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

Mechanics of Materials 13-1 Stress-Strain Curve for Mild Steel Professional Publications, Inc. FERC

Mechanics of Materials 13-2a Definitions • Hooke’s Law • Shear Modulus: • Stress: • Strain: • Poisson’s Ratio: • Normal stress or strain = � to the surface • Shear stress = || to the surface Professional Publications, Inc. FERC

Mechanics of Materials 13-2b Definitions Uniaxial Load and Deformation Thermal Deformation Professional Publications, Inc. FERC

Mechanics of Materials 13-3a Stress and Strain Thin-Walled Tanks Hoop Stress: Axial Stress: Professional Publications, Inc. FERC

Mechanics of Materials 13-3b Stress and Strain Transformation of Axes Professional Publications, Inc. FERC

Mechanics of Materials 13-3c Stress and Strain Five simplified steps to construct Mohr’s circle 1. Determine the applied stresses ( σ x , σ y , τ xy ). 2. Draw a set of σ - τ axes. 2 ( � x + � y ). 3. Locate the center: � c = 1 4. Find the radius (or τ max ): 5. Draw Mohr’s circle. Professional Publications, Inc. FERC

Mechanics of Materials 13-3d1 Stress and Strain For examples 1 and 2, use the following illustration. Example 1 (FEIM) The principal stresses ( σ 2 , σ 1 ) 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 Professional Publications, Inc. FERC

Mechanics of Materials 13-3d2 Stress and Strain The center of Mohr’s circle is at 2 ( � x + � y ) = 1 2 ( � 48000 kPa + 0) = � 24000 kPa � c = 1 Using the Pythagorean theorem, the radius of Mohr’s circle ( τ max ) is: 2 + (24000 kPa) 2 = 38419 kPa (30000 kPa) � max = � 1 = � c � � max = ( � 24000 kPa � 38419 kPa) = � 62419 kPa � 2 = � c + � max = ( � 24000 kPa + 38419 kPa) = 14418 kPa Therefore, (D) is correct. Professional Publications, Inc. FERC

Mechanics of Materials 13-3d3 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 In the previous example problem, the radius of Mohr’s circle ( τ max ) was 2 + (24000 kPa) (30000 kPa) 2 � max = = 38419 kPa (38400 kPa) Therefore, (C) is correct. Professional Publications, Inc. FERC

Mechanics of Materials 13-3e Stress and Strain General Strain Note that � x is no longer proportional to � x . Professional Publications, Inc. FERC

Mechanics of Materials 13-3f Stress and Strain Static Loading Failure Theory Maximum Normal Stress: A material fails if � � S t • Or � � S c • This is true of brittle materials. For ductile materials: Maximum Shear � � � 1 � � 2 � 1 � � 3 � 2 � � 3 � > S yt T max = max , , � � � 2 2 2 2 � � Distortion Energy (von Mises Stress) 2 + � 1 � � 3 2 + � 2 � � 3 � 2 � ( ) ( ) ( ) � > S yt � � = � 1 � 1 � � 2 2 � � Professional Publications, Inc. FERC

Mechanics of Materials 13-3g Stress and Strain Torsion • For a shaft of length ( L ), the total • For a body with radius r being angular displacement ( φ ) is: strained to an angle φ , the shear strain and stress are: � = r d � � = G � = Gr d � dz dz Torsional stiffness: • For a body with polar moment of inertia ( J ), the torque ( T ) is: T = G d � = GJ d � r 2 dA � dz dz A The shear stress is: � � z = Gr T GJ = Tr J • For a body, the general angular displacement ( φ ) is: T L GJ dz � � = 0 Professional Publications, Inc. FERC

Mechanics of Materials 13-3h Stress and Strain Hollow, Thin-Walled Shafts Professional Publications, Inc. FERC

