moocMechanics of Materials章节答案(mooc2023课后作业答案)
43 min readmoocMechanics of Materials章节答案(mooc2023课后作业答案)
1.2.1 Review Statics随堂测验
1、章节作业Determine the internal force in member EF.
A、答案答案716.025N
B、课后616.025N
C、章节作业516.025N
D、答案答案416.025N
1.2.2 Internal Loadings随堂测验
1、课后
A、章节作业0，答案答案 -2.5kN,课后 37.5kNm
B、0，章节作业 -5.5kN,答案答案 37.5kNm
C、0，课后 -2.5kN,章节作业 35.5kNm
D、0，答案答案 -5.5kN,课后 35.5kNm
1 _ Homework
1、
2、
3、
4、
Lecture 2
1.4.1 Axial Force Diagram随堂测验
1、
A、10kN
B、12kN
C、14kN
D、8kN
1.4.2 Average Normal Stress in Axially Loaded Member随堂测验
1、
A、14 MPa
B、13 MPa
C、12 MPa
D、11 MPa
2 _ Homework
1、
2、
3、
Lecture 3
1.5 Average Shear Stress随堂测验
1、
A、35.4MPa
B、45.4MPa
C、55.4MPa
D、65.4MPa
1.6 Allowable Stress随堂测验
1、
A、h1=1.8 in., h2=3.6in.
B、h1=2.8 in., h2=3.6in.
C、h1=1.8 in., h2=4.6in.
D、h1=2.8 in., h2=4.6in.
1.7 Design of Simple Connections随堂测验
1、
A、88.1MPa, 81.9MPa, 81.9MPa
B、98.1MPa, 81.9MPa, 81.9MPa
C、88.1MPa, 91.9MPa, 91.9MPa
D、98.1MPa, 91.9MPa, 91.9MPa
3 _ Homework
1、
2、
3、
Lecture 4
3.4-3.5 Strain Energy随堂测验
1、
A、
B、
C、
D、
Examination_1_ Chapter 1& Chapter 2
1、
A、24kN
B、-24kN
C、40kN
D、-40kN
2、
A、3kN
B、2kN
C、4kN
D、5kN
3、
A、5kN
B、4kN
C、6kN
D、2kN
4、
A、kL/
B、2kL/
C、kL/2
D、kL/3
5、
A、37.7 kN
B、27.7 kN
C、47.7 kN
D、57.7 kN
6、
A、FN=-800N, Vx=-300N, Vz=771N, Mx=2.11kNm, Ty=-600Nm, Mz=600Nm
B、FN=800N, Vx=300N, Vz=771N, Mx=2.11kNm, Ty=600Nm, Mz=600Nm
C、FN=-800N, Vx=-600N, Vz=771N, Mx=5.11kNm, Ty=-600Nm, Mz=600Nm
D、FN=-800N, Vx=-300N, Vz=871N, Mx=2.11kNm, Ty=-800Nm, Mz=600Nm
7、
A、16 kN/m
B、18 kN/m
C、20 kN/m
D、14 kN/m
8、
A、FN=10N, V=17.3N, M=1.60Nm
B、FN=10N, V=7.3N, M=1.60Nm
C、FN=10N, V=17.3N, M=3.60Nm
D、FN=20N, V=17.3N, M=1.60Nm
9、
A、
B、
C、
D、
10、
A、25.5MPa， 4.72MPa
B、35.5MPa， 4.72MPa
C、25.5MPa， 6.72MPa
D、35.5MPa， 6.72MPa
11、
A、FN=150N, FS=200N, M=10Nm
B、FN=200N, FS=150N, M=10Nm
C、FN=150N, FS=200N, M=20Nm
D、FN=200N, FS=200N, M=20Nm
12、
A、FN=20kN, FS=0, M=40kNm
B、FN=40kN, FS=0, M=40kNm
C、FN=20kN, FS=10, M=40kNm
D、FN=20kN, FS=0, M=20kNm
13、
A、FN=0, FS=1kN, M=22kNm
B、FN=0, FS=10kN, M=22kNm
C、FN=0, FS=1kN, M=32kNm
D、FN=1, FS=1kN, M=22kNm
14、
A、shear stress: 106MPa, bearing stress:141MPa
B、shear stress: 116MPa, bearing stress:141MPa
C、shear stress: 106MPa, bearing stress:151MPa
D、shear stress: 116MPa, bearing stress:151MPa
15、
A、137.2kN
B、127.2kN
C、117.2kN
D、147.2kN
16、
A、D=d
B、D=2d
C、D=3d
D、D=4d
17、
A、d=26.36mm, t=15.75mm.
B、d=16.36mm, t=10.75mm.
C、d=16.36mm, t=15.75mm.
D、d=26.36mm, t=10.75mm.
