History of group technology, role of G.T in CAD/CAM integration, part families,

History of group technology, role of G.T in CAD/CAM integration, part families,

1. SHEAR STRESSES: SHEAR STRESSES AT SECTION, SHEAR STRESS DISTRIBUTION IN RECTANGULAR AND CIRCULAR, BY Dr. D.GOVARDHAN PROFESSOR & HEAD DEPARTMENT OF AERONAUTICAL ENGINEERING INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) DUNDIGAL, HYDERABAD - 500 043 1 MECHANICS OF SOLIDS Don’t write or place any image in this area

2. 2 Course Title MECHANICS OF SOLODS Course Code AAEB04 Class B.TECH III SEM Section A & B Name of the Faculty Dr . D.GOVARDHAN Lecture hour - Date 07-09-2020 Course Outcome/s Explain the bending stresses and their distribution along the sections in simple and composite beams for evaluating its bending strength Topic Covered Bending Stresses In simple beams Topic Learning Outcome Discuss the bending stresses at various loading conditions in simple beams. MECHANICS OF SOLIDS

3. When a beam is loaded, the shear force at a section always comes into play, along with the bending moment. It has been observed that the effect of shear stress, as compared to the bending stress, is quite negligible, and is not of much importance. But sometimes, the shearing stress at a section assumes much importance in the design criterion. Shearing Stress at a Section in a Loaded Beam : Figure1: Shearing stress SHEARING STRESSES IN BEAMS Don’t write or place any image in this area

4. Consider a small portion ABDC of length dx of a simple supported beam carrying with uniformly distributed load as shown in figure 1(a). When a beam is loaded with a uniformly distributed load, the shear force and bending moment vary at every point along the length of the beam. Consider two sections AB and CD of this beam at a distance dx apart. Let M = Bending moment at AB , M + dM = Bending moment at CD, F = Shear force at AB, F + dF = Shear force at CD, and I = Moment of inertia of the section about its neutral axis . SHEARING STRESSES IN BEAMS Don’t write or place any image in this area

5. Now consider an elemental cylindrical strip of area dA at a distance y from the neutral axis as shown in figure 1(b). Let σ = Intensity of bending stress across AB at distance y from the neutral axis on a elementarily cylindrical strip of cross-sectional area of dA . According bending moment equation M/I = σ /y or σ = (M/I) × y ------( i ) SHEARING STRESSES IN BEAMS Don’t write or place any image in this area

6. 6 Similarly Intensity of bending stress across CD = (σ + dσ ) =(M + dM ) × y/I The force acting across AB on elemental area = Stress × Area = σ × dA = (M/I)× y × dA --------( iii) Similarly, force acting across CD = (σ + dσ ) × dA = ( M + dM ) x y x dA /I------(iv) At the two ends of elemental cylinder, the forces are different. They are acting along the same line, but are in opposite direction. Hence there will be unbalanced force on the elemental cylinder . ∴ Net unbalanced force on strip = [(M + dM ) /I] y dA - (M/I) ydA = ( dM /I) y dA The total unbalanced force (F) above the neutral axis is obtained by integrating the above equation from 0 and d/2. SHEARING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

7. 7 d/2 d/2 ∴ F = ∫ ( dM /I) y dA = ( dM /I) ∫ ( y x dA ) = ( dM /I) x Ay c ) ----( iv) where A = Area of the beam above neutral axis, and y c = distance between the centre of gravity of the area and the neutral axis. This unbalanced force acting on the beam may fail due to shear. Hence horizontal section must offer shear resistance and must be equal to shear force to avoid shearing. The shear stress given by equation (iv), is the horizontal stress at y c from neutral axis. But the principal of complementary shear stress, the horizontal shear stress is always accompanied by the vertical shear stress (τ) of the same SHEARING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

8. 8 Let τ = intensity of the horizontal stress shear stress which is also equal to vertical shear stress in transverse direction Let dx = length of the strip and b = width of the beam Shear force = shear stress x shear area = τ x b x dx ----------(iii) Equating two values of shear force from equations (ii) and (iii), Hence τ x b x dx = ( dM /I) x Ay c ) τ = ( dM / dx ) x Ay c / (I x b) = F x A.y c / (I x b )-------( iv) SHEARING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

