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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)
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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, . 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
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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;
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Ans:
smax = 2.34 MPa (C)
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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;
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Ans:
P = 128 kN
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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
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Ans:
sD = - 88.0 MPa, tD = 0
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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.
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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
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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.
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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
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840. Continued
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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
Normal Stress: The normal stress is the combination of axial and bending stress.Thus,
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.
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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
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841. Continued
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Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,
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
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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.
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a a
5 ft
1.5 ft
300 lb400 lb
2 in.
2.5 in.
A
B
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842. Continued
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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
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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,
Normal Stress: The normal stress is contributed by bending stress only.Thus,
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.
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843. Continued
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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
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Referring to Fig. a,
The cross-sectional area and moments of inertia about they and z axes of the crosssection are
The normal stress developed is the combination of axial and bending stress.Thus,
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.
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.
a
a
6 in.
6 kip
12 kip3 in.
A B
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Referring to Fig. a,
The cross-sectional area and the moment of inertia about the y and z axes of thecross section are
The normal stress developed is the combination of axial and bending stress.Thus,
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.
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.
a
a
6 in.
6 kip
12 kip3 in.
A B
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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
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Ans:
sA = 1.00 ksi (C), sB = 3.00 ksi (C)
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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
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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.
Internal Loadings: Considering the equilibrium of the free-body diagram of theposts upper segment, Fig.a,
Section Properties: The moment of inertia about they and z axes of the posts crosssection is
Referring to Fig.b,
Normal Stress: The normal stress is contributed by bending stress only.Thus,
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
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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.
855. Continued
Ans:
sA = 7.20 MPa (T), tA = 0.6 MPa
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798
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.
Internal Loadings: Considering the equilibrium of the free-body diagram of theposts upper segment, Fig.a,
Section Properties: The moment of inertia about theyand zaxes of the posts crosssection is
Referring to Fig.b,
Normal Stress: The normal stress is contributed by bending stress only.Thus,
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
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799
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.
856. Continued
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802
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.
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)
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803
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.
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
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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.
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.
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
Normal Stress: The normal stress is the combination of axial and bending stress.Thus,
For point E, . Then
Ans.
Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,
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
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823
875. Continued
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.
Ans:
sE = 802 kPa, tE = 69.8 kPa
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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
Normal Stress: The normal stress is the combination of axial and bending stress.Thus,
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
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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
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825
Shear Stress: The shear stress is contributed by transverse shear stress only.Thus,
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
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.
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834
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.
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,
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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.
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:
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