18、
A、shear stress: 45.86MPa, bearing stress:180MPa
B、shear stress: 55.86MPa, bearing stress:180MPa
C、shear stress: 45.86MPa, bearing stress:190MPa
D、shear stress: 55.86MPa, bearing stress:190MPa
19、
A、shear and bearing stresses : 100.09 MPa and 235.70 MPa
B、shear and bearing stresses : 110.09 MPa and 235.70 MPa
C、shear and bearing stresses : 100.09 MPa and 255.70 MPa
D、shear and bearing stresses : 110.09 MPa and 255.70 MPa
20、
A、19mm
B、20mm
C、18mm
D、21mm
4 _ Homework
1、
2、
Lecture 5
4.2 Elastic Deformation of an Axially Loaded Member随堂测验
1、
A、2.31mm, 2.64mm
B、2.31mm, 3.64mm
C、3.31mm, 2.64mm
D、3.31mm, 3.64mm
4.2 Problem Discussion_Displacement calculation随堂测验
1、
A、17.3 mm
B、16.3 mm
C、15.3 mm
D、18.3 mm
5 _ Homework
1、
2、
3、
4、
Lecture 6
4.2 Displacement Calculation_Energy Method随堂测验
1、
A、2.03mm
B、1.03mm
C、3.03mm
D、4.03mm
4.4 Statically Indeterminate Axially Loaded Member随堂测验
1、
A、Post: 5.34MPa, Tube: 2.79MPa
B、Post: 8.34MPa, Tube: 2.79MPa
C、Post: 5.34MPa, Tube: 5.79MPa
D、Post: 8.34MPa, Tube: 5.79MPa
2、
A、FA=8P/17, FC=9P/17
B、FA=9P/17, FC=8P/17
C、FA=2P/3, FC=P/3
D、FA=P/3, FC=2P/3
6 _ Homework
1、
2、
3、
4、
Lecture 7
4.5 Problem Discussion_Assembling Stress随堂测验
1、
A、FA=179.63kN, FD=20.4kN
B、FA=279.63kN, FD=20.4kN
C、FA=179.63kN, FD=50.4kN
D、FA=279.63kN, FD=50.4kN
4.6 Thermal Stress随堂测验
1、
A、904N
B、804N
C、704N
D、604N
Examination_2_Chapter 4
1、
A、0.262 in.
B、0.524 in.
C、0.131 in.
D、1.048 in.
2、
A、3.6 kN
B、7.2 kN
C、14.4 kN
D、28.8 kN
3、
A、0.0367 in.
B、0.0733 in.
C、0.1466 in.
D、0.2932 in.
4、
A、0.00129 in.
B、0.00257 in.
C、0.00514 in.
D、0.01028 in.
5、
A、T2=56℉ σal=σcu=12.8ksi
B、T2=112℉ σal=σcu=12.8ksi
C、T2=56℉ σal=σcu=25.6ksi
D、T2=112℉ σal=σcu=25.6ksi
6、
A、FAC = FAB = 10.0 lb FAD = 136 lb
B、FAC = FAB = 5.0 lb FAD = 136 lb
C、FAC = FAB = 10.0 lb FAD = 272 lb
D、FAC = FAB = 5.0 lb FAD = 272 lb
7、
A、σb=67MPa σr=16.8MPa
B、σb=33.5MPa σr=16.8MPa
C、σb=33.5MPa σr=8.4MPa
D、σb=67MPa σr=8.4MPa
8、
A、A
B、B
C、C
D、D
9、
A、50.6MPa, 33.8MPa
B、55.6MPa, 30.8MP ?
C、50.6MPa, 30.8MPa
D、55.6MPa, 33.8MPa
10、
A、CD:67.34MPa, AB、EF：412.65MPa
B、CD:77.34MPa, AB、EF：412.65MPa
C、CD:67.34MPa, AB、EF：512.65MPa
D、CD:77.34MPa, AB、EF：512.65MPa
11、
A、FC=1.04kN, FB=-0.08kN?
B、FC=2.04kN, FB=-0.08kN ?
C、FC=1.04kN, FB=-1.08kN ?
D、FC=2.04kN, FB=-1.08kNN
12、
A、165.7MPa, 165.7MPa
B、165.7MPa, 185.7MPa
C、185.7MPa, 165.7MPa
D、185.7MPa, 185.7MPa
13、
A、17.5MPa, 17.5MPa, -35MPa
B、18.5MPa, 18.5MPa, -37MPa
C、19.5MPa, 19.5MPa, -39MPa
D、20.5MPa, 20.5MPa, -41MPa
14、
A、1.242F, 1.242F
B、2.242F, 1.242F
C、1.242F, 2.242F
D、2.242F, 2.242F
15、The outer radius of the steel pipe is 20 mm and the inner radius is 15 mm. If it is close to the fixed wall before loading, what is the reaction force of the wall under the load shown?
