# MOS Set_1 - 4300:202 Set 1 Mechanics of Solids Dr A F...

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Unformatted text preview: 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb Chapter (1) Tension (Compression) and Shear I. Normal Stress/Strain in Tension and Compression Define Normal Stress σ: σ= P A …………..(1) = Force - intensity per unit area (e.g. in Ksi, psi, MPa, etc.) = Positive (+) for tensile stress = Negative (-) for compressive stress Where, P = P(x) = normal force at section located at x (Parenthesis to indicate function of) A = A(x) = cross-section area at the same location with equilibrium as a basis for deriving (1). (Overall to find RA = P and internal for either left or right “FBD” to get P(x) = P) 1 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb • Normal Strain ε : − We study geometry / kinematics − Let δ = u B = ( absolute ) displacement of point “B”. u C = ( absolute ) displacement of point“C”. u D = ( absolute ) displacement of point“D”. − Define Δu mn = relative displacement between two points, m and n. Δu BA = u B – u A ( definition ) , or Δu BA = u B – ( 0 ) = δ ( since u A − = ( here ) − − − − − (2a) 0 at support ) Define normal strain as : ε≡ Δu BA δ = ( here ) − − − − − (2b) L BA L 2 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb i.e., ε = change-of-length / unit length = (+) for elongation = (-) for shortening = nondimensional (e.g., in/in or m/m) - Generally, we can also consider any two points; e.g., “C” and “D” ∴ε = Δu DC u D -u C = L DC x D -x C - OR, for a small (differential) material element at x = xC and length ∆ x = x D -x C Δu du = − − − −(3) Δx → 0 Δ x dx ε = ε ( x ) = Lim - Eq.(3) is valid in “any” bar (prismatic/non-prismatic, etc.) ; i.e., if strain ε(x) is given, we then calculate displacement u(x) at any point by integration: u (x) = ∫ ε(x) dx + Co − − − −(4a) where Co = constant (found from support condition), or, using possibly definite integrals x u ( x ) – u ( x o ) = ∫ ε(x) dx − − − −(4b) x0 3 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb *Stress-Strain Relations ( σ vs. ε) - From experiments : Tension ~ for Ductile (e.g., metals) Compression ~ for Brittle (e.g., Concrete, Ceramics) - Real σ vs. ε relaxation, etc) curves are complex (e.g., nonlinear, plastic, creep, - Approximations at “low” (e.g. relative to the yield stress) stress levels are valid; i.e., linear/elastic (Hooke’s Law ); i.e., stress is linearly proportional to strain: σ = Eε − − − −(5a) where E = Young’s modulus (constant with stress units) - Also, lateral strains develop; i.e., longitudinally-extended bars will contract laterally (note the –ve sign) ε lat = - υ ε − − − −(5b) where υ = (constant) Poisson’s ratio Remark: Structural steel: E = 30, 000 ksi υ = 0.25 → 0.3 ( non dimensional constant ) Yield Stress = 30 Ksi << E ∴Typical operating stresses in real engineering structures are below yielding leading to small deformations. 4 4300:202 Mechanics of Solids Set # 1 Dr. A. F. Saleeb (b) Brittle Ceramic Material (Difference εB - εA= “small”) (a) Ductile Metal (εE>> εA) Time-Independent Behavior Time-Dependent Creep Behavior under Constant Load 5 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb Example: Given: Shown bar with E = 3 x 104 ksi, υ = 0.3 (constants along ABC) Non prismatic with circular cross sectional areas having different diameters (d1, d2) Find: (i) stresses (and strains) in parts AB, BC of bar (ii) displacements at B and C (i.e., u diagram shown) (iii) change in diameter d1 for segment BC. Solution: (i) Stresses and Strains Part BC: Pbc = P = 18.85k Abc = A1 = (at any cut in BC) π2π 2 d1 = (1.5 ) 4 4 6 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb From Eq. (1), σ1 = Pbc 18.85 = = 10.7 ksi π A bc (1.5)2 4 Part AB: Pab = P = 18.85k Aab = A2 = (at any cut in AB) π2 π 2 d 2 = (2) 4 4 From Eq. (1), σ2 = Pab 18.85 = = 6.0 ksi π2 A ab (2) 4 Also, from Eq. (5a) : ε1 = σ1 10.7 -4 = ≅ 3.57×10 4 E 3x10 (in/in along BC) ε2 = σ2 6 -4 = ≅ 2×10 4 E 3x10 (in/in along AB) (ii) Displacements Consider part AB: ∆uBA = uB – uA = uB (since uA = 0 at support) LAB = L2 = 18” Apply Eq. (2b) directly, ∴ ε2 = uB , 18 7 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb or, uB = 18(2x10-4) = 36×10-4 in (to right Consider part BC: ε1(x) = 3.57x10-4 (i.e., constant in BC) Apply Eq. (4b) with, xo = xB = 18” and, u(xo) = uB = 36x10-4 in x = xC = 36” and, u(x) = uC (required) 36 ∴ uC = uB + ∫ ε -4 1 -4 dx = 36x10 + 3.57×10 (36-18) 18 = 1×10-2 in (to right ) (iii) Change in diameter Let, ∆d1 = change in diameter d1 of segment BC use Eq. (5b): ∴ ε1Lat = -υ ε1 = -0.3(3.57x10-4); But Δd1 = ε1Lat d1 -4 Δd1 = -1.607×10 in. (i.e., decrease) 8 ) 