Mechanics of Materials 13-4a Beams Professional Publications, Inc. FERC

Mechanics of Materials 13-4b Beams Load, Shear, and Moment Relations Load: Shear: For a beam deflected to a radius of curvature ( ρ ), the axial strain at a � x = � y / � . distance ( y ) from the neutral axis is Professional Publications, Inc. FERC

Mechanics of Materials 13-4c1 Beams Shear and Bending Moment Diagrams Example 1 (FEIM): Draw the shear and bending moment diagrams for the following beam. Professional Publications, Inc. FERC

Mechanics of Materials 13-4c2 Beams � � R l + R r = 100 N � 16 m ( ) = 1600 N � m � � R l = (8) � R r (4) = 0 Therefore, R l = 533.3 N and R r = 1066.7 N � � � � From 0 m to 12 m, V = R l � 100 N x = 533.3 N � 100 N x ; 0 m < x < 12 m � � � � m m � � � � Shear is undefined at concentrated force points, but just short of x = 12 m � � � ) = 533.3 N � 100 N V (12 (12 m) = � 666.7 N � � m � � From 12 m to 16 m, V = V (12 � ) + R r � (100 N)( x � 12) V = 1600 N � 100 N � � x ; 12 < x � 16 m � � m � � So the shear diagram is: Professional Publications, Inc. FERC

Mechanics of Materials 13-4c3 Beams The bending moment is the integral of the shear. M = 533.3 x – 50 x 2 ; 0 m < x < 12 m � 1600 N � 100 N � 12 x 12 M = Sdx Sdx = � 800 + m x dx � � � + � � 0 12 0 � � x � � ) x � 50 N � � ( = � 800 N � m + 1600 N x 2 � � � � m � � � � 12 ) x � 50 N � � ) + 50 N � � 2 2 � 1600 N ( ( ) 12 m ( ( ) M = � 800 N � m + 1600 N x � 12 m � � � m m � � � � ) x � 50 N � � ( M = � 12800 N � m + 1600 N x 2 � � m � � 12 m < x � 16 m Or, let the right end of the beam be x = 0 m � � Then, S = � 100 N x ; � 4 m < x � 0 m � � m � � � 100 N � � x = � 50 N � � 0 0 M = Sdx x 2 � � = � � � � m m x x � � � � Professional Publications, Inc. FERC

Mechanics of Materials 13-4c4 Beams The bending moment diagram is: Professional Publications, Inc. FERC

Mechanics of Materials 13-4d1 Beams Example 2 (FEIM): The vertical shear for the section at the midpoint of the beam shown is (A) 0 2 P (B) 1 (C) P (D) none of these Drawing the force diagram and the shear diagram, Therefore, (A) is correct. Professional Publications, Inc. FERC

Mechanics of Materials 13-4d2 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. Professional Publications, Inc. FERC

Mechanics of Materials 13-4e Beams Bending Stress Shear Stress Deflection Note: Beam deflection formulas are given in the NCEES Handbook for any situation that might be on the exam. Professional Publications, Inc. FERC

Mechanics of Materials 13-4f Beams Example (FEIM): Find the tip deflection of the beam shown. E I is 3.47 × 106 N • m 2 , the load is 11 379 N/m, and the beam is 3.7 m long. From the NCEES Handbook: � � 11379 N 4 � 3.7 m ( ) � m � = w o L 4 � � = 0.077 m 8 E I = ( ) 3.47 � 10 8 ( 6 N � m 2 ) Professional Publications, Inc. FERC

Mechanics of Materials 13-5a Columns Beam-Columns (Axially Loaded Beams) Maximum and minimum stresses in an eccentrically loaded column: Professional Publications, Inc. FERC

Mechanics of Materials 13-5b Columns Euler’s Formula Critical load that causes a long column to buckle: r = the radius of gyration k = the end-resistant coefficient k l = the effective length l = slenderness ratio r Professional Publications, Inc. FERC

Mechanics of Materials 13-5c Columns Elastic Strain Energy: Strain energy per unit volume for tension: Professional Publications, Inc. FERC