A、8.0kN,8.0kN
B、11.2kN,4.8kN
C、3.6kN,12.4kN
D、12.0kN,4.0kN
16、Steel bolts with a diameter of 10 mm are surrounded by bronze sleeves. The outer diameter of the sleeve is 20 mm and the inner diameter is 10 mm. If the yield stress of steel is 640MPa, E = 200GPa; if the yield stress of bronze is 520MPa, E = 100GPa. What is the maximum elastic load P that can be applied to the component?
A、204kN
B、50kN
C、126kN
D、150kN
17、teel bolts with a diameter of 10 mm are surrounded by bronze sleeves. The outer diameter of the sleeve is 20 mm and the inner diameter is 10 mm. The elastic modulus of steel and bronze are 200 GPa and 100 GPa respectively. If the bolt is subjected to a pressure of P = 20KN, what is the average stress of steel and bronze?
A、80.2MPa,71.0MPa
B、102.0MPa,50.9MPa
C、121.0MPa,36.0MPa
D、138.2MPa,21.0MPa
7 _ Homework
1、
2、
3、
Lecture 8
5.1 Problem Discussion_Torque Diagram随堂测验
1、Point out the torque acting on section A.
A、150Nm
B、-150Nm
C、50Nm
D、-50Nm
5.2 The Torsion Formula随堂测验
1、
A、AB: 23.9MPa, BC:15.9MPa
B、AB: 33.9MPa, BC:15.9MPa
C、AB: 23.9MPa, BC:25.9MPa
D、AB: 33.9MPa, BC:25.9MPa
8 _ Homework
1、Plot torque diagram for the following shaft.
2、Plot torque diagram for the following shaft.
3、
Lecture 10
A.1 Centriod of an Area随堂测验
1、
A、0.99in, 1.99in
B、0.89in, 1.99in
C、0.99in, 1.89in
D、0.89in, 1.89in
A.2 Parallel Axis Theorem随堂测验
1、
A、24867.35cm
B、14867.35cm
C、34867.35cm
D、44867.35cm
A.3 Product of Inertia for an Area随堂测验
1、
A、-0.0417
B、-0.0517
C、-0.0617
D、-0.0717
10 _ Homework
1、Locate the centroid C of the cross-sectional area for the T-beam shown in Fig. (a).
2、
3、Determine the moments of inertia of the beam's cross-sectional area shown in Fig. (a) about the x and y centroidal axes.
4、
Lecture 12
6.1 Shear and Moment at Appointed Section随堂测验
1、Determine the shear force and bending moment on section C.
A、30 kN， 0
B、0 kN， 30 kN.m
C、30 kN， 30 kN.m
D、30 kN， 60 kN.m
6.1 Shear and Moment Diagrams随堂测验
1、Draw the shear and moment diagrams for the following beam and point out the maximum shear force and bending moment on the beam.
A、5kN, 5kNm
B、10kN, 10kNm
C、5kN, 10kNm
D、10kN, 5kNm
12 _ Homework
1、The dead-weight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.
2、Draw the shear and moment diagrams for the overhang beam.
3、Express the shear and moment in terms of x for 0 < x< 3m and 3m < x < 4.5 m, and then draw the shear and moment diagrams for the simply supported beam.
4、Draw the shear and moment diagrams for the simply supported beam.
Lecture 13
6.2 Graphical Method for Constructing Shear and Moment Diagrams随堂测验
1、Draw the shear and moment diagrams for the simply supported beam. Point out the maximum shear force and bending moment.
A、12kN, 22kNm
B、12kN, 20 kNm
C、4 kN, 22.4kNm
D、4 kN, 20 kNm
2、For the given shear force diagram, plot the bending moment diagram and point out the distributed load applied on the beam.
A、6kN/m
B、5kN/m
C、4kN/m
D、3kN/m
3、For the given moment diagram, plot the shear diagram and point out the distributed load applied on the beam.
A、downward q
B、downward 2q
C、upward q
D、upward 2q
13 _ Homework
1、Draw the shear and moment diagrams for the compound beam.
2、Draw the shear and moment diagrams for the simply supported beam.
3、Draw the shear and moment diagrams for the beam.
4、The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.
Lecture 15
6.4 The Flexure Formula_Transverse Bending随堂测验
1、
A、200 kN
B、100 kN
C、300 kN
D、400 kN
2、
A、7.0 MPa
B、6.0 MPa
C、8.0 MPa
D、9.0 MPa
15 _ Homework
1、Determine the moment M that will produce a maximum stress of 70 MPa on the cross-section.
2、Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M=6kN.m.
3、The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.
4、An A-36 steel strip has an allowable bending stress of 165 MPa. If it is rolled up, determine the smallest radius r of the spool if the strip has a width of 10 mm and a thickness of 1.5 mm. Also, find the corresponding maximum internal moment developed in the strip.
Lecture 24
12.2 Slope and Displacement by Integration随堂测验
1、
A、0.0905in. , 0.293in.
