Mechanics of Materials 13-1 Stress-Strain Curve for Mild Steel - - PowerPoint PPT Presentation

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13-1 Mechanics of Materials

Stress-Strain Curve for Mild Steel

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13-2a Mechanics of Materials

Definitions

• Hooke’s Law
• Shear Modulus:
• Stress:
• Strain:
• Poisson’s Ratio:
• Normal stress or strain =
• Shear stress = || to the surface

to the surface

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13-2b Mechanics of Materials

Definitions

Uniaxial Load and Deformation Thermal Deformation

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13-3a Mechanics of Materials

Stress and Strain

Thin-Walled Tanks Hoop Stress: Axial Stress:

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13-3b Mechanics of Materials

Stress and Strain

Transformation of Axes

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13-3c Mechanics of Materials

Stress and Strain

Five simplified steps to construct Mohr’s circle 1. Determine the applied stresses (σx, σy, τxy). 2. Draw a set of σ-τ axes. 3. Locate the center: 4. Find the radius (or τmax): 5. Draw Mohr’s circle. c = 1

2 ( x + y).

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13-3d1 Mechanics of Materials

Stress and Strain

For examples 1 and 2, use the following illustration. Example 1 (FEIM) The principal stresses (σ2, σ1) are most nearly (A) –62 400 kPa and 14 400 kPa (B) 84 000 kPa and 28 000 kPa (C) 70 000 kPa and 14 000 kPa (D) 112 000 kPa and –28 000 kPa

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13-3d2 Mechanics of Materials

Stress and Strain

The center of Mohr’s circle is at max = (30000 kPa)

2 +(24000 kPa) 2 = 38419 kPa

1 = c max = (24000 kPa 38419 kPa) = 62419 kPa 2 = c + max = (24000 kPa +38419 kPa) = 14418 kPa Therefore, (D) is correct. Using the Pythagorean theorem, the radius of Mohr’s circle (τmax) is: c = 1

2 ( x + y) = 1 2 (48000 kPa +0) = 24000 kPa

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13-3d3 Mechanics of Materials

Stress and Strain

Example 2 (FEIM): The maximum shear stress is most nearly (A) 24 000 kPa (B) 33 500 kPa (C) 38 400 kPa (D) 218 000 kPa Therefore, (C) is correct. In the previous example problem, the radius of Mohr’s circle (τmax) was max = (30000 kPa)

2 +(24000 kPa) 2

= 38419 kPa (38400 kPa)

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13-3e Mechanics of Materials

Stress and Strain

General Strain Note that x is no longer proportional to x.

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13-3f Mechanics of Materials

Stress and Strain

Static Loading Failure Theory Maximum Normal Stress: A material fails if

• Or
• St

Sc This is true of brittle materials. For ductile materials: Maximum Shear Distortion Energy (von Mises Stress) Tmax = max 1 2 2 , 1 3 2 , 2 3 2

• > Syt

2

• =

1 2

1 2

( )

2 + 1 3

( )

2 + 2 3

( )

2

• > Syt

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13-3g Mechanics of Materials

Stress and Strain

Torsion

• For a body with radius r being

strained to an angle φ, the shear strain and stress are: = r d dz = G = Gr d dz

• For a body with polar moment of

inertia (J), the torque (T) is: T = G d dz r

2dA A

• = GJ d

dz The shear stress is: z = Gr T GJ = Tr J

• For a body, the general angular

displacement (φ) is: = T GJ dz

L

• For a shaft of length (L), the total

angular displacement (φ) is: Torsional stiffness:

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13-3h Mechanics of Materials

Stress and Strain

Hollow, Thin-Walled Shafts

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13-4a Mechanics of Materials

Beams

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13-4b Mechanics of Materials

Beams

Load, Shear, and Moment Relations Load: Shear: For a beam deflected to a radius of curvature (ρ), the axial strain at a distance (y) from the neutral axis is x = y /.

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13-4c1 Mechanics of Materials

Beams

Shear and Bending Moment Diagrams Example 1 (FEIM): Draw the shear and bending moment diagrams for the following beam.

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13-4c2 Mechanics of Materials

Beams

Shear is undefined at concentrated force points, but just short of x = 12 m Rl +Rr = 100 N m

• 16 m

( ) = 1600 N

Rl = (8) Rr (4) = 0 Therefore, Rl = 533.3 N and Rr = 1066.7 N So the shear diagram is: From 0 m to 12 m, V = Rl 100 N m

• x = 533.3 N 100 N

m

• x; 0 m < x < 12 m

V(12

) = 533.3 N 100 N

m

• (12 m) = 666.7 N

V = 1600 N 100 N m

• x; 12 < x 16 m

From 12 m to 16 m, V =V(12

)+Rr (100 N)(x 12)

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13-4c3 Mechanics of Materials

Beams

The bending moment is the integral of the shear. M = Sdx

12

• +

Sdx

12 x

• = 800+

1600 N100 N m x

• 12
• dx

= 800 Nm+ 1600 N

( )x 50 N

m

• x

2

• 12

x

M = 533.3x – 50x2; 0 m < x < 12 m M = 800 Nm+ 1600 N

( )x 50 N

m

• x

2 1600 N

( ) 12 m ( ) + 50 N

m

• 12 m

( )

2

M = 12800 Nm+ 1600 N

( )x 50 N

m

• x

2

12 m < x 16 m Or, let the right end of the beam be x = 0 m Then, S = 100 N m

• x; 4 m < x 0 m

M = Sdx

x

• =

100 N m

• x
• x = 50 N

m

• x

2

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13-4c4 Mechanics of Materials

Beams

The bending moment diagram is:

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13-4d1 Mechanics of Materials

Beams

Example 2 (FEIM): The vertical shear for the section at the midpoint of the beam shown is (A) 0 (B) (C) P (D) none of these Drawing the force diagram and the shear diagram, Therefore, (A) is correct.

1 2 P

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13-4d2 Mechanics of Materials

Beams

Example 3 (FEIM): For the shear diagram shown, what is the maximum bending moment? The bending moment at the ends is zero, and there are no concentrated couples. (A) 8 kN • m (B) 16 kN • m (C) 18 kN • m (D) 26 kN • m Starting from the left end of the beam, areas begin to cancel after 2 m. Starting from the right end of the beam, areas begin to cancel after 4 m. The rectangle on the right has an area of 16 kN • m. The trapezoid on the left has an area of (1/2)(12 kN + 14 kN) (2 m) = 26 kN • m. The trapezoid has the largest bending moment. Therefore, (A) is correct.

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13-4e Mechanics of Materials

Beams

Bending Stress Deflection Shear Stress Note: Beam deflection formulas are given in the NCEES Handbook for any situation that might be on the exam.

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13-4f Mechanics of Materials

Beams

Example (FEIM): Find the tip deflection of the beam shown. EI is 3.47 × 106 N • m2, the load is 11 379 N/m, and the beam is 3.7 m long. From the NCEES Handbook: = woL

4

8EI = 11379 N m

• 3.7 m

( )

4

8

( ) 3.4710

6 Nm 2

( )

= 0.077 m

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13-5a Mechanics of Materials

Columns

Beam-Columns (Axially Loaded Beams) Maximum and minimum stresses in an eccentrically loaded column:

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13-5b Mechanics of Materials

Columns

Euler’s Formula Critical load that causes a long column to buckle: r = the radius of gyration k = the end-resistant coefficient kl = the effective length = slenderness ratio l r

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13-5c Mechanics of Materials

Columns

Elastic Strain Energy: Strain energy per unit volume for tension: