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.
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For the second frame, the moment distribution method is applied. Stiffness factors are calculated and distribution factors are used to solve for the unknown end moments. The bending moment diagram is drawn based on the calculated member end moments.
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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: