; -P(0.075) = -My

F = 3P

+ c Fx = (FR)x; -P - 2P = -F

*860. Determine the maximum allowable force P, if thecolumn is made from material having an allowable normalstress of .sallow = 100 MPa

100 mm

15 mm

15 mm

15 mm

75 mm

150 mm

150 mm

100 mm

100 mm

P

2P


25/30

822

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

a

Internal Loadings: Consider the equilibrium of the free-body diagram of the right

segment shown in Fig. b.

By = 196.2N554.94 sin45 - 20(9.81) - By = 0+ c Fy = 0;

Bx = 392.4N554.94 cos45 - Bx = 0:+ Fx = 0;

F = 554.94NFsin45(1) - 20(9.81)(2) = 0+ MB = 0;

875. The 20-kg drum is suspended from the hook mountedon the wooden frame. Determine the state of stress at pointEon the cross section of the frame at section aa. Indicate theresults on an element.


a M = 98.1N # m196.2(0.5) - M = 0+ MC = 0;

V = 196.2NV - 196.2 = 0+ c Fy = 0;

N = 392.4NN - 392.4 = 0:+ Fx = 0;

Section Properties: The cross -sectional area and the moment of inertia of the crosssection are

Referring to Fig. c,QE is


For point E, . Then

Ans.


Ans.

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

tE = VQA

It =

196.2 C31.25 A10- 6 B D

1.7578 A10- 6 B(0.05)= 69.8 kPa

sE =

392.4

3.75 A10- 3 B +98.1(0.0125)

1.7578 A10 - 6 B = 802 kPa

y = 0.0375 - 0.025 = 0.0125m

s =N

A ;My

I

QE = yA = 0.025(0.025)(0.05) = 31.25 A10- 6 B m3

I =1

12(0.05) A0.0753 B = 1.7578 A10- 6 B m4

A = 0.05(0.075) = 3.75 A10- 3 B m2

1 m

1 m

1 m

b

a

a

b

CB

A

30

1 m0.5 m0.5 m

50 mm

75 mm

25 mm

Section a a

E

75 mm

75 mm

25 mm

Section b b

FD


26/30

823

875. Continued


Ans:

sE = 802 kPa, tE = 69.8 kPa


27/30

824

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

a

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

a

Section Properties: The cross -sectional area and the moment of inertia about thecentroidal axis of the cross section are

Referring to Fig. c,QE is


For point F, . Then

Ans.= -695.24 kPa = 695 kPa (C)

sF = -422.75

5.625 A10 - 3 B -

130.8(0.0125)

2.6367 A10 - 6 B

y = 0.0375 - 0.025 = 0.0125 m

s =N

A ;My

I

QF = yA = 0.025(0.025)(0.075) = 46.875 A10- 6

B m3

I =1

12(0.075) A0.0753 B = 2.6367 A10- 6 B m4

A = 0.075(0.075) = 5.625 A10- 3 B m2

M = 130.8N # m130.8(1) - M = 0+ MC = 0;

N = 422.75N422.75 - N = 0+ c Fy = 0;

V = 130.8N130.8 - V = 0:+ Fx = 0;

Ax = 130.8NAx - 261.6sin30 = 0:+ Fx = 0;

Ay = 422.75NAy - 261.6cos30 - 20(9.81) = 0+ c Fy = 0;

FBD = 261.6NFBDsin30(3) - 20(9.81)(2) = 0+ MA = 0;

*876. The 20-kg drum is suspended from the hookmounted on the wooden frame. Determine the state of stressat point Fon the cross section of the frame at section .Indicate the results on an element.

b-b


1 m

1 m

1 m

b

a

a

b

CB

A

30

1 m0.5 m0.5 m

50 mm

75 mm

25 mm

Section a a

E

75 mm

75 mm

25 mm

Section b b

FD


28/30

825


Ans.

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

tA = VQA

It =

130.8 c46.875 A10- 6

Bd2.6367 A10- 6 B(0.075)

= 31.0 kPa

876. Continued



29/30

834


a

At point C,

Ans.

Ans.

At point D,

Ans.

Ans.tD = 0

sD = P

A

- Mc

I

=2(5.80)

1

-[2(5.80)](1)

0.333

= -23.2 ksi

tC = 0

sC = P

A =

2(5.80)

1 = 11.6 ksi

A = 2(0.25)(2) = 1 in2

I = 2 c 112

(0.25)(2)3 d = 0.333 in4FA = 11.60 kip

+ MB = 0 ; 12(3) + 10(8) - FA(10) = 0

885. The wall hanger has a thickness of 0.25 in. and isused to support the vertical reactions of the beam that isloaded as shown. If the load is transferred uniformly to eachstrap of the hanger, determine the state of stress at pointsCand D on the strap at A. Assume the vertical reactionF atthis end acts in the center and on the edge of the bracketas shown.

10 kip

A B

2 kip/ft

2 ft 2 ft 6 ft

2 in.

3.75 in.

2.75 in.

3 in.

1 in.

2 in.

2 in.

F

CD

1 in.

Ans:

sD = -23.2 ksi, tD = 0

sC = 11.6 ksi, tC = 0,


30/30


a

At point C:

Ans.

Ans.

At point D:

Ans.

Ans.tD = 0

sD = P

A -Mc

I =

2(5.20)

1 -

[2(5.20)](1)

0.333 = -20.8 ksi

tC = 0

sC = P

A =

2(5.20)

1 = 10.4 ksi

I = 2 c 112

(0.25)(2)3 d = 0.333 in4; A = 2(0.25)(2) = 1 in2+ MA = 0; FB(10) - 10(2) - 12(7) = 0; FB = 10.40 kip

886. The wall hanger has a thickness of 0.25 in. and isused to support the vertical reactions of the beam that isloaded as shown. If the load is transferred uniformly to eachstrap of the hanger,determine the state of stress at pointsCand D on the strap at B. Assume the vertical reaction F atthis end acts in the center and on the edge of the bracketas shown.

10 kip

A B

2 kip/ft

2 ft 2 ft 6 ft

2 in.

3.75 in.

2.75 in.

3 in.

1 in.

2 in.

2 in.

F

CD

1 in.

Ans:

ej cargas combinadas

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