B、0.1905in. , 0.293in.
C、0.0905in. , 1.293in.
D、0.1905in. , 1.293in.
24 _ Homework
1、
2、
3、
4、
Lecture 9
5.4 Angle of Twist随堂测验
1、
A、1度
B、2度
C、3度
D、4度
5.5 Statically Indeterminate Torque Loaded Members随堂测验
1、
A、Me/17, Me/17
B、Me/15, Me/15
C、Me/13, Me/13
D、Me/19, Me/19
5.6 Strain Energy in Torsion & Noncircular Shaft随堂测验
1、
A、64/3GIp
B、32/3GIp
C、16/3GIp
D、8/3GIp
Examination_3_Chapter 5
1、
A、3.45ksi
B、6.9ksi
C、13.8ksi
D、27.6ksi
2、
A、11.2MPa
B、22.4MPa
C、44.8MPa
D、89.6MPa
3、
A、3.67ksi
B、7.33ksi
C、14.66ksi
D、29.32ksi
4、
A、d=12.5mm
B、d=25mm
C、d=50mm
D、d=100mm
5、
A、0.1428°
B、0.2875°
C、0.575°
D、1.15°
6、
A、0.216°
B、0.432°
C、0.864°
D、1.728°
7、
A、2.44MPa
B、4.89MPa
C、9.77MPa
D、19.54MPa
8、
A、7.325ksi
B、14.65ksi
C、29.3ksi
D、58.6ksi
9、
A、48.63psi
B、97.25psi
C、194.5psi
D、389psi
10、
A、34.1MPa
B、68.2MPa
C、136.4MPa
D、272.8MPa
9 _ Homework
1、
2、
3、
Lecture 11
A.4 Moment of Inertia for an Area about Inclined Axes随堂测验
1、
A、13.4(10+6)mm4, 1.4(10+6)mm4
B、15.4(10+6)mm4, 1.4(10+6)mm4
C、13.4(10+6)mm4, 2.4(10+6)mm4
D、15.4(10+6)mm4, 2.4(10+6)mm4
Examination 4_Appendix A
1、
A、11.5775×10-6m4
B、23.1554×10-6m4
C、46.31×10-6m4
D、96.62×10-6m4
2、
A、0.1907×10-3m4
B、0.3814×10-3m4
C、0.7628×10-3m4
D、1.5256×10-3m4
3、
A、14.4004×10-6m4
B、28.8007×10-6m4
C、57.6014×10-6m4
D、115.2028×10-6m4
4、
A、2.6066×10-3m4
B、5.2132×10-3m4
C、10.4264×10-3m4
D、20.8528×10-3m4
5、
A、1.3184×10-6m4
B、2.6367×10-6m4
C、5.2734×10-6m4
D、10.5468×10-6m4
6、
A、0.0455×10-3m4
B、0.0909×10-3m4
C、0.1818×10-3m4
D、0.3636×10-3m4
7、
A、A
B、B
C、C
D、D
8、
A、A
B、B
C、C
D、D
9、
A、A
B、B
C、C
D、D
10、
A、A
B、B
C、C
D、D
11、
A、A
B、B
C、C
D、D
12、
A、A
B、B
C、C
D、D
13、
A、A
B、B
C、C
D、D
14、
A、A
B、B
C、C
D、D
15、
A、A
B、B
C、C
D、D
16、
A、A
B、B
C、C
D、D
17、
A、A
B、B
C、C
D、D
18、
A、A
B、B
C、C
D、D
19、
A、A
B、B
C、C
D、D
20、
A、A
B、B
C、C
D、D
21、
A、A
B、B
C、C
D、D
22、
A、A
B、B
C、C
D、D
23、
A、A
B、B
C、C
D、D
11 _ Homework
1、Determine the principal moments of inertia for the beam's cross-sectional area shown in Fig. A-15 with respect to an axis passing through the centroid C.
Lecture 14
6.2 Construction of Moment Diagram by Principle of Superposition随堂测验
1、Plot bending moment diagram by the priciple of superposition. Point out the maximum value of bending moment.
A、
B、
C、
D、
6.2 Internal Loading Diagrams of Inclined or Curved Beams随堂测验
1、
A、
B、
C、
D、
2、
A、1.06 kN, 1.5kNm
B、2.06 kN, 1.5kNm
C、1.06 kN, 2.5kNm
D、2.06 kN, 2.5kNm
6.2 Internal Loading Diagrams of Frames随堂测验
1、Plot the shear and moment diagrams for the following frame. Please specify the maximum values of shear force and bending moment.
