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Deformation of members under axial loading
1.
Third Edition LECTURE
RODS: AXIAL LOADING AND DEFORMATION • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 3 by Dr. Ibrahim A. Assakkaf SPRING 2003 Chapter ENES 220 – Mechanics of Materials 2.8 Department of Civil and Environmental Engineering University of Maryland, College Park LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 1 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Uniform Member – Consider a homogeneous rod as shown in the figure of the next viewgraph. – If the resultant axial stress σ = P/A dos not exceed the proportional limit of the material, then Hooke’s law can applied, that is σ = Eε σ = axial stress, ε = axial strain E = modulus of elasticity
2.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 2 ENES 220 ©Assakkaf Deformations of Members under Axial Loading Uniform Member L δ P LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 3 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Uniform Member – From Hooke’s law, it follows that σ = Eε (Hooke' s law) P = Eε A δ But ε = , therefore, L P δ PL = E ⇒δ = A L EA
3.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 4 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Uniform Member – The deflection (deformation),δ, of the uniform member subjected to axial loading P is given by PL δ= (1) EA LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 5 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Multiple Loads/Sizes – The expression for the deflection of the previous equation may be used only if the rod or the member is homogeneous (constant E) and has a uniform cross sectional area A, and is loaded at its ends. – If the member is loaded at other points, or if it consists of several portions of various cross sections, and materials, then
4.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 6 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Multiple Loads/Sizes – It needs to be divided into components which satisfy individually the required conditions for application of the formula. – Denoting respectively by Pi, Li, Ai, and Ei, the internal force, length, cross-sectional area, and modulus of elasticity corresponding to component i,then n n PL δ = ∑δ i = ∑ i i i =1 i =1 Ei Ai LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 7 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Multiple Loads/Sizes n n Pi Li δ = ∑δ i = ∑ i =1 i =1 Ei Ai E1 E2 E3 L1 L2 L3
5.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 8 ENES 220 ©Assakkaf Deformations of Members under Axial Loading Multiple Loads/Sizes – The deformation of of various parts of a rod or uniform member can be given by n n Pi Li δ = ∑δi = ∑ (2) i =1 i =1 Ei Ai E1 E2 E3 L1 L2 L3 LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 9 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 – Determine the deformation of the steel rod shown under the given loads. Assume that the modulus of elasticity for all parts is – 29×106 psi 2 Area = 0.9 in 45 kips 30 kips 75 kips Area = 0.3 in2 12 in 12 in 16 in
6.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 10 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 (cont’d) – Analysis of internal forces 1 2 3 45 kips 30 kips 75 kips 30 kips P3 → + ∑ Fx = 0; − P3 + 30 = 0 ∴ P3 = 30 kips LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 11 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 (cont’d) – Analysis of internal forces 1 2 3 45 kips 30 kips 75 kips P2 45 kips 30 kips → + ∑ Fx = 0; − P2 − 45 + 30 = 0 ∴ P2 = −15 kips
7.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 12 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 (cont’d) – Analysis of internal forces 1 2 3 45 kips 30 kips 75 kips 45 kips P1 30 kips 75 kips → + ∑ Fx = 0; − P + 75 − 45 + 30 = 0 1 ∴ P = 60 kips 1 LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 13 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 (cont’d) – Deflection • Input parameters L1 = 12 in L2 = 12 in L3 = 16 in A1 = 0.9 in 2 A2 = 0.9 in 2 A3 = 0.3 in 2 From analysis of internal forces, P = 60 kips = 60,000 lb 1 P2 = −15 kips = −15,000 lb P3 = 30 kips = 30,000 lb
8.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 14 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 4 (cont’d) – Carrying the values into the deformation formula: 3 Pi Li 1 P L1 P2 L2 P3 L3 δ =∑ = 1 + + i =1 Ei Ai E A1 A2 A3 1 60,000(12) − 15,000(12) 30,000(16) + + 29 × 106 0.9 0.9 0.3 δ = 0.0759 in LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 15 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Relative Deformation A δA A PL δ B/ A = δ B − δ A = AE L C D δB B P
9.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 16 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Relative Deformation – If the load P is applied at B, each of the three bars will deform. – Since the bars AC and AD are attached to the fixed supports at C and D, their common deformation is measured by the displacement δA at point A. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 17 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Relative Deformation – On the other hand, since both ends of bars AB move, the deformation of AB is measured by the difference between the displacements δA and δB of points A and B. – That is by relative displacement of B with respect to A, or PL (3) δ B/ A = δ B − δ A = EA
10.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 18 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 5 The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 19 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 5 (cont’d) SOLUTION: • Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. • Evaluate the deformation of links AB and DC or the displacements of B and D. • Work out the geometry to find the deflection at E given the deflections at B and D.
