Theory of Structures II Lecture Notes PDF

Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 Example (2): Find the maximum deflection at free end of the beam shown below. Use the virtual work m...

Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 Example (2): Find the maximum deflection at free end of the beam shown below. Use the virtual work method. Take 𝐸 = 200 𝐺𝑃𝑎 and 𝐼 = 500 ∗ 106 𝑚𝑚4 Solution: Real Moment 𝑥 𝑥2 𝑀 = −𝑤. 𝑥. = −(12) = −6𝑥 2 2 2 Virtual Moment 𝑚 = −1. 𝑥 = −𝑥 Virtual work equation 𝐿 𝑀. 𝑚 1. ∆𝐵 = ∫ 𝑑𝑥 𝐸𝐼 0 10 (−6𝑥 2 )(−𝑥) ∆𝐵 = ∫ 𝑑𝑥 𝐸𝐼 0 10 1 6𝑥 4 ∆𝐵 = [ ] 𝐸𝐼 4 0 15 ∗ 103 ∆𝐵 = 𝐸𝐼 1  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 15 ∗ 103 ∆𝐵 = = 0.15 𝑚 200 ∗ 106 ∗ 500 ∗ 106 ∗ 10−12 ∆𝐵 = 150 𝑚𝑚 Example (3): Determine the slope at point B of the beam shown in Figure below. Take 𝐸 = 200 𝐺𝑃𝑎 and 𝐼 = 60 ∗ 106 𝑚𝑚4 Solution: Real Moment For segment AB (𝟎 ≤ 𝒙𝟏 ≤ 𝟓 𝒎) 𝑀1 = −3𝑥1 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟓 𝒎) 𝑀2 = −3 (5 + 𝑥2 ) Virtual Moment For segment AB (𝟎 ≤ 𝒙𝟏 ≤ 𝟓 𝒎) 𝑚𝜃1 = 0 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟓 𝒎) 𝑚𝜃2 = 1 Virtual Work Equation 2  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 𝐿 𝑀. 𝑚𝜃 1. 𝜃𝐵 = ∫ 𝑑𝑥 𝐸𝐼 0 5 5 (0)(−3 𝑥1 ) (1)(−3)(5 + 𝑥2 ) 𝜃𝐵 = ∫ 𝑑𝑥1 + ∫ 𝑑𝑥2 𝐸𝐼 𝐸𝐼 0 0 −112.5 𝜃𝐵 = 𝐸𝐼 −112.5 𝜃𝐵 = 200 ∗ 106 ∗ 60 ∗ 106 ∗ 10−12 𝜃𝐵 = −0.00938 𝑟𝑎𝑑 Example (4): Determine the displacement at point D of the beam shown in Figure below. Take 𝐸 = 200 𝐺𝑃𝑎 and 𝐼 = 300 ∗ 106 𝑚𝑚4 Solution: Real Moment ∑ 𝑴𝑪 = 𝟎 → 𝟔 𝑹𝒂 + 𝟏𝟐𝟎 − 𝟑𝟎 ∗ 𝟑 = 𝟎 → 𝑹𝒂 = −𝟓 𝒌𝑵 3  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 ∑ 𝑭𝒚 = 𝟎 → 𝑹𝒄 = 𝟑𝟓 𝒌𝑵 For segment CD (𝟎 ≤ 𝒙𝟏 ≤ 𝟒. 𝟓 𝒎) 𝑀1 = 0 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟑 𝒎) 𝑀2 = 𝑅𝑐. 𝑥2 = 35 𝑥2 For segment AB (𝟎 ≤ 𝒙𝟑 ≤ 𝟑 𝒎) 𝑀3 = 𝑅𝑎. 