A、qa,
B、qa/2,
C、qa,
D、qa/2,
Examination 5 _ Shear and moment diagrams
1、
A、VB = -45 kN, MB = -63 kN ? m
B、VB = -22.5 kN, MB = -63 kN ? m
C、VB = -45 kN, MB = -31.5 kN ? m
D、VB = -22.5 kN, MB = -31.5 kN ? m
2、
A、V∣x =15 ft = 2.24 kip, M∣x =15 = -1.95 kip · ft
B、V∣x =15 ft = 1.12 kip, M∣x =15 = -1.95 kip · ft
C、V∣x =15 ft = 2.24 kip, M∣x =15 = -0.975 kip · ft
D、V∣x =15 ft = 1.12 kip, M∣x =15 = -0.975 kip · ft
3、
A、V = 1050 - 150x ,M = -75x2 + 1050x - 3200
B、V = 1050 - 75x ,M = -75x2 + 1050x - 3200
C、V = 1050 - 75x ,M = -37.5x2 + 1050x - 3200
D、V = 1050 - 150x ,M = -37.5x2 + 1050x - 3200
4、
A、V = -315 lb, M = { -630x + 2400} lb · ft
B、V = -315 lb, M = { -315x + 1200} lb · ft
C、V = -630 lb, M = { -630x + 2400} lb · ft
D、V = -630 lb, M = { -315x + 1200} lb · ft
5、
A、V = { 1050 - 150 x} lb, M = { -37.5x2 + 525x - 2000} lb ? ft
B、V = { 525 - 75 x} lb, M = { -75x2 + 1050x - 4000} lb ? ft
C、V = { 525 - 75 x} lb, M = { -37.5x2 + 5250x - 2000} lb ? ft
D、V = { 1050 - 150 x} lb, M = { -75x2 + 1050x - 4000} lb ? ft
6、
A、V = { 30.0 - 2 x} kip, M = { -x2 + 30.0x - 216} kip ? ft
B、V = { 15.0 - x} kip, M = { -0.5x2 + 15.0x - 108} kip ? ft
C、V = { 15.0 - x} kip, M = { -x2 + 30.0x - 216} kip ? ft
D、V = { 30.0 - 2 x} kip, M = { -0.5x2 + 15.0x - 108} kip ? ft
7、
A、V = 31.2 N, M = { 31.2x + 200} N? m
B、V = 31.2 N, M = { 15.6x + 100} N? m
C、V = 15.6 N, M = { 15.6x + 100} N? m
D、V = 15.6 N, M = { 31.2x + 200} N? m
8、
A、V = -1.0 kip, M = { -45 - 1.0x} kip ? ft
B、V = -1.0 kip, M = { -22.5 - 0.5x} kip ? ft
C、V = -0.5 kip, M = { -45 - 1.0x} kip ? ft
D、V = -0.5 kip, M = { -22.5 - 0.5x} kip ? ft
9、
A、V = -16 kip, M = [8(18 - x)] kip ? ft
B、V = -16 kip, M = [16(36 - x)] kip ? ft
C、V = -8 kip, M = [16(36 - x)] kip ? ft
D、V = -8 kip, M = [8(18 - x)] kip ? ft
10、
A、A
B、B
C、C
D、D
11、
A、A
B、B
C、C
D、D
12、
A、A
B、B
C、C
D、D
13、The internal force on the right section of beam section C is ( )
A、
B、
C、
D、
14、The maximum shear force and bending moment on the beam shown in the figure are ( )
A、30kN, 29.36kNm
B、30kN, 20kNm
C、15kN, 29.36kNm
D、15kN, 20kNm
15、Is the shear moment diagram of the beam shown in the figure correct?
A、correct
B、section B bending moment error
C、Section B shear error
D、Section A bending moment error
16、Is the shear force and bending moment diagram of the beam shown in the figure correct?
A、correct
B、Wrong bending moment value of Section D
C、Wrong shear value of Section D
D、The bending moment of section C is correct
17、The concentrated couple value of the beam corresponding to the shear diagram in the figure is
A、0
B、5kNm
C、10kNm
D、15kNm
18、The concentrated couple value of the beam corresponding to the shear diagram in the figure is
A、0
B、0.2kNm
C、0.5kNm
D、0.8kNm
19、The shear force on the beam corresponding to the bending moment diagram is
A、0
B、1 kN
C、2kN
D、3kN
20、The distributed load on the beam corresponding to the bending moment diagram in the figure is
A、0
B、q
C、2q
D、0.5q
21、The maximum shear force and bending moment on the beam shown in the figure are:
A、5kN, 5kNm
B、5kN, 10kNm
C、10kN, 5kNm
D、10kN, 10kNm
22、The maximum shear force and bending moment on the beam shown in the figure are:
A、80kN, 160kNm
B、50kN, 160kNm
C、80kN, 100kNm
D、50kN, 100kNm
23、The maximum shear force and bending moment on the beam shown in the figure are:
A、
B、
C、
D、
24、The maximum shear force and bending moment on the beam shown in the figure are:
A、38.5kN, 77.01kNm
B、32.625kN, 77.01kNm
C、38.5kN, 29.43kNm
D、32.625kN, 29.43kNm
25、As shown in the figure, the simply supported beam is subjected to various external loads. When the distributed external force acts on section AB, and its concentration is w = 1 ? kn / m, the maximum shear force and bending moment on the beam are
A、3.2kN, 6.01kNm
B、3.2kN, 5.67kNm
C、1.83kN, 6.01kNm
D、1.83kN, 5.67kNm
14 _ Homework
1、
2、Draw the shear and moment diagrams for the compound beam.