11.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 20 Deformations of Members under ENES 220 ©Assakkaf Axial Loading SOLUTION: Example 5 (cont’d) Free body: Bar BDE ∑MB = 0 0 = −(30 kN × 0.6 m ) + FCD × 0.2 m FCD = +90 kN tension ∑ MD = 0 0 = −(30 kN × 0.4 m ) − FAB × 0.2 m FAB = −60 kN compression LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 21 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 5 (cont’d) Displacement of B: PL δB = AE (− 60×103 N)(0.3m) = (500×10-6 m2 )(70×109 Pa) = −514×10−6 m δ B = 0.514 mm↑
12.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 22 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 5 (cont’d) Displacement of D: PL δD = AE (90 ×103 N)(0.4 m) = (600×10-6 m2 )(200×109 Pa ) = 300 ×10−6 m δ D = 0.300 mm ↓ LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 23 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 5 (cont’d) Displacement of D: BB′ BH = DD′ HD 0.514 mm (200 mm ) − x = 0.300 mm x x = 73.7 mm EE ′ HE = DD′ HD δE = (400 + 73.7 )mm 0.300 mm 73.7 mm δ E = 1.928 mm δ E = 1.928 mm ↓
13.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 24 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Nonuniform Deformation – For cases in which the axial force or the cross-sectional area varies continuously along the length of the bar, then Eq. 1 – (PL / EA) is not valid. – Recall that in the case of variable cross section, the strain depends on the position of point Q, where it is computed from dδ ε= dx LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 25 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Nonuniform Deformation dδ ε= L dx x P
14.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 26 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Nonuniform Deformation – Solving for dδ and substituting for ε dδ = εdx – But ε = σ / E, and σ = P/A. therefore σ P (4) dδ = εdx = dx = dx E EA LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 27 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Nonuniform Deformation – The total deformation δ of the rod or bar is obtained by integrating Eq. 4 over the length L as L Px δ =∫ dx (5) 0 EAx
15.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 28 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 6 – Determine the deflection of point a of a homogeneous circular cone of height h, density ρ, and modulus of elasticity E due to its own weight. a h LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 29 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 6 (cont’d) – Consider a slice of thickness dy – P = weight of above slice – = ρg (volume above) y h 1 2 δy r P = ρg πr y 3 1 ρg πr 2 y Pdy 3 ρg dδ = = 2 = ydy EA E (πr ) 3E
16.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 30 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 6 (cont’d) ρg dδ = ydy 3E h h ρg ρg y 2 δ =∫ ydy = 0 3E 3E 2 0 ρgh 2 δ= 6E LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 31 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Normal Stresses in Tapered Bar – Consider the following tapered bar with a thickness t that is constant along the entire length of the bar. h1 h2 x h(x) L
17.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 32 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Normal Stresses in Tapered Bar h1 h2 x h(x) L – The following relationship gives the height h of tapered bar as a function of the location x x hx = h1 + (h2 − h1 ) (6) L LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 33 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Normal Stresses in Tapered Bar h1 h2 x h(x) L – The area Ax at any location x along the length of the bar is given by x Ax = thx = t h1 + (h2 − h1 ) (7) L
18.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 34 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Normal Stresses in Tapered Bar h1 h2 x h(x) L – The normal stress σx as a function of x is given P P σx = = Ax x (8) t h1 + (h2 − h1 ) L LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 35 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 7 – Determine the normal stress as a function of x along the length of the tapered bar shown if – h1 = 2 in – h2 = 6 in – t = 3 in, and – L = 36 in x h1 h(x) h2 – P = 5,000 lb L
19.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 36 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 7 (cont’d) – Applying Eq. 8, the normal stress as a function of x is given by P P σx = = Ax x t h1 + (h2 − h1 ) L 5000 5000 σx = = x x 32 + (6 − 2) ) 6 + 36 3 15000 σx = 18 + x LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 37 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Example 7 (cont’d) x (in) σ (psi) – Max σ = 833.3 psi 0 833.3 3 714.3 • At x =0 6 625.0 9 555.6 – Min σ = 277.8 psi 12 500.0 • At x =36 in 15 454.5 18 416.7 21 384.6 24 357.1 h1 h2 x h(x) 27 333.3 30 312.5 33 294.1 L 36 277.8
20.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 38 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Deflection of Tapered Bar – Consider the following tapered bar with a thickness t that is constant along the entire length of the bar. h1 dx h2 x h(x) L LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 39 Deformations of Members under ENES 220 ©Assakkaf Axial Loading Deflection of Tapered Bar – Recall Eq. 5 L Px δ =∫ dx 0 EAx – Substitute Eq. 7 – into Eq. 5, therefore L PL 1 δ= ∫ h1L + (h2 − h1 )x dx Et 0 (9)
21.
LECTURE 3. RODS:
AXIAL LOADING AND DEFORMATION (2.8) Slide No. 40 ENES 220 ©Assakkaf Deformations of Members under Axial Loading Deflection of Tapered Bar Integrating Eq. 9, the deflection of a tapered bar is given by PL 1 δ= h − h ln[(h2 − h1 )L ] (10) Et 2 1
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