Exam2

# Exam2 - CE 215 Spring 2010 Exam 2 closed book/notes one...

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Unformatted text preview: CE- 215 Spring 2010 Exam 2, closed book/notes, one sheet of notes, calculator allowed Student ID (Last four digits): ________ NAME: ________________________________ σ = Eε ε T = α ΔT r dφ rφ = rθ γ max = rθ = dx L 4 4 ρ πr πd τ = G ρθ = τ max Ip = = r 2 32 τ = Gγ () δ T = ε T L = α ΔT L () γ max = γ = ρθ = ρ γ r max τ max = Grθ TL φ = ∑ φi = ∑ i i i =1 i = 1 Gi I pii n n T dx φ = ∫ dφ = ∫ 0 0 GI p x L L () φ = ∫ dφ = 0 L ∫ GI ( x ) 0 p L T x dx () κ= 1 dθ dθ = = ρ ds dx εx = − y = −κ y ρ σ x = Eε x = − Ey = − Eκ y ρ σ1 = − I= bh3 , 12 S= bh2 6 Mc1 M =− I S1 σ2 = − Mc2 M =− I S2 S1 = I I , S2 = c1 c2 τ max = + Vh2 3V = 8I 2A τ xy = τ yx 1 1 σx = 1 σx + σy 2 1 σx − σy 2 σx + σy 2 cos 2θ + τ xy sin 2θ − τx y = − 11 σx − σy 2 sin 2θ + τ xy cos 2θ σx + σy = σx + σy tan 2θ p = 2τ xy σy = σx − σy 2 cos 2θ − τ xy sin 2θ 2 σx − σy σ 1, 2 = σx + σy 2 ⎛ σx − σy ⎞ 2 ±⎜ ⎟ + τ xy 2⎠ ⎝ σ1 + σ 2 = σ x + σ y cos 2θ p = 1 σx − σy 2R sin 2θ p = 1 τ xy 2R τ max ⎛ σx − σy ⎞ 2 =⎜ ⎟ + τ xy 2⎠ ⎝ 2 τ max = σ1 − σ 2 2 σ aver = σx + σy 2 Problem 1 (50 pts) A metal sheet is cut at 55O as shown and then put together by weld spots s 0.7 MPa equally spaced at distance s between them. The 0.7 MPa s sheet is 1 cm (0.01 m) thick and is subjected to a s stress of 0.7 MPa in one direction as shown. 55o Each weld spot has an allowable load in lowa ll tension/compression, Ftalnsionble = 15 kN , and an allowable load in shear, Fsaheowable = 10 kN . Find the e ar minimum spacing, s, such that the allowable force on the weld spots in tension and in shear will not be exceeded. Solution The cut is oriented at θ = −35O with respect to the horizontal axis, x. The transformation equations for stress yield the following normal and shear stresses at the cut: σ cut = σx + σy 2 + σx − σy 2 cos 2θ + τ xy sin 2θ = 0.7 + 0 0.7 − 0 + cos( −70O ) + 0 sin( −70O ) = 0.47 MPa 2 2 τ cut = − σx − σy 2 sin 2θ + τ xy cos 2θ = − 0.7 − 0 sin( −70O ) + 0 cos( −70O ) = 0.329 MPa 2 Let the thickness of the sheet be t (t=0.01 m in this case). Then, assuming two welds (one on each side of the sheet), σ cut t = τ cut t = lowa 2 Ftalnsionble e s 2F allowable shear →s= →s= lowa 2 Ftalnsionble e σ cut t ll 2 Fsaheowable ar = = 2 × 15 × 103 = 6.39m = 639 cm 0.47 × 106 × 0.01 2 × 10 × 103 = 6.088 m = 608.8 cm 0.329 × 106 × 0.01 s τ cut t Thus the minimum spacing is 6.09 m. In practice, this is a long distance, for s, so at least three weld spots would be placed for rigidity. we ld s po ts Problem 2 (50 points) At a material element the state of plane stress is as shown in the figure. (a) Draw the Morh’s circle representing this state of stress. (b) Using the Mohr’s circle determine the principal stresses σ1 and σ2. (c) draw the material element in the principal directions, i.e. in a properly oriented element. !x = -48.0 kPa "xy = -20,7 kPa Solution (a) Center is at (-48.0+0)/2 = -24 kPa Point A is at (-48.0,-20.7) The above yield the circle (b) σ1 = center + R, σ2 = center - R A Σ2 ￿60 ￿40 ￿20 Σ1 R = 20.7 2 + ( 48 − 24)2 = 31.7 kPa These yield σ1 = 7.7 kPa, σ2 = -55.7 kPa τ xy 1 ArcTan[ ] = 20.39O (from A to σ2). 2 σ x − center This yields the following material element in the principal directions (c) θ P = !1=7.7 kPa !2 = -55.7 kPa 20.39o ...
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