Stress&strain part 2
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The document describes solutions to multiple problems involving calculating stress, strain, deflection, and elongation of rods and beams under various loads. The problems involve determining: (1) the elongation of a steel rod supported by a polystyrene cylinder and plate under a 3.2 kN load, (2) the total deformation of a composite steel and brass rod under separate loads, (3) the necessary force Q to make the deflection at the end of an aluminum rod zero under a 4 kN load P.
Stress&strain part 2
- 1. Prob. 2.15: A 3-mm thick hollow polystrene cylinder (𝑬 = 𝟑 𝑮𝑷𝒂) and a rigid circular plate (only part of which is shown) are used to support a 250-mm-long steel rod 𝐀𝐁 (𝑬 = 𝟐𝟎𝟎 𝑮𝑷𝒂) of 6-mm diameter. If a 𝟑. 𝟐 𝑲𝑵 load P is applied at B, determine (a)the elongation of rod AB, (b) the deflection of point B, (c) the average normal stress in rod AB. 𝑺𝒕𝒓𝒆𝒔𝒔 & 𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠 𝑷𝒂𝒓𝒕 𝟐
- 2. Solution: ΔL = 𝑭𝑳 𝑬𝑨 For the rod AB: ΔL1 = 𝟑𝟐𝟎𝟎×𝟎.𝟐𝟓 𝟐𝟎𝟎∗𝟏𝟎𝟗∗𝜋(0.003)2 = 1.414*𝟏𝟎−𝟒 m The elongation of rod AB = 0.1414 mm For the cylinder: The part of which thickness is 3 mm will be affected by the stress resulting from the force F=3.2 KN The area of this part (A) = 𝝅((𝟎. 𝟎𝟐𝟓)𝟐 − (𝟎. 𝟎𝟐𝟐)𝟐 A = 4.429*𝟏𝟎−𝟒 𝒎𝟐 ΔL2 = −𝟑𝟐𝟎𝟎∗𝟎.𝟎𝟑 𝟑∗𝟏𝟎𝟗∗𝟒.𝟒𝟐𝟗∗𝟏𝟎−𝟒 = -7.225*𝟏𝟎−𝟓 m UB = ΔL1 – ΔL2 = 1.414*𝟏𝟎−𝟒 - (-7.225*𝟏𝟎−𝟓 ) UB = 2.14 mm Normal stress for rod AB: σ = 𝑭 𝑨 = 𝟑𝟐𝟎𝟎 𝝅(𝟎.𝟎𝟎𝟑)𝟐 = 𝟏𝟏𝟑𝟏𝟕𝟔𝟖𝟒𝟖. 𝟒 𝑵 𝒎𝟐
- 3. Prob. 2.17: Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel (𝐄 = 𝟐𝟎𝟎 𝐆𝐏𝐚) and rod BC of brass (𝐄 = 𝟏𝟎𝟓 𝐆𝐏𝐚). Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B. Solution: UB = ΔL1 = −(𝟑𝟎𝟎𝟎𝟎+𝟒𝟎𝟎𝟎𝟎)𝑳 𝑬𝑨 = − 𝟑𝟎𝟎𝟎𝟎+𝟒𝟎𝟎𝟎𝟎 (𝟎.𝟑) 𝟐𝟎𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏𝟓)𝟐 UB = -1.485*𝟏𝟎−𝟒 m ΔL2 = −(𝟑𝟎𝟎𝟎𝟎)𝑳 𝑬𝑨 = − 𝟑𝟎𝟎𝟎𝟎 (𝟎.𝟐𝟓) 𝟏𝟎𝟓∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟐𝟓)𝟐 = −𝟑. 𝟔𝟑𝟕 ∗ 𝟏𝟎−𝟓 𝒎 Total deflection of rod ABC = ΔL1 + ΔL2 = 1.8487*𝟏𝟎−𝟒 𝒎
- 4. Prob. 2.19: Both portions of the rod ABC are made of an aluminum for which 𝐄 = 𝟕𝟎 𝐏𝐚 . Knowing that the magnitude of P is 4 KN, determine (a) the value of 𝐐 so that the deflection at A is zero, (b) the corresponding deflection of B. Solution: We will call the portion BC: part (1) And the portion AB: part (2) ΔL = 𝑭𝑳 𝑬𝑨 For part (1): F = Q-P = Q-4000 ΔL1 = − 𝑸−𝟒𝟎𝟎𝟎 (𝟎.𝟓) 𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐
- 5. For part (2): ΔL2 = 𝟒𝟎𝟎𝟎 (𝟎.𝟒) 𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏)𝟐 Since the deflection at A = 0, then ΔL1 = ΔL2 − 𝑸−𝟒𝟎𝟎𝟎 (𝟎.𝟓) 𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐 = 𝟒𝟎𝟎𝟎 (𝟎.𝟒) 𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏)𝟐 (-Q+4000) (0.5) = (4000) (0.4) (9) -0.5Q+2000 = 14400 0.5Q = - 16400 Q = - 32800 N Q = 32.8 KN Prob. 2.20: The rod ABC is made of an aluminum for which 𝐄 = 𝟕𝟎 𝐆𝐏𝐚. Knowing that 𝐏 = 𝟔 𝐊𝐍 and 𝐐 = 𝟒𝟐 𝐊𝐍, determine the deflection of (a) point A, (b) point B. (𝑺𝒆𝒆 𝒕𝒉𝒆 𝒑𝒓𝒆𝒗𝒊𝒐𝒖𝒔 𝒇𝒊𝒈𝒖𝒓𝒆) Solution: We will call the portion BC: part (1) And the portion AB: part (2) ΔL = 𝑭𝑳 𝑬𝑨 For part (1): F = Q-P = -42+6 -36 KN
- 6. ΔL1 = − 𝟑𝟔𝟎𝟎𝟎 (𝟎.𝟓) 𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐 = UB = 9.09𝟏𝟎−𝟓 𝒎 ΔL2 = 𝟔𝟎𝟎𝟎 𝟎.𝟒 𝟕𝟎∗𝟏𝟎𝟗∗𝝅 𝟎.𝟎𝟏 𝟐 = 1.09*𝟏𝟎−𝟒 𝒎 UA = ΔL1 + ΔL2 = 9.09𝟏𝟎−𝟓 + 1.09*𝟏𝟎−𝟒 = 1.81*𝟏𝟎−𝟓 𝒎 Prob. 2.28: The length of the 2-mm-diameter steel wire CD has been adjusted so that with no load applied, a gap of 1.5 mm exists between the end B of the rigid beam ACB and a contact point E. Knowing that 𝐄 = 𝟐𝟎𝟎 𝐆𝐏𝐚, determine where a 20-kg block should be placed on the beam in order to cause contact between B and E.
- 7. Solution: