2.002 Mechanics and Materials II, Solutions for Homework 2

MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II SOLUTIONS FOR HOMEWORK NO. 2 Problem 1 (15 points) The axial stiffness of the structure is: kaxial = Fx σxx A A = =E δx xx L L (1) where A is the area of the cross section and L is the length of the beam. The bending stiffness is defined as: Fy , (2) kbending = δy where Fy is the concentrated tip load, and δy is the related tip deflection. According to the 3 yL . Thus, the bending stiffness is: beam theory, we know that δy = F3EI kbending = Fy Fy 3EI = Fy L3 = 3 . δy L (3) 3EI Combining Eq.1 and Eq.3, the ratio of a slender cantilever’s bending stiffness to its axial stiffness is: 3EI 3I/A l2 kbending 3 = LA = = 3( )2 (4) 2 kaxial L L EL � where l2 ≡ I/A. Since L  l2 , we find the above ratio is extremely small. For a solid circular cross-section beam with diameter d, we have: πd4 64 (5) πd2 4 Substitution of the A and I into Eq. 4, we have (6) I= and A= 3I/A 3 d kbending = = ( )2 2 kaxial L 16 L Problem 2 (45 points) Part A: No forces and moments are applied to the beam. And at any x, 0 < x < l, we have: � N = Fx = σxx dA = 0, 1 (7) (8) � and M= σxx (−y)dA = 0 (9) The stress is a function of y, σxx = σxx (y), and does not change along the x-direction. The deformation is composed of two parts: the thermal part and the mechanical part, thus the total stain at each point can be expressed as: (total) = (mechanical) + (thermal) (10) (mechanical) = σ(y)/E (11) where and ∆T ∗ y ) (12) h/2 Meanwhile, since the beam deflects laterally with a constant curvature, κ(thermal) , the total strain of the beam can also be expressed as: (thermal) = α∆T = α(∆T0 − total = 0 − κ(thermal) y (13) where 0 is the total strain for the mid-surface axial. ∂u0 (x) = 0(thermal) → total = 0(thermal) − κ(thermal) y ∂x From Eq.s 10, 11, 12 and Eq. 14, we have: 0 = 0(thermal) − κ(thermal) y = (mechanical) + (thermal) = σ(y)/E + α(∆T0 − (14) ∆T ∗ y ) h/2 (15) Thus, the stress can be expressed as: σxx (y) = E[0(thermal) − κ(thermal) y − α(∆T0 − ∆T ∗ y )] h/2 (16) Substituting Eq. 16 into Eq. 8, we have: � h/2 ∆T ∗ y )]dy = 0 Eb [0(thermal) − κ(thermal) y − α(∆T0 − h/2 −h/2 (17) ∗ y The integration of the linear terms in the above equation, −κ(thermal) y and ∆T , will be h/2 zero, since the integral range is symmetric (from −h/2 to h/2). Thus we have: � h/2 Eb(0(thermal) − α∆T0 ) dy = 0 → 0(thermal) = α∆T0 (18) −h/2 Another easier way to get this relation is that: 0(thermal) = α∆T (y = 0) → 0(thermal) = α∆T0 2 (19) Substituting Eq. 16 into Eq. 9 we have: � h/2 ∆T ∗ y Eb [0(thermal) − κ(thermal) y − α(∆T0 − )](−y)dy = 0 h/2 −h/2 �  0, we get: Considering that 0(thermal) = α∆T0 , and y 2dA ≡ I = κ(thermal) = 2∆T ∗ α h Part B: Substitution of Eq. 18 and Eq. 21 into the expression of the stress, Eq. 16, we get: 2∆T ∗ α ∆T ∗ y y − α(∆T0 − )] = 0 σxx (y) = E[α∆T0 − h h/2 (20) (21) (22) The axial stress in the thermally-loaded cantilever is zero everywhere. Part C: The tip deflection caused by the thermal load is: 1 ∆T ∗ α 2 L v∆T (x = L) = κ(thermal) L2 = 2 h The tip deflection caused by the reaction force Rtip is: Rtip L3 3EI The actual tip deflection is zero. By applying superposition, we get: vRtip (x = L) = 0 = v∆T (x = L) + vRtip (x = L) → ∆T ∗ α 2 Rtip L3 3EIα∆T ∗ L =− → Rtip = − h 3EI hL (23) (24) (25) Part D: As shown in Part B, the thermal load does not generate stress, thus the stress field of this tip- restrained thermally-loaded cantilever beam is the same as the beam is only subjected to a tip load of Rtip . Since the sign of the tip-load is negative, Rtip is downwards and the upper part of the beam is in tension and the lower pare is in compression. The bending moment is expressed as: M(x) = Rtip (L − x) (26) The maximum tensile stress is at (x = 0, y = h/2), and value is expressed as: σmax Rtip L h2 M h2 3 =− = Eα∆T ∗ =− I I 2 (27) Part E: The problem requests that σmax < σy = 350MP a, substitution of Eq. 25 into the above relation, we have: 3 2σy (28) σmax = Eα∆T ∗ < σy → ∆T ∗ < 2 3Eα With E = 210GP a and α = 12 × 10−6 /◦ K, we find the largest bottom/top difference in temperature change, ∆T ∗ = 92.6◦ K. So the largest bottom/top difference in temperature change is 2 ∗ ∆T ∗ = 185.2◦ K 3 * 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Fig 01: The total deflection 0.9 1 x/L Problem 3 (20 points) The lattice-based prediction of the density of iron is: ρF e = × 10−23 g 2 × 55.847 6.02 = 7.88Mg/m3 0.28663 × 10−27 m3 (29) For Sodium chloride, there is one half atom (4 × 1/8) of both sodium (Na) and chlorine (Cl) in each cell. The atomic mass for sodium is 22.99g/mol and 35.453g/mol for chlorine. ρN aCl = 0.5 × (22.99+35.453) 6.02 r03 m3 × 10−23 g = 2.17Mg/m3 (30) Solving the above equation, we got the estimated lattice spacing r0 = 0.2818nm. Problem 4 (20 points) . S0 = Er0 (31) Metal E (GPa) a0 (˚ A) S0 (N/m) Ni 214 3.1517 67.446 Al 70 4.0496 28.347 Pt 172 3.9231 67.477 Pd 124 3.8902 48.238 Cu 124 3.6151 44.827 Au 82 4.0786 33.445 Ag 76 4.0862 31.055 We can see that Au and Ag have the similar value in S0 and Ni and Pt have the similar value in S0 . Conclusion: elements in the same column of the periodic table of elements tend to have similar atomic bond stiffness (S0 ) 4 Ni log(E) Pt S o Cu Pd 2 10 Au Ag Al 0.5 0.6 10 10 log(a) Fig 02: E vs. ao