𝑥3 + 120 = 120 − 5 𝑥3 Virtual Moment ∑ 𝑴𝑪 = 𝟎 → 𝟔 𝑹𝒂 + 𝟏(𝟒. 𝟓) = 𝟎 → 𝑹𝒂 = −𝟎. 𝟕𝟓 𝒌𝑵 ∑ 𝑭𝒚 = 𝟎 → 𝑹𝒄 = 𝟏. 𝟕𝟓 𝒌𝑵 For segment CD (𝟎 ≤ 𝒙𝟏 ≤ 𝟒. 𝟓 𝒎) 𝑚1 = −1. 𝑥1 = − 𝑥1 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟑 𝒎) 𝑚2 = −1(4.5 + 𝑥2 ) + 𝑅𝑐. 𝑥2 = −4.5 + 0.75𝑥2 = 0.75𝑥2 − 4.5 For segment AB (𝟎 ≤ 𝒙𝟑 ≤ 𝟑 𝒎) 𝑚3 = 𝑅𝑎. 𝑥3 = −0.75 𝑥3 4  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 Virtual Work Equation 𝐿 𝑀. 𝑚 1. ∆𝐷 = ∫ 𝑑𝑥 𝐸𝐼 0 4.5 3 3 𝑀1. 𝑚1 𝑀2. 𝑚2 𝑀3. 𝑚3 ∆𝐷 = ∫ 𝑑𝑥1 + ∫ 𝑑𝑥2 + ∫ 𝑑𝑥3 𝐸𝐼 𝐸𝐼 𝐸𝐼 0 0 0 4.5 3 3 0 (𝑥1 ) (35 𝑥2 ). (0.75 𝑥2 − 4.5) (120 − 5 𝑥3) (−0.75𝑥3 ) ∆𝐷 = ∫ 𝑑𝑥1 + ∫ 𝑑𝑥2 + ∫ 𝑑𝑥3 𝐸𝐼 𝐸𝐼 𝐸𝐼 0 0 0 0 472.5 371.25 843.75 ∆𝐷 = − − =− 𝐸𝐼 𝐸𝐼 𝐸𝐼 𝐸𝐼 −843.75 ∆𝐷 = = −0.0141 𝑚 200 ∗ 106 ∗ 300 ∗ 106 ∗ 10−12 ∆𝐷 = −14.1 𝑚𝑚 Example (5): Determine the horizontal displacement of point C on the frame shown in Figure below. Take 𝐸 = 200 𝐺𝑃𝑎 and 𝐼 = 235 ∗ 106 𝑚𝑚4 Solution: 5  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 Real Moment ∑ 𝑴𝑨 = 𝟎 → 𝟐. 𝟒 𝑹𝒄 − 𝟔𝟎 ∗ 𝟑 ∗ 𝟏. 𝟓 = 𝟎 → 𝑹𝒄 = 𝟏𝟏𝟐. 𝟓 𝒌𝑵 ∑ 𝑭𝒚 = 𝟎 → 𝑹𝒂 = −𝟏𝟏𝟐. 𝟓 𝒌𝑵 ∑ 𝑭𝒙 = 𝟎 → 𝑯𝒂 = 𝟔𝟎 ∗ 𝟑 = 𝟏𝟖𝟎 𝒌𝑵 For segment AB (𝟎 ≤ 𝒙𝟏 ≤ 𝟑 𝒎) 𝑥1 𝑀1 = 𝐻𝑎. 𝑥1 − 60 ∗ 𝑥1 ∗ = 180𝑥1 − 30𝑥1 2 2 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟐. 𝟒 𝒎) 𝑀2 = 𝑅𝑐. 𝑥2 = 112.5 𝑥2 Virtual Moment 6  Theory of Structures II Prof. Dr. Abdulkhaliq A. Jaafer Lecture 1 Third Stage-University of Misan Lecture 2 ∑ 𝑴𝑨 = 𝟎 → 𝟐. 𝟒 𝑹𝒄 − 𝟏 ∗ 𝟑 = 𝟎 → 𝑹𝒄 = 𝟏. 𝟐𝟓 𝒌𝑵 ∑ 𝑭𝒚 = 𝟎 → 𝑹𝒂 = −𝟏. 𝟐𝟓 𝒌𝑵 ∑ 𝑭𝒙 = 𝟎 → 𝑯𝒂 = 𝟏 𝒌𝑵 For segment AB (𝟎 ≤ 𝒙𝟏 ≤ 𝟑 𝒎) 𝑚1 = 𝐻𝑎. 𝑥1 = 𝑥1 For segment BC (𝟎 ≤ 𝒙𝟐 ≤ 𝟐. 𝟒 𝒎) 𝑚2 = 𝑅𝑐. 𝑥2 = 1.25 𝑥2 Virtual Work Equation 𝐿 𝑀. 𝑚 1. ∆𝐶ℎ = ∫ 𝑑𝑥 𝐸𝐼 0 3 2.4 (180𝑥1 − 30𝑥1 2 )( 𝑥1 ) (112.5 𝑥2 )(1.25 𝑥2 ) ∆𝐶ℎ = ∫ 𝑑𝑥1 + ∫ 𝑑𝑥2 𝐸𝐼 𝐸𝐼 0 0 1012.5 648 1660.5 ∆𝐶ℎ = + = 𝐸𝐼 𝐸𝐼 𝐸𝐼 1660.5 ∆𝐶ℎ = = 0.0353 𝑚 200 ∗ 106 ∗ 235 ∗ 106 ∗ 10−12 ∆𝐶ℎ = 35.3 mm 7