P4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections


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1 4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections Shear stress and shear strain. Equality of shear stresses on perpendicular planes. Hooke s law in shear. Normal and shear stresses on inclined sections. Maimum stresses on a bar in tension. Introduction to stress elements. 1
2 Shear stress Shear stress acts tangential to the surface of a material. Top view Side view Average shear stress τ = V / A Greek letter τ (tau) V = shear force A = area on which it acts (Disregard friction in the calculations to err on the conservative side.) 2
3 Bolts in Single shear Average (normal) bearing stress σ B σ b = total bearing force ( ) projected area of curved bearing surface ( A) A = plate thickness t bolt diameter d t d d t 3
4 Forces applied to the bolt by the plates These forces must be balanced by a shear force in the bolt. This results in a shear stress τ. τ V τ = = A ( π d 2 / 4) 4
5 Bolts in Double shear /2 /2 /2 V τ = = A /2 ( / 2) ( π d 2 / 4) 5
6 Eample of bearing stress and shear stress (based on Eample 15, page 36, Gere, 6th ed. 2004) A steel strut S is used as a brace for a boat hoist. It transmits a force to the deck of a pier. The strut has a hollow square crosssection with wall thickness t. A pin through the strut transmits the compressive force from the strut to two gussets G that are welded to the base plate B. Calculate: (a) the bearing stress between the strut and the pin (b) the shear stress in the pin (c) the bearing stress between the pin and the gussets (d) the bearing stress between the anchor bolts and the base plate (e) the shear stress in the (Gere 2004) anchor bolts 6
7 = 50 kn, t = 10 mm diameter of pin, d pin = 20 mm thickness of gussets, t G = 15 mm diameter of anchor bolts d bolt = 12 mm thickness of base plate t B = 10 mm (Gere 2004) (Gere 2004) (a) bearing stress between strut and pin σ = 2( t d pin = 125 Ma ) /2 (b) shear stress in pin τ = ( ( π / 2) / 4) = 2 d pin /2 S /2 pin G G / Ma 7
8 (c) bearing stress between pin and gussets σ = = 83.3Ma 2( t G d pin ) (Gere 2004) (d) bearing stress between anchor bolts and base plate σ = cosθ 4 ( t B d bolt ) = 79.8 Ma (Gere 2004) (e) shear stress in anchor bolts τ = ( cosθ / 4) ( π / 4) = 2 d bolt 84.7 Ma 8
9 These are eamples of direct shear  the shear stresses are a result of the direct action of a shear force trying to cut through the material. Xsection τ V M Shear force diagram τ ma = 3V / 2A Shear stresses can also arise in an indirect manner during tension, torsion, and bending. 9
10 Equality of shear stresses on perpendicular planes τ 2 τ 3 c b a τ 1 τ 4 Forces Vertical direction: τ 1 (bc) = τ 3 (bc) Horizontal direction: τ 2 (ac) = τ 4 (ac) So τ 1 = τ 3 and τ 2 = τ 4 10
11 τ 2 (ac) τ 3 (bc) b τ 1 (bc) a τ 4 (ac) Moments about [ τ 1 (bc) ] (a) = [ τ 2 (ac) ] (b) about [ τ 3 (bc) ] (a) = [ τ 4 (ac) ] (b) So τ 1 = τ 2 and τ 3 = τ 4 Combining this with the previous result gives τ 1 = τ 2 = τ 3 = τ 4 = τ, which called pure shear. 11
12 Shear strain Shear stresses have no tendency to elongate or shorten; instead they produce a change in shape. τ γ/2 γ γ γ/2 This change in shape is quantified by the angle γ, the shear strain. The angle is measured in radians, not degrees. Greek letter γ (gamma) 12
13 Hooke s Law in shear τ = G γ G is the shear modulus of elasticity (or modulus of rigidity ). Units are N / m 2 = a. It can be shown that the elastic constants E and G are related by: G = E 2 ν ( 1+ ) Since 0 < ν < 0.5 for most materials, G is typically onethird to onehalf E. For mild steel, E = 210 Ga and G = 81 Ga. For aluminium alloy, E = 72 Ga and G = 28 Ga. 13
14 Normal and shear stresses on inclined sections To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an inclined (as opposed to a normal ) section through the bar. Inclined section Normal section Because the stresses are the same throughout the entire bar, the stresses on the sections are uniformly distributed. Inclined section Normal section 14
