MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS This the fourth and final tutorial on bending of beams. You should judge our progress b completing the self assessment exercises. On completion of this tutorial ou should be able to do the following. Define the relationship between bending moment and shear stress in a beam. Define complementar shear stress. Define shear flow. Determine how shear stress is distributed over the cross section of a beam. Determine how shear stress is distributed over the flanges and webs of a beam. Define and shear centre. Calculate the position of the shear centre. It is assumed that students doing this tutorial alread understand the basic principles of moments, shear force, stress and moments of area. Students must also be able to perform basic differentiation and calculus from their maths studies. You will find other good tutorial material at the following web addresses (currentl available in 007) http://www.me.mtu.edu/~mavable/spring0/chap6.pdf http://gaia.ecs.csus.edu/~ce11/steel-shear.pdf http://www.ae.msstate.edu/~masoud/teaching/exp/a14.8_ex1.html http://ctsm.umd.edu/assakkaf/courses/enes0/lectures/lecture15.pdf D.J.DUNN 1
1. THE RELATIONSHIP BETWEEN BENDING MOMENT AND SHEAR FORCE Consider a beam subject to bending and transverse shear. At some distance along the x direction further consider a short length x. Over this length the bending moment increases b dm and the shear force increases b df. Figure 1 If we take this section out of the beam we must add the forces and moments that were previousl exerted b the material in order to maintain equilibrium. Equilibrium of forces and moments exists at all points so it is convenient to look at the equilibrium of moments at the bottom right corner. F x + M = M + M F = M/x and in the limit as x 0 we have F = dm/dx D.J.DUNN
. COMPLEMENTARY SHEAR STRESS If a transverse shear force F acts at a point on a beam in the vertical () direction, transverse shear stress is created. Studies of two dimensional stress and strain in elastic materials (see the tutorial on Mohr s circle of stress and strain) show that shear stress cannot exist in just one direction but it is alwas accompanied b an equal and opposite shear stress on the plane normal to it. It follows that if a shear stress exists on the cross sectional area of a beam, there must be an equal shear stress acting in the x direction. This is called the complementar shear stress. SHEAR STRESS DISTRIBUTION Figure Consider a simple cantilever beam. Bending Moment M = F x M F x The direct stress at distance from the centre is σ I I F x da F x b d The force acting on the area da is df = da = b d = or I I If we integrate between and the m, we get the force acting normal to the plane. This must be balanced b a force F x at level on the x plane F x F x I m da D.J.DUNN Figure Suppose the cross section does not have a uniform width b. The shear stress on the x plane is found b dividing F x b the area b x τ x F b I F da or b I m m bd This shear stress must be complimentar to the shear stress on the plane so it follows that m F τ x τ da b I m da is the first moment of area A Figure 4 τ x F A Note the term A refers to the shaded area on the diagram. If the shape is rectangular it b I is eas to calculate the shear stress. The equation indicates that is constant at all values of x if I and A are constant.
4. SHEAR FLOW and SHEAR FORCE The shear force should be equal to the integration of the shear stress over the section. df = da = b d top F τ b d The product b is called the shear flow and denoted q. F bottom For standard sections such as I, T, U and L, there is vertical shear flow in the flange and horiontal shear flow in the web in the horiontal direction. For this reason t is often used for the thickness of the flange or web. F q d and the same applies to an direction such as F q d The importance of this work will be clear later in the tutorial. The units of q are N/m. If the cross section is not a uniform shape, e.g. a Tee section, t or b is not constant and the maximum shear stress occurs where /b is a maximum. Figure 5 top q d bottom WORKED EXAMPLE No.1 Derive an equation for the shear stress and shear flow distribution on a rectangular cross section. Compare the average and maximum shear stress. Show