3、For the given shear force diagram, construct moment and load diagrams. There is no concentrated moment applied on the beam.
4、For the given moment diagram, construct shear and load diagrams.
5、Construct the shear and moment diagrams of the frame.
6、
Lecture 16
6.5 Unsymmetric Bending随堂测验
1、
A、
B、
C、
D、
16 _ Homework
1、The member has a square cross section and is subjected to the moment M = 850N. m as shown. Determine the stress at each corner and sketch the stress distribution. Set θ= 30°.
2、If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520N.m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut's cross-sectional area must be determined. Also, specify the orientation of the neutral axis.
3、The 65-mm-diameter steel shaft is subjected to the two loads. If the journal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximum bending stress developed in the shaft.
4、
Lecture 19
8.1 Thin-Walled Vessels随堂测验
1、
A、2.88 MPa, 0.452
B、3.88 MPa, 0.452
C、2.88 MPa, 0.552
D、3.88 MPa, 0.552
8.2 State of Stress Caused by Combined Loadings随堂测验
1、
A、σ=6532 MPa， τ=1220 MPa
B、σ=653.2 MPa， τ=122 MPa
C、σ=6532 MPa， τ=122 MPa
D、σ=653.2 MPa， τ=1220MPa
19 _ Homework
1、
2、
3、
4、
Lecture 17
7.3 Shear Stresses in Beams随堂测验
1、
A、97.2 MPa
B、87.2 MPa
C、77.2 MPa
D、67.2 MPa
17 _ Homework
1、
2、
3、
4、
Lecture 20
9.3 Principle Stresses and Maximum In-Plane Shear随堂测验
1、
A、σ1=27.8 MPa，σ2=0，σ3=-112.8 MPa, τmax=70.3MPa
B、σ1=37.8 MPa，σ2=0，σ3=-112.8 MPa, τmax=80.3MPa
C、σ1=27.8 MPa，σ2=0，σ3=-212.8 MPa, τmax=70.3MPa
D、σ1=37.8 MPa，σ2=0，σ3=-212.8 MPa, τmax=80.3MPa
20 _ Homework
1、
2、
3、
4、
Lecture 18
7.5 Shear Flow in Thin-Walled Members随堂测验
1、
A、a=106mm, s=44.3mm
B、a=116mm, s=44.3mm
C、a=106mm, s=54.3mm
D、a=116mm, s=54.3mm
Examination 6 _ Chapter 6 & Chapter 7
1、
A、(σt)max = 3.72 ksi, (σc)max = 1.78 ksi
B、(σt)max = 7.44 ksi, (σc)max = 1.78 ksi
C、(σt)max = 7.44 ksi, (σc)max = 3.56 ksi
D、(σt)max = 3.72 ksi, (σc)max = 3.56 ksi
2、
A、F = 21 kip
B、F = 10.5 kip
C、F = 42 kip
D、F = 5.25 kip
3、
A、F = 9.12kN
B、F = 2.28 kN
C、F = 18.24 kN
D、F = 4.56 kN
4、
A、F = 753 lb
B、F = 376.5 lb
C、F = 1506 lb
D、F = 188.25 lb
5、
A、d = 2.56in.
B、d = 1.28 in.
C、d = 5.12 in.
D、d = 0.64 in.
6、
A、d = 11/2in.
B、d = 11/4in.
C、d = 11/8in.
D、d = 11/16in.
7、
A、τmax = 1.955MPa
B、τmax = 3.91 MPa
C、τmax =7.82 MPa
D、τmax = 15.64 MPa
8、
A、τmax = 5.625 MPa
B、τmax = 11.25 MPa
C、τmax = 22.5 MPa
D、τmax = 45.0 MPa
9、
A、Pmax = 0.64 kip
B、Pmax = 1.28 kip
C、Pmax = 2.56 kip
D、Pmax = 5.12 kip
10、
A、w = 11.3 kip/ft, τmax = 531 psi
B、w = 11.3 kip/ft, τmax = 1062 psi
C、w = 22.6 kip/ft, τmax = 531 psi
D、w = 22.6 kip/ft, τmax = 1062 psi
11、The load of the overhanging beam is shown in the figure. If the [σ] = 10MPa, [τ] = 1.5MPa, the beam diameter d = 200mm, a = 1m, determine the maximum load that the beam can bear.
A、7.85kN
B、8.85kN
C、9.85kN
D、10.85kN
12、Determine the maximum normal stress and maximum shear stress on the cross section of the beam shown in the figure.
