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) = crosssection 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 = changeoflength / 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/nonprismatic, 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 *StressStrain 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., longitudinallyextended 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) TimeIndependent Behavior TimeDependent 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(2x104) = 36×104 in (to right Consider part BC:
ε1(x) = 3.57x104 (i.e., constant in BC) Apply Eq. (4b) with,
xo = xB = 18”
and,
u(xo) = uB = 36x104 in
x = xC = 36”
and,
u(x) = uC (required)
36 ∴ uC = uB + ∫ ε 4 1 4 dx = 36x10 + 3.57×10 (3618) 18 = 1×102 in (to right ) (iii) Change in diameter
Let,
∆d1 = change in diameter d1 of segment BC
use Eq. (5b): ∴ ε1Lat = υ ε1 = 0.3(3.57x104); 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)BoltedConnection Dr. A. F. Saleeb (ii) PunchPlate 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) xface of element
q m (p n) = positive (negative) yface 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 shockmount 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. II 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 barbolt
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 • Boltshear Failure Mode. • Combined Platebolt
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 ...
View
Full
Document
 Spring '11
 oldham
 Shear, Strain, Stress

Click to edit the document details