9. 9 By finding the value of shear stresses at several section on a beam, it is possible to determine the distribution of the shear stress along the depth of a beam. Such a diagram helps in obtaining the value of shear stress at any section along the depth of the beam. For subject point of view, the distribution of shear stress over the following sections are important: 1 . Rectangular sections, 2. Triangular sections, 3. Circular sections, 4. I-sections, 5. T-sections and 6. Miscellaneous sections . DISTRIBUTION OF SHEARING STRESS Don’t write or place any image in this area

10. DISTRIBUTION OF SHEARING STRESS - RECTANGULAR SECTION: Consider a beam of rectangular section ABCD of width and depth as shown in figure 2 (a). The shear stress on a layer JK of beam, at a distance y from the neutral axis, τ =F x Ay c / (I x b) -------------------------- ( i ) Where F = Shear force at the section, A = Area of section above y (shaded area AJKD ), y = Distance of the shaded area from the neutral axis, ∴ A y = Moment of the shaded area about the neutral axis, I = Moment of inertia of the whole section about its neutral axis, and b = Width of the section. Figure 2 : Rectangular Section DISTRIBUTION OF SHEARING STRESS OVER A RECTANGULAR SECTION Don’t write or place any image in this area

11. 11 The area of the shaded portion AJKD, A = b( d/2 − y) ----(ii) Distance between the centre of gravity of the area and the neutral axis. ( y c ) = y + 1/2 x (d /2− y) = y + d/4 –y/2 = y/2 + d/4 = 1/2 x (d /2+ y) ------(iii) Substituting the above values of A and y in equation ( i ), τ = F x Ay c / (I x b) = F x b( d/2 − y) 1/2 x (d /2+ y) / (I x b) = F x b( d/2 − y) 1/2 x (d /2+ y) / (I x b ) = F x b/2 ( d 2 /4− y 2 ) / (I x b) = [F /2I ] x [( d 2 /4− y 2 ) ] ----------------(iv ) From the above equation, that τ increase as y decreases. At a point, where y = d/2, τ = 0;and where y = zero , τ is maximum. BENDING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

12. 12 The variation of τ with respect to y is a parabola. At neutral axis, the value of τ is maximum. Thus substituting y = 0 and I = bd 3 /12 in the above equation, (iv) τ max = 3F /2bd =1.5 τ av (∵ τ av = F/ Area = F/ bd ) Now draw the shear stress distribution diagram as shown in Figure 2(b). BENDING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

13. 13 Problem-1: A wooden beam 100 mm wide, 250 mm deep and 3 m long is carrying a uniformly distributed load of 40 kN /m. Determine the maximum shear stress and sketch the variation of shear stress along the depth of the beam. SOLUTION. Given: Width (b) = 100 mm ; Depth (d) = 250 mm ; Span (l) = 3 m = 10 3 mm Uniformly Distributed Load (w) = 40 kN /m = 40 N/mm. The shear force at one end of the beam, F = wl /2 = 40 x(3 x10 3 ) /2 N = 60 × 10 3 N and Area of beam section, A = b · d = 100 × 250 = 25 000 mm 2 ∴ Average shear stress across the section, τ av = F /A = 60 X10 3 /25000 = 2.4 N/mm 2 = 2.4 MPa and Maximum shear stress, τ max = 1.5 × τ = 1.5 × 2.4 = 3.6 MPa Ans. The diagram showing the variation of shear along the depth of the beam is shown in Fig. 3 (b) BENDING STRESSES IN SIMPLE BEAMS Don’t write or place any image in this area

14. SHEAR STRESSES: SHEAR STRESSES AT SECTION, SHEAR STRESS DISTRIBUTION IN RECTANGULAR AND CIRCULAR, BY Dr. D.GOVARDHAN PROFESSOR & HEAD DEPARTMENT OF AERONAUTICAL ENGINEERING INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) DUNDIGAL, HYDERABAD - 500 043 14 MECHANICS OF SOLIDS Don’t write or place any image in this area

