1 / 28

Simple Stresses and Strains Problems: Solutions & Formulas

Master simple stress and strain problems with detailed solutions and essential formulas for mechanical engineering students. Calculate elongation, shortening, stress, strain, and more. Enhance your understanding of Strength of Materials.

jordanc
Download Presentation

Simple Stresses and Strains Problems: Solutions & Formulas

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. e – Lesson Module for C-16 Curriculum State Board of Technical Education & Training Andhra Pradesh • Year/Semester : III Semester • Branch : Mechanical engg • Subject : M-302, Strength of Materials • Topic : Simple stresses and Strains • Sub Topic : Simple problems on simple stress and strains • Duration : 50 mins Prepared by : S V V RAMANA, Sr. Lecturer in M.E Government Polytechnic, Gannavaram Guided by : D V S S N V PRASAD BABU SL/ME Government Polytechnic, Gannavaram C-16-M-302-10

  2. Recap The important formulae are Stress Strain Young modulus Deformation C-16-M-302-10

  3. OBJECTIVES On the completion of his period, you would be able to solve problems on • Simple Stress and Strains C-16-M-302-10

  4. Problem – 1 Calculate the elongation of a mild steel rod of 30 mm in diameter and 3 m in length when subjected to an axial pull of 60 kN. Take for steel E = 2 X 1011 N/m2. Solution: Given data: Diameter of the rod (d) = 30mm = 30 x 10-3m Length of the rod l = 3m Axial pull (P) = 60 kN = 60 x 103 N C-16-M-302-10

  5. Young’s modulus for mild steel (E) = 2 x 1011 N/m2 Cross-sectional area of the rod (A) = C-16-M-302-10

  6. we know, Young’s modulus (E) = strain (e) = Also e = Elongation of the rod (δl) = e.l = 4.244x10-4x3 = 1.273x10-3m=1.273mm C-16-M-302-10

  7. Problem - 2 A square steel rod 20mm x 20 mm in section is to carry an axial load (compressive) of 200 kN. Calculate the shortening in a length of 60 mm. Taken E = 2.1 x 108 KN/m2 Solution: Given data: Cross-sectional area of the rod (A) = 20 x 20 = 400 mm2 = 400 x 10-6 m2 C-16-M-302-10

  8. Length of the rod (l) = 60mm = 60 x 10-3m Axial load (P) = 200 kN = 200 x 103 N Young’s modulus (E) = 2.1 x 108 kN/m2 = 2.1 x 1011 N/m2 Young’s Modulus (E) = C-16-M-302-10

  9. Strain (e) = Also shortening of rod (δl) = e.l = 2.38 x 10-3x60x10-3 = 1.428 x 10-4m  δl= 0.1428mm C-16-M-302-10

  10. Problem – 3 A mild steel rod of 3m long having a cross-sectional area 5cm2 is subjected to an axial pull of 100kN. If ‘E’ for steel = 2 x 1011N/m2, find (i) the stress (ii) the strain (iii) elongation of the rod and (iv) work done C-16-M-302-10

  11. Solution: Length of the rod (l) = 3m Cross –sectional area of the rod (A) = 5 cm2 = 5 x 10-4 m2 Axial pull (P) = 100 kN = 100x103N Young’s Modulus (E) = 2 x 1011 N/m2 C-16-M-302-10

  12. (i) (ii) we have  e = 0.001 C-16-M-302-10

  13. (iii)Elongation of the rod (δl) we know, Strain (e) =  Elongation (δl) = e l = 0.001x3= 3 x 10-3 m = δl = 3 mm (iv) Work done (w) Work done (w) = = 1.5 N-m C-16-M-302-10

  14. Problem – 4 A hollow cast-iron cylinder 5m long. 400 mm outer diameter, and thickness of 50mm is subjected to a central load on the top when standing straight. The stress produced is 80000 kN/m2. Assume Young’s modulus for cast-iron as 1.5x1011 N/m2. C-16-M-302-10

  15. Find. i) Magnitude of the load. ii) Linear strain produced and iii) Total decrease in length C-16-M-302-10

  16. Solution: Given data: Outer diameter (d0) = 400mm = 0.4m Thickness (t) = 50mm Length (l) = 5m stress produced (σ) = 80000 kN/m2 = 8 x 107 N/m2 Young’s modulus (E) = 1.5 x 1011 N/m2 Inner diameter of the cylinder (d1) = d0 -2t = 400 -2 x 50 = 300mm = 0.3m C-16-M-302-10

  17. (i) Magnitude of the Load (P) when A = Cross – sectional area of hollow cylinder A = P = 8 x 107x 0.054 = 4392000 N= 4392 kN C-16-M-302-10

  18. (ii) Linear Strain Produced (e) (iii) Total Decrease in Length (δl) δI = e.I = 0.00053x5 = 2.65 x 10-3 m = 2.65 mm C-16-M-302-10

  19. Problem – 5 A steel bar 20mm in diameter, 25 cm long was tested to destruction. During the test, the following observations were recorded. Load at elastic limit = 75 kN Extension at elastic limit = 0.25mm Load at upper yield point = 80 kN Maximum load = 140 kN Breaking load = 120kN Diameter at neck = 15mm Final length = 30 cm C-16-M-302-10

  20. Determine: (i) Modulus of elasticity (ii) Upper yield Stress (iii) Ultimate stress (iv) stress at breaking point (v) Percentage elongation (vi) Percentage reduction in cross- sectional area. C-16-M-302-10

  21. Solution: Given data: Diameter of the bar (d) = 20 mm = 20 x 10-3 m Length of the bar (l) = 25 cm = 25 x 10 -2 m Load at elastic limit (P) = 75 kN = 75 x 103 N Elongation at elastic limit = 0.24 mm = 0.24 x 10-3 m Load at upper yielding point = 80 kN = 80 x 103 N C-16-M-302-10

  22. Maximum load (ultimate load) = 140 kN = 140 x 103 N Load at breaking point = 120 kN = 120 x 103 N Diameter at neck = 15 mm = 15 x 10-3 m Final length = 30 cm = 30 x 10-2 m • Modulus of Elasticity we know, Modulus of elasticity (E) = Stress with in elastic limit Strain C-16-M-302-10

  23. Stress with in elastic limit Strain at the elastic limit (e) C-16-M-302-10

  24. Modulus of elasticity (young’s modulus) (ii) Upper Yield Stress Upper yield stress = (iii) Ultimate Stress C-16-M-302-10

  25. (iv) Stress at Breaking Point Stress at breaking point (Breaking stress) (v) Percentage Elongation: C-16-M-302-10

  26. (vi) Percentage Reduction in Cross-sectional Area Percentage reduction in cross-sectional area = 3.142x10-4 m2 a2 = Cross sectional area at neck = 1.767 x 10-4 m2 Percentage reduction in Cross – sectional area = 43.76% C-16-M-302-10

  27. Summary We solved simple problem on • Simple Stresses and strains C-16-M-302-10

  28. Frequently asked questions • Calculate the elongation of a steel rod of 25mm diameter and 2mm in length when subjected to an axial pull of 40kN. Take young's modulus of steel is 2x105 N/mm2 • A mould steel rod of 10mm dia is subjected to a compressive load of 50kN. It resulted a decrease of 1.5mm length in the total length of 2m. Calculate 1. Stress 2. Strain 3.Young’s modulus C-16-M-302-10

More Related