15 2D view of the normal section (but don t forget the thickness perpendicular to the page) y σ = / A Area A 2D view of the inclined section 15
16 Specify the orientation of the inclined section pq by the angle θ between the ais and the normal to the plane. y p n m θ Normal section θ = 0 Top face θ = 90 q Left face θ = 180 Bottom face θ = 270 or 90 y N θ V The force can be resolved into components: Normal force N perpendicular to the inclined plane, N = cos θ Shear force V tangential to the inclined plane V = sin θ 16
17 If we know the areas on which the forces act, we can calculate the associated stresses. y σ θ y area A area A τ θ area ( A / cos θ) area ( A / cos θ) Force N cosθ σ θ = = = = Area Area A/ cosθ A 2 σ σ θ = σ cos θ = ( 1+ cos 2θ ) 2 Force V sinθ τθ = = = = Area Area A/ cosθ σ τθ = σ sinθ cosθ = ( sin 2θ ) 2 A 2 cos θ sinθ cosθ 17
18 Sign convention y τ θ σ θ θ Normal stresses σ θ positive for tension. Shear stresses τ θ positive when they tend to produce counterclockwise rotation of the material. Note that these equations are derived from statics only and are therefore independent of the material (linear or nonlinear, elastic or inelastic). 18
19 lot σ θ and τ θ versus θ. σ θ = σ at θ = 0 This is σ ma. σ θ = σ /2 at θ = ± 45 σ θ = 0 at θ = ± 90 No normal stresses on sections cut parallel to the longitudinal ais. τ θ = 0 at θ = 0, 90 τ θ = τ ma = ± σ /2 at θ = /+ 45 τ ma = σ /2 19
20 Eample of stresses on inclined sections (based on Eample 211, page 114, Gere, 6th ed. 2004) A compression bar with a square cross section of width b must support a load = 36 kn. The bar is constructed from two pieces of material that are connected by a glued joint (known as a scarf joint) along plane pq which is at an angle α = 40º to the vertical. The material is a structural plastic with σ allow (compression) = 7.6 Ma τ allow = 4.1 Ma The glued joint has σ allow (compression) = 5.2 Ma τ allow = 3.4 Ma (Gere 2004) Determine the minimum width b of the bar. 20
21 σ = / A = / b 2 b = / σ Smallest σ governs the design. (Gere 2004) Values of σ based on allowable stresses in the plastic: Maimum compressive stress is 7.6 Ma = σ Maimum shear stress is 4.1 Ma = τ ma = σ /2 at θ = ±45º This gives σ = 2τ ma = 8.2 Ma 21
22 Values of σ based on allowable stresses in the glued joint: y p θ = 50º q α = 40º τ θ σ θ θ (/A) σ σ σ σ θ σ = 2 2σ θ = 1+ cos 2θ 2( 5.2) = 1+ cos 2( 50) ( 1+ cos 2θ ) = 12.6 Ma τ θ σ σ σ σ = 2 2τ θ = sin 2θ 2( 3.4) = sin 2( 50) ( sin 2θ ) = 6.9 Ma 22
23 Smallest σ = 6.9 Ma (shearing of glue joint) b = / σ b b b = = = ( ( m 72.2 mm 3 6 ) ) (Gere 2004) 23
24 Introduction to stress elements Stress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are infinitesimal, but are drawn to a large scale. y σ = / A Area A y y σ z σ = A / σ σ = A / 24
25 Maimum stresses on a bar in tension a b σ σ = σ ma = / A a No shear stresses 25
26 a b b σ /2 τ ma = σ /2 σ /2 θ = 45 Angle σ θ τ θ θ = 45 σ /2 σ /2 θ = 135 σ /2 σ /2 θ = 45 σ /2 σ /2 θ = 225 σ /2 σ /2 In case b (θ = 45 ), the normal stresses on all four faces are the same, and all four shear stresses have equal and maimum magnitude. 26
27 If the bar is loaded in compression, σ will have a negative value and stresses will be in the opposite directions. σ /2 σ σ =  / A τ ma = σ /2 a b σ /2 Even though the maimum shear stress in an aially loaded bar is only half the maimum normal stress, the shear stress may cause failure if the material is much weaker in shear than in tension. 27
28 Eamples Wood block in compression fails by shearing on 45 planes Mild steel loaded in tension. Visible slip bands (Lüders bands) appear on the sides of the bar at approimately 45 to the ais of loading when the yield stress is reached. These indicate that the material is failing in shear along planes of maimum shear stress (cupandcone failure). Note that uniaial stress (simple tension or compression in one direction) is just a special case of a more general stress state known as plane stress. 28
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