that the shear force is obtained b integrating the shear flow over the section. SOLUTION Let D/ = C. In this case the shaded area is B (C ) and = (C + )/ b = B I = BD /1 A = BD F B C C 1 F C C 6 F C 6 F C F τ 1 BI B D AD 4AC A C Plotting against we find the shear stress varies as a parabola from ero at the bottom to a maximum at the centroid and ero at the top. The maximum shear stress occurs at = 0 Figure 6 τ max F/A The transverse shear force on the section is F and the mean shear stress is F/A so τ τ x / The shear flow is easil obtained since the width is constant. q F top q bottom d F 4C C C 1 C F 4C C C C F C C 4C C F B 1 4C C τ C - C C max F mean D.J.DUNN 4
5 OTHER SECTIONS 5.1 CIRCULAR SECTION A circular section can be solved with some difficult in the same wa. For a circular section of radius R: τ 4 F R 4 F 4 τ 4 max τ mean π R π R 5. NON UNIFORM SECTIONS If the shape is hollow or does not have a constant width b, the problem is more complex. For example consider a triangular section. F A τ I b Because b is a function of, the maximum shear stress does not occur at the centroid but at the point shown. 6 SUDDEN DISCONTINUITIES Figure 7 With T, I, U and L sections, the width b or t suddenl changes at the junctions of the web and flange so the shear stress suddenl changes as the ratio of the widths. This is best illustrated with a worked example. WORKED EXAMPLE No. Determine the shear stress at the junction of the top flange and web for the section shown when a shear force of 40 kn acts verticall down on the section. SOLUTION Figure 8 First calculate the second moment of area. The tabular method is used here. Divide the shape into three sections A, B and C. First determine the position of the centroid from the bottom edge. Area A A 600 mm 45 mm 7 000 mm B 00 mm 5 mm 7500 mm C 400 mm 5 mm 000 mm Totals 100 mm 65000 mm D.J.DUNN 5
For the whole section the centroid position is =65000/100 = 8.07 mm Now find the second moment of area about the base using the parallel axis theorem. BD /1 A I = BD /1 +A A 60 x 10 /1=5000 mm 4 600 x 45 =115000 10000 mm 4 B 10 x 0 /1=500 mm 4 00 x 5 =187500 10000 mm 4 C 40 x 10 /1= mm 4 400 x 5 =10000 1 mm 4 Total = 144 mm 4 The total second moment of area about the bottom is 144 mm 4 Now move this to the centroid using the parallel axis theorem. I = 144 - A =144-100 x 8.07 = 41800 mm 4 Now calculate the stress using the well known formula B = M/I Top edge = distance from the centroid to the edge = 50 8.07 = 1.9 mm FA τ At the junctions of the flange and web, this will change suddenl as the value of b changes. I b Consider the junction of the top flange with the web. Figure 9 A = 600 mm I gg = 41800 mm 4 = 16.9 mm FA 40 x 10 x 600 x 10 x 16.9 x 10 970.8 x 10 τ I b 1 41800x10 x b b Where b = 10 mm the shear stress is = 97 MPa Where b = 60 mm the shear stress is = 16.18 MPa 6 D.J.DUNN 6
WORKED EXAMPLE No. Determine the shear stress distribution for the T section shown when the transverse shear force is 00 kn SOLUTION Figure 10 First find b taking first moments of area about the base s s. Area = (150 x 50) (10 x 0) = 7600 mm 1 st Mom of Area = 150 x 50 / 10 x 0 / = 149000 mm = 149000/7600 = 164. mm Next find I gg I ss = (150 x 50 /) (10 x 0 /) = 54 x 10 6 mm4 I gg = I ss - A = 54 x 10 6 7600 x 164. = 48.75 x 10 6 mm4 Now examine the shear stress. The bottom half is the easiest so start there. τ FA I b A 00 x 10 x A 6 48.75 x10 x b b t 0.00 9 164. x 0 x 8.15 x 10 Figure 11 D.J.DUNN 7
Evaluate at various values of and we get mm 0 40 80 10 160 164. MPa 55.4 5.1 4. 5.9.9 0 The top half is more difficult. For = 65.66 to = 85.66 mm b = 150 mm. A 150 x 85.66 x 65.66 85.66 FA 00 x 10 x A τ b t 0.150 I b 6 48.75 x10 x b Evaluate at various values of and we get mm 65.66 75 85.7 MPa 6..5 0 Finall for = 0 to = 65.66 mm 85.66 A 150 x 85.66 x 10 x Figure 1 65.66- FA 00 x 10 x A τ b t 0.00 I b 6 48.75 x10 x b Evaluate at various values of and we get mm 0 0 40 65.66 MPa 55.4 54.6 5.1 46.5 x 65.66 The distribution is like this. Figure 1 D.J.DUNN 8
