{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW12+Solutions

# HW12+Solutions - HOMEWORK SOLUTIONS Week 12 Due Friday 2...

This preview shows pages 1–5. Sign up to view the full content.

HOMEWORK SOLUTIONS Week 12: Due Friday 2 December 1. Figure 1 shows the stresses acting on an element of material. Find the three principal stresses σ 1 , σ 2 , σ 3 and the absolute maximum shear stress τ max . 12 12 8 8 6 6 17 17 all stress components in ksi. Figure 1 Solution The stress components are σ x = - 17 ksi ; σ y = 12 ksi ; τ xy = 6 ksi ; σ z = 8 ksi , and τ yz = τ zx = 0 , so σ z is a principal stress which we can denote as σ 3 . The points X ( σ x , τ xy ) and Y ( σ y , - τ xy ) are located as (-17,6) and (12,-6) respectively, as shown in Figure 1.1. O τ σ R N σ 2 σ 1 X ( - 17,6) Y (12, - 6) C σ 3 Figure 1.1 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The centre of the Mohr’s circle is therefore at the point (-2.5,0), since c = ( σ x + σ y ) 2 = - 17 + 12 2 = - 2 . 5 ksi. We note that σ 3 = 8 ksi lies inside the circle, so this is a case where the absolute maximum shear stress is the radius of the original circle. To find this radius, we use Pythagoras’ theorem on the triangle CNX obtaining R = τ max = 6 2 + 14 . 5 2 = 15 . 69 ksi . [Note that CN = 17 - 2 . 5 = 14 . 5 ksi.] The remaining two principal stresses are σ 1 = c + R = - 2 . 5 + 15 . 69 = 13 . 19 ksi ; σ 2 = c - R = - 2 . 5 - 15 . 69 = - 18 . 19 ksi . Summary:- σ 1 = 13 . 19 ksi ; σ 2 = - 18 . 19 ksi ; σ 3 = 8 ksi ; τ max = 15 . 69 ksi . 2
2. The principal stresses at a given point are σ 1 = 10 MPa, σ 2 = - 100 MPa. Sketch Mohr’s circle for this state of stress and determine the normal stress on a plane inclined at an angle θ to the principal plane 1. Hence find the range of values of θ for which the normal stress is tensile. Solution The corresponding circle is shown in Figure 2.1. Its radius is R = 10 - ( - 100) 2 = 55 MPa and the centre is at ( c, 0) , where c = 10 - 100 2 = - 45 MPa. O σ C 10 - 100 P - 45 φ τ Figure 2.1 The point P on the Mohr’s circle corresponding to a plane inclined at an angle θ counterclock- wise from the principal plane 1 is located as shown in Figure 2.1, where φ = 2 θ . Thus, σ P = - 45 + 55cos2 θ . For σ P > 0 , we require 55cos2 θ > 45 and hence cos2 θ > 45 55 = 0 . 818 . This in turn requires that - 35 . 1 o < 2 θ < 35 . 1 o and hence - 17 . 55 o < θ < 17 . 55 o . 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. A glass wire of 1 mm diameter is tested in tension and found to fracture at a tensile force of 40 N. A similar unbroken wire is used to suspend a mass of 2 kg from the ceiling. The mass is now subjected to a moment T about the vertical axis, causing the wire to be loaded in torsion as well as tension. Find (i) the value of T at which the wire will break and (ii) the angle θ between the fracture surface and the cross-sectional (horizontal) plane. Would your answer to (i) be different if the wire were made of copper? Solution Glass is a brittle material which we assume fractures when the maximum tensile stress reaches a critical value, which is therefore σ fracture = 40 π × 1 2 / 4 = 50 . 93 MPa . When the wire is used to support a mass of 2 kg, there will be a tensile stress σ y = 2 × 9 . 81 π × 1 2 / 4 = 24 . 98 MPa . The wire is now subjected to a torque which contributes a shear stress τ xy , whilst σ y remains unchanged and σ x = 0 . This means that the Mohr’s circle will have the form of Figure 3.1.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}