15. Consider a beam of triangular cross-section ABC of base b and height h as shown in Figure 4 (a). The shear stress on a layer JK at a distance y from the neutral axis, τ =F x A 1 y c / (I x b 1 ) -----------------( i ) Where F = Shear force at the section, A 1 y c = Moment of the shaded area about the neutral axis and I = MI of the triangular section about its neutral axis. Figure 4: Triangular section The width of the strip JK, b 1 =b × (x/h) ∴ Area of the shaded portion AJK, A 1 = (1/2) b 1 × x = (1/2) b × (x/h) × x = (b/2h) × x 2 ∴ Distance between the centre of gravity of the area and the neutral axis. ( y c ) = (2/3)h - (2/3)x = (2/3)(h − x) Substituting the values of b 1 A 1 and y c in equation ( i ), τ =F x Ay c / (I x b) = F × (b/2h) × x 2 × (2/3)(h − x) / Ib 1 = F × (b/2h) × x 2 × (2/3)(h − x) / I [b × (x/h)] ∴ τ = [F /3I]× (h − x) × x = [F/3I] × (h x − x 2 ) ------------------------(ii) DISTRIBUTION OF SHEARING STRESS OVER AE TRIANGULAR SECTION

16. BENDING STRESSES IN SIMPLE BEAMS 16 Thus the variation of τ with respect to x is parabola and at a point and τ = 0 , when x = 0 or x = h, At neutral axis, where x = 2/3h, τ NA = [F/3I] × [h × (2/3)h) − (2/3h) 2 ] τ NA = [F/3I] × [ (2/3) h 2 − (4/9)h 2 ] = [F/3I] × [(2/9) h 2 ] = [F/3I] × [(2/9) h 2 ] = (2F/27I) × h 2 Substituting the value of I = bh 3 /36 in above equation, we have τ NA = (2F/27) (bh 3 /36 )× h 2 = (8/3) (F/ bh ) =4/3) (F/A), where A = area of the triangle ∴ τ NA = 4/3) τ ave = 1.33 τ ave Now for maximum intensity, differentiating the equation (ii) and equating to zero, [F/3I] × (h − 2x) =0 or h -2x =0 or x= h/2------------------------(iii) Now substituting this value of x in equation (ii), τ max = 3F/ bh = (3/2) ×(F/A) Hence τ max = 1.5 × τ ave

17. BENDING STRESSES IN SIMPLE BEAMS 17 SOLUTION. Given: Base width (b) = 100 mm ; Height (h) = 150 mm and shear force (F) = 13.5 kN =13.5 x10 3 N The area of beam section, A = bx h /2 = 100 x150/ 2 mm 2 = = 7500 mm 2 ∴ Average shear stress across the section, τ av = F /A = 13.5 x10 3 /7500 N/mm 2 = 1.8 N/mm 2 = 1.8 MPa and maximum shear stress, τav = 1.5 × τav = 1.5 × 1.8 = 2.7 MPa Ans. EXAMPLE 16.2. A beam of triangular cross section having base width of 100 mm and height of 150 mm is subjected to a shear force of 13.5 kN. Find the value of maximum shear stress and sketch the shear stress distribution along the depth of beam. The diagram showing the variation of shear stress along the depth of the beam is shown in Fig. 16.5(b).

18. DISTRIBUTION OF SHEARING STRESS OVER A CIRCULAR SECTION 18 Consider a circular section of diameter d as shown in Figure 6(a). The shear stress on a layer JK at a distance y from the neutral axis, τ =F x A 1 y c / (I x b 1 ) -----------------( i ) Where F = Shear force at the section, A 1 y c = Moment of the shaded area about the neutral axis and I = Moment of Inertia of the triangular section about its neutral axis. Where F = Shear force at the section, A y = Moment of the shaded area about the neutral axis, r = Radius of the circular section, I = Moment of inertia of the circular section and b = Width of the strip JK. In a circular section, width of the strip JK, b = 2√( r 2 − y 2 ) and Fig. 16.6. Circular section. Area of the shaded strip, A = 2√( r 2 − y 2 ) dy

