Problem Number (1)
A 3-mm thick hollow polystyrene cylinder E = 3GPa and
a rigid circular plate (only part of which is sho...
Problem Number (2)
Two solid cylindrical rods are joined at B and loaded as
shown. Rod AB is made of steel E = 200GPa and ...
Both portions of the rod ABC are made of an aluminum
for which E = 70 GPa. Knowing that the magnitude of P is
4KN, determi...
The rod ABC is made of an aluminum for which E =
70GPa. Knowing that P = 6KN and Q = 42 KN, determine the
deflection of (a...
Each of the links AB and CD is made of steel
(E = 200GPa) and has a uniform rectangular cross section of
6 * 24 mm. Determ...
∴ Ξ” 𝐸 = (2.5)(βˆ’1.736𝑃)(10)βˆ’8
βˆ’ (1.5)(1.0416𝑃)(10)βˆ’8
= βˆ’2.7776𝑃(10)βˆ’8
π‘š
For maximum deflection |Ξ” 𝐸 | = 0.25π‘šπ‘š
∴ 2.7776𝑃(10...
Problem Number (6)
The length of the 2-mm diameter steel wire CD has been
adjusted so that with no load applied, a gap of ...
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Deflection and member deformation

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Transcript of "Deflection and member deformation"

  1. 1. Problem Number (1) A 3-mm thick hollow polystyrene cylinder E = 3GPa 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 3.2KB 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. Solution: βˆ†πΏ = 𝐹 Γ— 𝐿 𝐴 Γ— 𝐸 = 3200 Γ— 0.25 3.14 Γ— 9 Γ— 10βˆ’6 Γ— 200 Γ— 109 = 1.4 Γ— 10βˆ’4 π‘š Deflection of B = 3200 Γ— 0.03 3.14((25)2 βˆ’ (22)2) Γ— 10βˆ’6 Γ— 3 Γ— 109 + 1.4 Γ— 10βˆ’4 = 0.214 π‘šπ‘š 𝜏 = 𝐹 𝐴 = 3200 3.14 Γ—9 Γ— 10βˆ’6 = 113.2 π‘€π‘ƒπ‘Ž
  2. 2. Problem Number (2) Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel E = 200GPa and rod BC of brass E = 105GPa. Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B. Solution: Assume that the force 40KN is directed to downward at point B βˆ†πΏ = 30 Γ— 103 Γ—0.25 3.14 Γ—15 Γ—15 Γ— 10βˆ’6 Γ—200 Γ— 109 + 70 Γ— 103 Γ—0.3 3.14 Γ—25 Γ—25 Γ— 10βˆ’6 Γ—105 Γ— 109 = 0.393 π‘šπ‘š Deflection of Point B = 70 Γ— 103 Γ—0.3 3.14 Γ—25 Γ—25 Γ— 10βˆ’6 Γ—105 Γ— 109 = 0.102 mm Problem Number (3)
  3. 3. Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4KN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B. Solution: βˆ†πΏ 𝐡𝐢 = βˆ†πΏ 𝐴𝐡 (𝑄 βˆ’ 4000) Γ— 0.5 3.14 Γ— 0.03 Γ— 0.03 Γ— 70 Γ— 109 = 4000 Γ— 0.4 3.14 Γ— 0.01 Γ— 0.01 Γ— 70 Γ— 109 Then, Q = 32800 N Then, Deflection of B = (32800βˆ’4000) Γ—0.5 3.14 Γ—0.03 Γ—0.03 Γ—70 Γ— 109 = 0.0728 mm Problem Number (4)
  4. 4. The rod ABC is made of an aluminum for which E = 70GPa. Knowing that P = 6KN and Q = 42 KN, determine the deflection of (a) point A, (b) point B. Solution: Deflection of A = βˆ†πΏ 𝐴𝐡 βˆ’ βˆ†πΏ 𝐡𝐢 = 6000 Γ—0.4 3.14 Γ— 0.01 Γ—0.01 Γ—70 Γ— 109 βˆ’ (42000βˆ’6000) Γ—0.5 3.14 Γ—0.03 Γ—0.03 Γ—70 Γ— 109 = 0.01819 π‘šπ‘š Deflection of B = (42000βˆ’6000)Γ—0.5 3.14 Γ—0.03 Γ—0.03 Γ—70 Γ— 109 = 0.091 π‘šπ‘š Problem Number (5)
  5. 5. Each of the links AB and CD is made of steel (E = 200GPa) and has a uniform rectangular cross section of 6 * 24 mm. Determine the largest load which can be suspended from point E if the deflection of E is not to exceed 0.25 mm. Solution: βˆ‘MB = P(375 + 250) – FDC (250) = 0 ∴ 𝐹 𝐷𝐢 = 2.5𝑃 (π‘‡π‘’π‘›π‘ π‘–π‘œπ‘›) βˆ‘Fy = FDC – FBA – P = 0 ∴ 𝐹 𝐡𝐴 = 1.5𝑃 (π‘‡π‘’π‘›π‘ π‘–π‘œπ‘›) ∴ Ξ”CD = 𝐹 𝐷𝐢 𝐿 𝐷𝐢 𝐸 𝐷𝐢 𝐴 𝐷𝐢 = 2.5𝑃 (200)(10)βˆ’3 (200)(10)9(6)(24)(10)βˆ’6 = 1.736𝑃 (10)βˆ’8 π‘š (π·π‘œπ‘€π‘›π‘€π‘Žπ‘Ÿπ‘‘) ∴ Ξ”BA = 𝐹 𝐡𝐴 𝐿 𝐴𝐡 𝐸𝐴𝐡 𝐴 𝐴𝐡 = 1.5𝑃 (200)(10)βˆ’3 (200)(10)9(6)(24)(10)βˆ’6 = 1.0416𝑃 (10)βˆ’8 π‘š (π‘ˆπ‘π‘€π‘Žπ‘Ÿπ‘‘) From geometry of the deflected structure: ∴ Ξ” 𝐸 = ( 250 + 375 250 ) Ξ”C βˆ’ ( 375 250 )Ξ”B
  6. 6. ∴ Ξ” 𝐸 = (2.5)(βˆ’1.736𝑃)(10)βˆ’8 βˆ’ (1.5)(1.0416𝑃)(10)βˆ’8 = βˆ’2.7776𝑃(10)βˆ’8 π‘š For maximum deflection |Ξ” 𝐸 | = 0.25π‘šπ‘š ∴ 2.7776𝑃(10)βˆ’8 = 0.25(10)βˆ’3 ∴P)max = 9.57 KN
  7. 7. Problem Number (6) The length of the 2-mm diameter steel wire CD has been adjusted so that with no load applied, a gap of 1.5mm 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. Solution:

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