A、297.4MPa， 39.6MPa
B、297.4MPa， 29.6MPa
C、197.4MPa， 39.6MPa
D、197.4MPa， 29.6MPa
13、If the moment of inertia of the cross-section beam to the neutral axis is Iz=2910000mm4，yC=65mm, and C is the centroid, determine the maximum tensile stress, maximum compressive stress and maximum shear stress on the beam.
A、45.6MPa， -67MPa, 4.4MPa
B、45.6MPa， -80MPa, 5.4MPa
C、55.6MPa， -67MPa, 5.4MPa
D、55.6MPa， -80MPa, 4.4MPa
14、The geometric dimensions L, B and h of the beam shown in the figure are known. If the allowable normal stress of the beam is [σ], determine the maximum shear stress in the beam when σ max = [σ].
A、τmax=h[σ]/3l
B、τmax=2h[σ]/3l
C、τmax=2h[σ]/l
D、τmax=h[σ]/l
15、Determine the maximum normal stress and maximum shear stress on the cross section of the beam shown in the figure.
A、25.13MPa，2.04MPa
B、25.13MPa，1.04MPa
C、15.13MPa，2.04MPa
D、15.13MPa，1.04MPa
16、As shown in the figure, the cross-section of the beam is welded by three steel plates. If the shear force acting on the cross-section is FS=18 kN, determine the force that must be borne by the weld per unit length. Dimensions in mm.
A、47.7N
B、37.7N
C、27.7N
D、17.7N
17、The bending moment on the cross-section of the beam is M=60 kN·m, and the cross-section size is shown in the figure. Determine the maximum normal stress on this section. (dimensions in the figure are in mm)
A、59.3MPa
B、49.3MPa
C、39.3MPa
D、29.3MPa
18、The diamond shaped beam is shown bending around the z-axis. Given that the shear force on the cross-section is FS, determine the location of the maximum shear stress on the cross-section. (it can be calculated by shear stress formula of rectangular section beam)
A、At neutral axis
B、h / 2 from neutral axis
C、h / 4 from neutral axis
D、h / 8 from neutral axis
19、The cross section of the beam is shown in the figure. If the compressive strain of the upper longitudinal fiber is ε=－0.0003, and the tensile strain of the lower longitudinal fiber is ε=0.0006. Determine the total normal internal force of the shadow part on the section. E = 200 GPa for known materials.
A、700kN
B、800kN
C、900kN
D、1000kN
20、The solid shaft is supported by smooth bearings A and B. the bearing only provides the reaction force in the vertical direction. If the diameter of the circular shaft d = 90mm, determine the maximum normal stress and maximum shear stress on the circular shaft.
A、158MPa, 4.09MPa
B、158MPa, 8.09MPa
C、258MPa, 4.09MPa
D、258MPa, 8.09MPa
21、The solid shaft is supported by smooth bearings A and B. the bearing only provides the reaction force in the vertical direction. If the allowable normal stress of the material is [σ] = 180MPa and the allowable shear stress is [σ] = 80MPa, determine the minimum diameter d of the circular shaft.
A、8.63cm
B、2.03cm
C、10.63cm
D、4.03cm
22、The cross-section steel beam is subjected to distributed load and the section size is shown in the figure. The normal stress and shear stress at the junction A of web and flange on section A are ( )
A、A
B、B
C、C
D、D
23、Two boards are nailed together to form a simply supported timber beam as shown in the figure. The bearing load is P = 20KN. If the maximum shear force that the nail can bear is 2.5kN and the cross section size is a = 5cm, determine the maximum spacing S of nails.
A、0.15m
B、0.16m
C、0.17m
D、0.18m
18 _ Homework
1、
2、
3、
4、
Lecture 26
13.3 Columns Having Various Types of Supports随堂测验
1、
A、604 kN
B、504 kN
C、704 kN
D、804 kN
26 _ Homework
1、
2、
3、
4、
Lecture 27
13.4 Critical Stress随堂测验
1、
A、AB: 2.125 in. , BC: 2 in.
B、AB: 3.125 in. , BC: 2 in.
C、AB: 2.125 in. , BC: 3 in.
D、AB: 3.125 in. , BC: 3 in.
Examination 9_ Chapter 13
1、
A、P = 8.8 kip
B、P = 17.6 kip
C、P = 35.2 kip
D、P = 70.4 kip
2、
A、Pcr = 81.25 kN
B、Pcr = 162.5 kN
C、Pcr = 325 kN
D、Pcr = 650 kN
3、
A、P = 2.42 kip
B、P = 4.84 kip
C、P = 9.68 kip
D、P = 19.36 kip
4、
A、P = 31.15 kip
B、P = 62.3 kip
C、P = 124.6 kip
D、P = 249.2 kip
5、
A、d=9/16 in.