19. BENDING STRESSES IN SIMPLE BEAMS 19 ∴ Moment of this area about the neutral axis = 2y 2√( r 2 − y 2 ) dy -------------------( i ) Moment of the whole shaded area about the neutral axis may be found out by integrating the above equation between the limits y and r, i.e. r Moment of Area Ay = ∫ y 2y 2√( r 2 − y 2 ) dy since ∫ bydy ... (∵b = 2√( r 2 − y 2 ) ) ...(ii) r r = - ∫ y (-2y )√( r 2 − y 2 ) dy = (2/3) [ ( r 2 − y 2 ) 3/2 ] y = - (2/3) [( r 2 − r 2 ) 3/2 - ( r 2 − y 2 ) 3/2 ] = - (2/3) [ - ( r 2 − y 2 ) 3/2 ] = (2/3) ( r 2 − y 2 ) 3/2 Substituting this value of Ay and b in our original formula for the shear stress, i.e., τ = F x A 1 y c / (I x b) = F × [(2/3) ( r 2 − y 2 ) 3/2 ] / I × 2√( r 2 − y 2 ) = F × [(2/3) ( r 2 − y 2 ) 3/2 ] / [I × 2√( r 2 − y 2 )] = F × [(1/3) ( r 2 − y 2 ) ] / I = F × [(1/3I) ( r 2 − y 2 ) ------------------------------(ii) The equation (ii), shows that shear stress distribution across the circular section is parabolic.

20. BENDING STRESSES IN SIMPLE BEAMS 20 Also it is clear from this equation that the with increase of y, the shear stress decreases and when y = R, the shear stress =0. Similarly the shear stress is maximum when y = 0 ie at neutral axis. From the equation (ii), the maximum value of shear stress is givesn by substituting y=0at neutral axis. Hence τ max = F × [(1/3I) × r 2 = F × [(r 2 /3I) -----------------(iii) Substituting I = (π/64) ×d 4 in the equation (iii) τ max = F × r 2 / [3 × (π/64) ×d 4 )] = 16 F/ (3πd 2 ) = (4 /3) F/ (πr 2 ) τ max = 4 /3 τ ave

21. DISTRIBUTION OF SHEARING STRESS OVER AN I-SECTION 21 Now we shall discuss two important cases: ( i ) when y is greater than d/2 and (ii) when y is less than d/2 Consider a beam of an I-section as shown in figure 8 (a) Let B = Overall width of the section, D = Overall depth of the section, d = Depth of the web, and b = Thickness of the web. The shear stress on a layer JK at a distance y from the neutral axis, τ =F× A y / I b ---------------------------( i )

22. BENDING STRESSES IN SIMPLE BEAMS 22 When y is greater than d/2, It means that y lies in the flange as shown in figure 9(a). In this case, shaded area of the flange (A) = (D/2-y)B and y c = y +1/2(D/2-y) Now substituting these values of A and y in equation ( i ) , in our original equation ( i ) of shear force, τ = F x A 1 y c / (I x b 1 ) -----------------( i ) = F x (D/2-y)B x [y +1/2(D/2-y)] / (I x B) = F x [(D/2-y)B x 1/2(D/2+y)] / (I x B) = F x [(D 2 /4-y 2 )] / 2I -----------------(ii) Thus τ the increases as y decreases also the variation of τ with respect to y is a parabolic curve. At the upper edge of the flange, where y = D/2, , shear stress is zero and at the lower edge where y = d/2, the shear stress, τ = F x [(D 2 /4-d 2 /4] / 2I = (F/8I) [(D 2 -d 2 ] (ii) When y is less than 2d: It means that y lies in the web as shown in Fig. 9(b). In this case, the value of A y for the flange = (D/2-y)B x [y +1/2(D/2-y)] ie y=d/2 A= (D/2-y)B x [y +1/2(D/2-y)] = B[(D 2 /4-y 2 )] =(B/8)[D 2 - d 2 ] The value of A y c for the web above AB = b(d/2-y) x [y +1/2(d/2-y)]

