Homework Help is Here – Start Your Trial Now!
Engineering
Civil Engineering
5. A load of 6,900 N acts on the machine part shown in Fig. 5a. The machine part has a uniform thickness of 15 mm (i.e., 15-mm thickness in the z direction). Determine the normal and shear stresses acting at points H and K, which are shown in detail in Fig. 5b. For eachpoint, show these stresses on a stress element. 10 mm I 50 mm "IS" 135 mm 40° 50 mm 40 mm Fig. 5a D 40 mm HO 15 mm Big Eb 40 m

5. A load of 6,900 N acts on the machine part shown in Fig. 5a. The machine part has a uniform thickness of 15 mm (i.e., 15-mm thickness in the z direction). Determine the normal and shear stresses acting at points H and K, which are shown in detail in Fig. 5b. For eachpoint, show these stresses on a stress element. 10 mm I 50 mm "IS" 135 mm 40° 50 mm 40 mm Fig. 5a D 40 mm HO 15 mm Big Eb 40 m

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
See similar textbooks
icon
Related questions
Question
Please solve the question I'm providing u solution for reference Change value and provide solutions
(15mm) (40 mm) 12 A= (15 mm) (40 mm) = 600 mm² Equilibrium: EF, A₂ - FBC + (6, 100 N) cos 40° = 0 EF, A, +(6, 100 N) sin 40° 0 EMA (6, 100 N) (185 mm) sin 40° - Fac (90 mm) 0 FBC = 8,059.842 N A, 3, 386.971 N A, -3,921.004 N Internal forces at H: EF-F+ (6, 100 N) cos 40° = 0 EF, Vn+(6, 100 N) sin 40° 0 EMH-MH + (6, 100 N) (135 mm) sin 40° 0 ..M=529, 335.597 N-mm. Axial stress at H: 4,672.871 o axial=600mm=7.788 MPa(T) Shear stress at H: Qu (15 mm) (10 mm) (20 mm-5 mm)2, 250 mm³ (3,921.004 N) (2, 250 mm³) TH= = 7.352 MPa (80, 000 mm¹) (15 mm) Bending stress at H: bend (820,335.597 mm)(10 mm) 66.167 MPa(C) (by inspection) 80,000 = Summary of stresses at H: = 7.788 MPa - 66.167 MPa = -58.4 MPa a=0 MPa Try=7.35 MPa Internal forces at K: EF-Fac+VK=0 2. Vr = 8,059.842 N EF= F = 0 EMK MK-FBc (40 mm) = 0 . M 322, 393.697 N-mm = Axial stress at K: axial0 MPa Shear stress at K: QK (15 mm)(15 mm) (20 mm -7.5 mm) = 2,812.5 mm³ = (8,059.842 N) (2,812.5 mm³) TK = 18.890 MPa (80,000 mm¹) (15 mm) Bending stress at K: Obend = (322,393.607 X-mm)(mm) (5 = 20.150 MPa (T) (by inspection) 20.150 MPa (T) 80,000 m Summary of stresses at K: 0,0 MPa 0,0 MPa +20.150 MPa = 20.2 MPa Tay 18.89 MPa I₁ = = 80,000 mm ..FH 4,672.871 N ..VH-3,921.004 N 7.352 MPa. - 58.379 MPa 135 mm 18.89 MPa 20.15 MPa 135 40 mm Fac 6300 6,100 N P12.10 Element H M₂ P12.10 Element K X X
Transcribed Image Text:(15mm) (40 mm) 12 A= (15 mm) (40 mm) = 600 mm² Equilibrium: EF, A₂ - FBC + (6, 100 N) cos 40° = 0 EF, A, +(6, 100 N) sin 40° 0 EMA (6, 100 N) (185 mm) sin 40° - Fac (90 mm) 0 FBC = 8,059.842 N A, 3, 386.971 N A, -3,921.004 N Internal forces at H: EF-F+ (6, 100 N) cos 40° = 0 EF, Vn+(6, 100 N) sin 40° 0 EMH-MH + (6, 100 N) (135 mm) sin 40° 0 ..M=529, 335.597 N-mm. Axial stress at H: 4,672.871 o axial=600mm=7.788 MPa(T) Shear stress at H: Qu (15 mm) (10 mm) (20 mm-5 mm)2, 250 mm³ (3,921.004 N) (2, 250 mm³) TH= = 7.352 MPa (80, 000 mm¹) (15 mm) Bending stress at H: bend (820,335.597 mm)(10 mm) 66.167 MPa(C) (by inspection) 80,000 = Summary of stresses at H: = 7.788 MPa - 66.167 MPa = -58.4 MPa a=0 MPa Try=7.35 MPa Internal forces at K: EF-Fac+VK=0 2. Vr = 8,059.842 N EF= F = 0 EMK MK-FBc (40 mm) = 0 . M 322, 393.697 N-mm = Axial stress at K: axial0 MPa Shear stress at K: QK (15 mm)(15 mm) (20 mm -7.5 mm) = 2,812.5 mm³ = (8,059.842 N) (2,812.5 mm³) TK = 18.890 MPa (80,000 mm¹) (15 mm) Bending stress at K: Obend = (322,393.607 X-mm)(mm) (5 = 20.150 MPa (T) (by inspection) 20.150 MPa (T) 80,000 m Summary of stresses at K: 0,0 MPa 0,0 MPa +20.150 MPa = 20.2 MPa Tay 18.89 MPa I₁ = = 80,000 mm ..FH 4,672.871 N ..VH-3,921.004 N 7.352 MPa. - 58.379 MPa 135 mm 18.89 MPa 20.15 MPa 135 40 mm Fac 6300 6,100 N P12.10 Element H M₂ P12.10 Element K X X
5. A load of 6,900 N acts on the machine part shown in Fig. 5a. The machine part has a uniform thickness of 15 mm (i.e., 15-mm thickness in the z direction). Determine the normal and shear stresses acting at points H and K, which are shown in detail in Fig. 5b. For eachpoint, show these stresses on a stress element. 10 mm 50 mm I 135 mm 40° 50 mm 40 mm B Fig. 5a D 40 mm HO 15 mm Fig. 5b 40 m
Transcribed Image Text:5. A load of 6,900 N acts on the machine part shown in Fig. 5a. The machine part has a uniform thickness of 15 mm (i.e., 15-mm thickness in the z direction). Determine the normal and shear stresses acting at points H and K, which are shown in detail in Fig. 5b. For eachpoint, show these stresses on a stress element. 10 mm 50 mm I 135 mm 40° 50 mm 40 mm B Fig. 5a D 40 mm HO 15 mm Fig. 5b 40 m
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Basics of life cycle costing
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Structural Analysis
Structural Analysis
Civil Engineering
ISBN:
9781337630931
Author:
KASSIMALI, Aslam.
Publisher:
Cengage,
Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
Civil Engineering
ISBN:
9780134610672
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Principles of Foundation Engineering (MindTap Cou…
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781337705028
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
Civil Engineering
ISBN:
9780073398006
Author:
Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:
McGraw-Hill Education
Sustainable Energy
Sustainable Energy
Civil Engineering
ISBN:
9781337551663
Author:
DUNLAP, Richard A.
Publisher:
Cengage,
Traffic and Highway Engineering
Traffic and Highway Engineering
Civil Engineering
ISBN:
9781305156241
Author:
Garber, Nicholas J.
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS