280 likes | 284 Views
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
E N D
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
Recap The important formulae are Stress Strain Young modulus Deformation C-16-M-302-10
OBJECTIVES On the completion of his period, you would be able to solve problems on • Simple Stress and Strains C-16-M-302-10
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
Young’s modulus for mild steel (E) = 2 x 1011 N/m2 Cross-sectional area of the rod (A) = C-16-M-302-10
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
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
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
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
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
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
(i) (ii) we have e = 0.001 C-16-M-302-10
(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
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
Find. i) Magnitude of the load. ii) Linear strain produced and iii) Total decrease in length C-16-M-302-10
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
(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
(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
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
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
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
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
Stress with in elastic limit Strain at the elastic limit (e) C-16-M-302-10
Modulus of elasticity (young’s modulus) (ii) Upper Yield Stress Upper yield stress = (iii) Ultimate Stress C-16-M-302-10
(iv) Stress at Breaking Point Stress at breaking point (Breaking stress) (v) Percentage Elongation: C-16-M-302-10
(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
Summary We solved simple problem on • Simple Stresses and strains C-16-M-302-10
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