SELF ASSESSMENT EXERCISE No.1 1. Calculate the maximum shear stress and the shear stress at the junction of the bottom flange and web for the same case as worked example No. (104 MPa, MPa and 88.4 MPa). A T section beam has a transverse shear force of 80 kn at a point on its length. The flange and web are 1 mm thick. The flange is 80 mm wide and the overall depth is 100 mm. Determine the maximum shear stress and the shear stresses at the junction of the flange and web. (94. MPa, 86 MPa and 1.9 MPa) 7. SHEAR DISTRIBUTION IN THE FLANGES AUTHOR S NOTE. I admit that I don t full understand wh the horiontal shear stress in the flanges is determined in the manner described below. Rigid proofs should be sought elsewhere. I have attempted to place a simple explanation for shear flow in the flanges. The result is valid and used in the section following on shear centre. The shear stress distribution in non uniform sections such as T and U sections cannot be quite as simple as indicated previousl for the following reasons. The shear stress at a free surface can onl be parallel to the surface and not normal. It follows that the vertical shear stress in the outer regions of the flanges cannot be the same as at the junction of the flange and web. Figure 14 We should also consider how the shear force in the vertical direction is applied along the flange. Assuming a perfectl elastic material the force will be spread along the flange and some of it will cause bending. This will induce shear stress in the and directions. Figure 15 The result is that a shear flow q w is set up in the web and q f in the flanges as shown when the transverse shear force is vertical (F ). In the flanges the flow varies linearl with. D.J.DUNN 9
Figure 16 In the flange the shear flow is in the direction so there is a shear force F as shown. As the total force in the direction is ero, the forces in the two halves are equal and opposite. Here follows the bit I have a problem with because the formula was derived for vertical shear and I can t understand wh it applies to horiontal shear in the flange. We have defined shear flow as q = b for the web. To calculate the shear flow in the flange, b becomes the thickness t. end F A end τ A I t td t d τ q end F F d end - I N/m I F τt I t end - N/m Figure 17 WORKED EXAMPLE No.4 Figure 18 Determine the shear stress and shear flow distribution in the T section shown when a transverse shear force of 1 kn is applied verticall. Calculate the horiontal force in the flange and determine the fraction of the shear force carried b the web. D.J.DUNN 10
SOLUTION First determine the position of the centroid b taking first moments of area about the base. A (40 x 10 x 45) ( 40 x 10 x 0).5mm A (40 x 10) ( 40 x 10) Next determine I BD 40 x 10 10x40 I ( A ) 40 x 10(45.5) 10 x 40(.5 0) 1 1 1 I = 65.8 x 10 + 115.8 x 10 = 181.7 x 10 mm 4 Next calculate the shear stress in the flange for = 7.5 to 17.5 17.5 τ x The first moment of area is A = 17.5 τ x F A I b 1000 0 181.7 x 10 x 40 bd 40 17.5 d 017.5 17.5.75 x 10 17.5 F I A N/mm q = b = 40[.75 x 10 - (17.5 )] Next calculate the shear flow in the web for = -.5 to 7.5 To make life easier, take the area between = -.5 and bd 10 d 5. 5 The first moment of area is A = -.5 -.5 -.5 F A -1000 x 5 τx.5 I b 181.7 x 10 x 10 Evaluating the shear stress and plotting we get the following result..5 -.5.75 x 10 -.5 b Figure 19 Now let s examine the distribution in the flange. Consider the shaded area on the diagram. D.J.DUNN 11
Figure 0 For = 5 to 0 mm F end 1000 τ 1.5 0-68.8 x 10 I 181.7 x 10 At = 5 this is 1.0 N/mm and at = 0 mm it is ero. 0 q τt 688 x 10 0 and at = 5 mm this is 10. N/m This is a linear variation from 0 at = 0 mm and a maximum of 1.0 N/mm at = 5 mm Figure 1 The force in the direction is found is the area under the q graph so F = 10. x 15/ =77.4 N There is an equal and opposite force in the left side. Now consider the shear flow in the web below the flange. q 7.5 x 10.5 q = x 10 -.5 F 7.5 q d 7.5 x 10.5 -.5 7.5.5 84. N Next consider the vertical shear flow in the flange. q = b = 40 x.75 x 10 - (17.5 )] = 110 x 10 - (17.5 )] F 17.5 - q d 110 x 10 7.5 17.5 17.5 155.8 N The total vertical force is 84. + 155.8 = 9991 N This is quite close to the expected figure of 1000 N The web carries about 84% of the shear force and the flange about 16 % 7..5 Important note the thinner the flange, the closer the figure becomes to 100% in the web. D.J.DUNN 1
8. SHEAR CENTRE Consider a cantilever beam with a point load that acts verticall but not through the centroid. The beam bends but in addition it twists as shown. Figure This produces additional torsional stress in the beam. In the case of a smmetrical beam like that shown, the solution is simpl to appl the load so that it acts through the centroid. In other cases we must appl the load so that it acts through some other point that results in no twisting and this point is called the centre of shear or shear centre. The shear centre is that point through which the loads must act if there is to be no twisting, or torsion. The shear centre is alwas located on the axis of smmetr; therefore, if a member has two axes of smmetr, the shear centre will be the intersection of the two axes. If there is onl one axis of smmetr, the shear centre is somewhere on that axis. Here are some examples of sections that are smmetrical in two axis. Figure A U section is a good example of one where the shear centre is difficult to find. As the following example shows, it occurs off the section altogether. Note that when the sections are made from thin sheets of material, the problems are easier to resolve and most of this work concerns the shear stress in thin sections. D.J.DUNN 1
WORKED EXAMPLE No.5 Determine the position of the shear centre in terms of dimension a for the U section shown made from thin metal sheet of thickness t. The force is applied verticall. Figure 4 The centre of shear for the U channel is to the left of the section as drawn. First calculate the position of the centroid. This must be on the horiontal centre line so we need to calculate the position from the vertical edge. Area A A at a/ a t/ B at a/t a t/ C at t/ at Total 4at a t + at For the section = ( a t + at )/4at if t is small the t term ma be ignored so =a t /4at = a/4 Next we calculate the shear stress in the section due to transverse shear force F FA τ I b I is the second moment of area of the section about the axis. For the vertical section I = t(a) /1 = (/)a t For the flanges we ma approximate with I = A x a where A = a t so I = x a t Adding we get I = (/)a t + x a t 8 FA FA I a t τ I b 8a tb SHEAR DISTRIBUTION IN THE FLANGE Area A = (a-)t = a Figure 5 FA τ and in this case b = t the thickness of the flange. 8a tb F(a - ) F(a - ) τ and the shear flow is q = t = 8a t 8a D.J.DUNN 14
The maximum shear stress and shear flow occurs when = 0 and are: F F τ max q max 8at 8a The minimum values are ero at = a In between the variation is linear. The distribution in the bottom flange is the same but negative. SHEAR DISTRIBUTION IN THE WEB Figure 6 Next find the expression for the shear stress distribution in the vertical section. is harder to find. Area A A at a a t B (a-)t (a+)/ (a - )t/ Total Area = at + (a-)t = t(a-) Total 1 st moment = a t + (a - )t/ = (a - )t/ for the shaded section is (a - )t/ t(a-) = (a - )/ (a-) τ FA Fta a Fa Fa q τ t It 8a t a 16a t 16a 9F 9F F F When = 0 τ q When = a τ q 16at 16a 8at 8a The shear stress distribution is like this. The stress on the top and bottom flanges falls linearl to ero at the edges. I 8 a t Figure 7 F a F The forces in the flange are the area under the graphs. F' x 8a 16 These forces acts horiontall and are equal and opposite in direction in the top and bottom flanges. D.J.DUNN 15
Now we need the force in the web. This is twice the force in one half so F' a 0 F q d x 16a a 0 (a F ) d 8a a F 8a a x 8 F' F This is not totall unexpected since we have assumed ver thin sections. The forces are like this. a 0 Figure 8 Balancing moment about the centroid we have: a F e 4 a e 4 a e 8 16 a 8 Fa a 4 16 Fa F a 4 D.J.DUNN 16
SELF ASSESSMENT EXERCISE No. 1. A U section is made from thin plate of thickness t = mm with outer dimensions B =10 mm and D = 100 mm as shown. A transverse shear force of 60 kn is applied. Calculate the position of the shear centre. Figure 9 Answers =4.14 mm, I = 1. x 10 6 mm 4 Flange force = 9.9 kn each and web force = 60 kn The shear centre is 48.8 mm to the left of the section.. Repeat the problem but with t = 10 mm and comment on the differences. Answers =46.5 mm, I = 5. x 10 6 mm 4 Flange force =.6 kn each and web force = 59.6 kn The shear centre is 50. mm to the left of the section. The problems with thicker material is the distribution and force in the junction areas and what precise values to use for the dimensions. D.J.DUNN 17