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb II. Shear Stress and Strain: • Shear Stress, τ: - It acts parallel (or tangential) to the cross section. - Two types: (i) Direct or simple shear (e.g. connections) (ii) Indirect Shear due to bending, torsion, etc. (studied later) - For direct/simple shear, we have the average shear stress, τ given by: τ= Where V As ………… (1) V = shear force acting on area As As = “appropriate” shear area 9 4300:202 Set # 1 Mechanics of Solids (i)Bolted-Connection Dr. A. F. Saleeb (ii) Punch-Plate System (iii) Shear Test (Double Shear) Section: mn (or p q) V: Π( d 2 4 mn P P/2 As: mn P Π ⋅ dt ) b.h • State of Pure Shear Stress: (Conjugate Shear) - Consider a material element shown (m n p q) whose sides/faces are parallel to the directions x, y, z. axes. Define m n (p q) = positive (negative) x-face of element q m (p n) = positive (negative) y-face of element 10 4300:202 Mechanics of Solids Set # 1 Dr. A. F. Saleeb (a face defined by direction of its outward normal) - From equilibrium, equal shear stresses τ develop on four faces; i.e., conjugate shear - Signs: τ = positive if acting on a positive (negative) face in the positive (negative) direction of the coordinate axis parallel to the face (all τ - arrows are +ve in figure) • Shear Strain, γ: (Distortion/ Shape Change) γ = change in the original 90 degree between two perpendicular faces. = in radians (nondimensional) = (+) if decrease (increase) for positive (negative) faces 11 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb o From experiments, we determine τ vs γ curves for different materials. o Assuming linear elastic material response, τ = Gγ ……………. (2) where G = shear modulus (stress units) = E 2(1+υ) ……………. (3) Example: Given: P, dimensions (b, d, h) of the shock-mount system shown. G = shear modulus for rubber Assume rigid steel bar and tube 12 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb Required: (a) τ = τ(r) = shear stress at any radius r in rubber part (sec. I-I shown) (b) δ = downward displacement of bar Solution: (a) Consider FBD in equilibrium: ∑ F vertical = 0 ∴ P = (2 ⋅ Π ⋅ r ⋅ h) ⋅ τ That is, τ is now variable depending on radius r; i.e. τ = τ(r) τ= P 2⋅Π⋅h ⋅r ………….. (i) (b) From Hook’s law in shear, we have γ = γ(r) = τ(r) P = G 2⋅Π⋅h ⋅r ⋅G 13 ………….. (ii) 4300:202 Set # 1 Mechanics of Solids - Since Dr. A. F. Saleeb γ is variable for d ≤ r ≤ b in the rubber region (recall 2 2 assumption of rigid steel parts), we need to proceed from a differential (“small”) element dr at location r. Integrate with “appropriate” limits. δ r = b/2 ∫ dδ = ∫ 0 γ.dr …………..(iii) r =d/2 Substitute from (ii) into (iii) b/2 P 1 δ= ∫ dr 2 ⋅ Π ⋅ G ⋅ h d/2 r Recall, 1 ∫ r dr = ln ( r ) ∴δ = P ⎛b⎞ ln ⎜ ⎟ 2⋅Π⋅G ⋅h ⎝ d ⎠ 14 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb III. Allowable Stresses/Loads − To safeguard against failure (e.g. due to yielding, brittle fracture, excessive deformations, buckling, etc.), we use a safety factor, n. − Allowable stress σ all is then determined from the “failure” (e.g. yield) stress as σ all = − σy n (e.g. failure = yield) ……….. (4) Different design codes use “n” defined with respect to stresses or loads, and considering important requirements of strength, stiffness and stability. Example Find Qall based on the following allowable stresses in the bar-bolt connection shown. − t tension in bar, σ all = 118 MPa − shear in bolt, τ all =80 MPa − b bearing between bolt and bar, σ all =240 MPa (see schematics of some of the “failure modes” considered in the figures that follow) 15 4300:202 Mechanics of Solids Set # 1 Dr. A. F. Saleeb Solution • Considering Tension in Bar Part BC: Pbc = Q, A = b ⋅ t, Pbc t = σ all A (1) t ∴ Q = Pbc = σall ⋅ A = (118MPa)(20×35mm2 ) = 82.6 kN σ bc = Part AB: Critical section at bolt, A net = (b − d) ⋅ t Q 2 t σ ab =Pab /A net =σ all Pab = ∴ Q(2) = 2Pab = 2(118MPa) ⋅ ((35 − 15)(20)mm 2 ) = 94.4 kN 16 4300:202 Set # 1 Mechanics of Solids Dr. A. F. Saleeb • Considering shear in Bolt 1Q Q V= ( )= 22 4 Π As = d 2 4 V τ bolt = = τ all As ∴ Q(3) = 4V = 4( Π (15) 2 mm 2 )(80MPa) = 56.55 kN 4 • Considering Bearing A b (for bearing) = d ⋅ t; Fb = Q 2 σ bearing = ⇒ Fb b = σ all Ab Q(4) = 2Fb = 2(240MPa)(20×15mm 2 ) = 144 kN • Conclusion (1) (2) (3) (4) Qall = min ( Q , Q , Q , Q ) = 56.55 kN, i.e., bolt shear governs 17 4300:202 Mechanics of Solids Set # 1 • Bolt-shear Failure Mode. • Combined Plate-bolt bearing Failure Mode. • Tensile Failure Mode on the Net Section of Flat Bar (Plate). 18 Dr. A. F. Saleeb 4300:202 Mechanics of Solids Set # 1 19 Dr. A. F. Saleeb 4300:202 Mechanics of Solids Set # 1 20 Dr. A. F. Saleeb ...
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