23. BENDING STRESSES IN SIMPLE BEAMS 23 = b x [(d/2-y) x 1/2(D/2+y)] = b[(d 2 /4-y 2 )] / 2 -----------------(ii) Total A y c = (B/8)[D 2 - d 2 ] + b[(d 2 /4-y 2 )] / 2 substituting the value of A y c from the above equation, in our original equation of shear stress on a layer at a distance y from the neutral axis, i.e., τ = F x Ay c / (I x b 1 ) =F {(B/8)[D 2 - d 2 ] + b[(d 2 /4-y 2 )] / 2 } Ib = F {(B/8)[D 2 - d 2 ] + (b/2)[(d 2 /4-y 2 )]}/ Ib Thus it is observed that, in the web also τ the increases as y decreases, also the variation of τ with respect to y is a parabolic curve. At the neutral axis, where y =0, the shear stress is maximum and maximum the shear stress, τ max = F {(B/8)[D 2 - d 2 ] + (b/2)[(d 2 /4-0 2 )]}/ Ib = F {(B/8)[D 2 - d 2 ] + b(d 2 /8)}/ Ib Now, shear stress at the junction of the top of the web and bottom of the flange = F.B[D 2 - d 2 ] / 8Ib ie y= d/2 NOTES:1 From the shear stress distribution diagram, the most of the shear stress is taken up by the web. Hence it is an important factor in the design of various important structures.

24. BENDING STRESSES IN SIMPLE BEAMS 24 Problem-12 : T-shaped cross-section of a beam shown in Fig.12 is subjected to a vertical shear force of 100 kN. Calculate the shear stress at important points and draw shear stress distribution diagram. Moment of inertia about the horizontal neutral axis is 113.4 × 106 mm 4 . OLUTION. Given: Shear stress (F) = 100 kN = 100 × 10 3 N and Moment of inertia (I) = 113.4 × 106 mm 4 . Position of the neutral axis. The distance between the centre of gravity of the section and bottom of the web, y = [(200 x 50) 225] +[(200 x50) 100] /(200 x50) +(20 x50) = 162.5 mm ∴ Distance between the centre of gravity of the section and top of the flange, y C = (200 + 50) – 162.5 = 87.5 mm The shear stress at the top of the flanges is zero. Now let us find out the shear stress at the junction of the flange and web by considering the area of the *flange of the section. The area of the upper flange, A = 200 × 50 = 10000 mm 2 y = 87.5- 50 /2=62.5 2 mm B = 200 mm ∴ Shear stress at the junction of the flange and web, τ = 3 2 6 · 10000 62.5 100 10 N/mm · (113.4 10 ) 200 A y F I B × × =×× × × = 2.76 N/mm2 = 2.76 MP

25. BENDING STRESSES IN SIMPLE BEAMS 25 B = 200 mm ∴ Shear stress at the junction of the flange and web, τ = Fx A y /I B = = 100x10 3 x 10000 x 62.5 / (113.4 10 6 ) 200 = 2.76 N/mm 2 = 2.76 Mpa The shear stress at the junction suddenly increases from 2.76 MPa to 2.76 × 200 11.04 50 = MPa . Now let us find out the shear stress at the neutral axis, where the shear stress is maximum. Considering the area of the T-section above the neutral axis of the section, we know that * A y = 37.5 3 [(200 50) 62.5] (37.5 50) mm 2 ⎡ ⎤ ×× + ×× ⎢ ⎥ ⎣ ⎦ = 660.2 × 103 mm 3 and b = 50 mm ∴ Maximum shear stress, τmax = 3 3 2 6 · 660.2 10 100 10 N/mm · (113.4 10 ) 50 A y F I b × × =×× × × = 11.64 N/mm2 = 11.64 MPa Now draw the shear stress distribution diagram across the section as shown in Fig. 16.13(b).

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27. 27 BENDING STRESSES IN ANGLE AND CHANNEL SECTIONS, DESIGN OF SIMPLE BEAM SECTIONS, BEAMS OF UNIFORM STRENGTH. BY Dr. D.GOVARDHAN PROFESSOR & HEAD DEPARTMENT OF AERONAUTICAL ENGINEERING INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) DUNDIGAL, HYDERABAD - 500 043 27 MECHANICS OF SOLIDS Don’t write or place any image in this area