B、d=9/8 in.
C、d=9/4 in.
D、d=9/2 in.
6、
A、P = 21.4 kN
B、P = 42.8 kN
C、P = 85.6 kN
D、P = 171.2 kN
7、
A、Pcr = 0.1625 MN
B、Pcr = 0.325 MN
C、Pcr = 0.65 MN
D、Pcr = 1.30 MN
8、
A、d=9/4in.
B、d=9/8 in.
C、d=9/16 in.
D、d=9/32 in.
9、The plane static analysis frame is shown in the figure. It is known that the diameter of each rod is d, the length is a, and the elastic modulus is E. Try to calculate the ultimate load F that the structure can bear without instability of CD rod in the plane.
A、A
B、B
C、C
D、D
10、It is known that the material of each member of the plane rigid frame shown in the figure is the same, and the elastic modulus is E. if the CD member is a slender member and the cross section is a rectangular section, the flexibility of the CD member and the ultimate load F of the structure are ( ).
A、A
B、B
C、C
D、D
11、As shown in the figure, the structure is composed of vertical bar AB and inclined bar BC, both of which are slender bars with sectional bending stiffness EI, and bear the action of horizontal load F at node B. a = 4m and b = 3m, and try to calculate the allowable load of the structure.
A、A
B、B
C、C
D、D
12、Four equilateral angle steels 125 × 125 × 12 are used to form two kinds of cross-section columns as shown in (a) and (b). If the constraint at the rod end is the same and δ = 16 mm is known, the ratio of the critical load of the column in the two cases is ( ).
A、1/2
B、1/2.2
C、1/3.3
D、1/3
13、The square plane frame composed of five round section steel bars, the diameter of the bars is d = 40 mm, the elastic modulus of the material is E = 200 GPa, and the length of the bar is a = 1 m. Try to calculate the minimum load when the structure reaches the critical state. If F force acts inward, what is the minimum load?
A、124kN, 350.7kN
B、114kN, 300.7kN
C、124kN, 300
中国大学Mechanics of Materials
中国大学的Mechanics of Materials，也称为材料力学，是一门研究材料所受力学作用的学科。在机械工程、土木工程、材料科学和航空航天工程等学科中都有重要的应用。
课程内容
中国大学的Mechanics of Materials通常包括以下内容：
- 应力与应变
- 弹性力学
- 塑性力学
- 蠕变
- 断裂力学
- 复合材料力学
在应力与应变方面，学生会学习到应力、应变、弹性模量、泊松比等概念。在弹性力学方面，学生会了解梁和柱的弯曲、剪切、和轴向变形。在塑性力学方面，学生会学习到材料的塑性流动、本构方程和塑性失效的条件。在蠕变方面，学生会了解材料在高温下的蠕变现象。在断裂力学方面，学生会学习到断裂韧性、KIC值和疲劳寿命等概念。在复合材料力学方面，学生会学习到复合材料的力学性质、应力分析和破坏机理。
实验内容
中国大学的Mechanics of Materials课程通常包括实验。在实验中，学生会学习到如何使用力学测试仪器测量材料的性能。实验内容包括以下几个方面：
- 材料拉伸实验
- 材料压缩实验
- 梁弯曲实验
- 剪切实验
- 蠕变实验
- 疲劳实验
在材料拉伸实验中，学生会学习到如何测量材料的应力-应变曲线和弹性模量。在材料压缩实验中，学生会学习到如何测量材料的屈服强度和压缩模量。在梁弯曲实验中，学生会学习到如何测量梁的挠度和应力分布。在剪切实验中，学生会学习到如何测量材料的剪切模量和屈服强度。在蠕变实验中，学生会学习到如何测量材料的蠕变率和蠕变寿命。在疲劳实验中，学生会学习到如何测量材料的疲劳寿命和疲劳极限。
应用领域
中国大学的Mechanics of Materials课程在工程学科中有广泛的应用。以下是几个应用领域：
- 结构设计
- 材料选择
- 航空航天工程
- 制造工程
- 汽车工程
- 地震工程
在结构设计领域，Mechanics of Materials用于确定材料的最大承载能力和结构的稳定性。在材料选择领域，Mechanics of Materials用于评估不同材料的性能，以便选择最适合的材料。在航空航天工程中，Mechanics of Materials用于设计飞机、导弹和航天器的结构。在制造工程中，Mechanics of Materials用于控制零件的尺寸和形状。在汽车工程中，Mechanics of Materials用于设计汽车的结构、底盘和悬挂系统。在地震工程中，Mechanics of Materials用于评估建筑物的抗震性能。
结论
中国大学的Mechanics of Materials课程是工程学科中的重要基础课程。通过学习此课程，学生可以了解材料的力学性质和材料力学的基本原理，同时也可以学习到如何测量材料的性能和如何应用材料力学于工程实践中。因此，这门课程对于工程学生来说是非常重要的。
