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WORK HARD IN SILENCE; LET SUCCESS MAKE THE NOISE

# Note for Structural Analysis-1 - SA-1 By Engineering Kings

• Structural Analysis-1 - SA-1
• Note
• Civil Engineering
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#### Note for Structural Analysis-1 - SA-1 By Engineering Kings

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FMCET S.NO 1 2 3 4 5 6 7 CE6501 – STRUCTURAL ANALYSIS FMCET 2 16 MARKS Determine the vertical displacement of joint C of the steel truss shown in fig. The cross sectional area of each member is A = 400 mm2 and E = 2*105 N/mm2. Using the principle of virtual work, determine the vertical and horizontal deflection components of joint C of the truss in fig. A = 150*10-6 m2 and E = 200*106 kN/m2 Determine the vertical and horizontal displacements of the point C of the pin-jointed frame shown in fig. The cross sectional area of AB is 100 sqmm and of AC and BC 150 mm2 each. E= 2 x 10 5 N/mm2. (By unit load method) Using the principle of least work, analyze the portal frame shown in Fig. Using the method of virtual work, determine the horizontal displacement of support D of the frame shown in fig. The values of I are indicated along the members. Take E = 200 x 106 KN/m2 and I = 300 x 10-6 m4. Using the method of virtual work, determine the horizontal displacement of support D of the frame shown in fig. The values of I are indicated along the members. Take E = 200 x 106 KN/m2 and I = 300 x 10-6 m4. Using the method of virtual work, determine the horizontal displacement of support D of the frame shown in fig. The values of I are indicated along the members. Take E = 200 x 106 KN/m2 and I = 300 x 10-6 m4. DEPARTMENT OF CIVIL ENGINEERING/ FMCET PAGE NO 11 15 18 20 23 24 26

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FMCET CE6501 – STRUCTURAL ANALYSIS FMCET 4 Where Mx = moment at a section X due to the applied loads Mx = moment at a section X due to unit load applied at the point i and in the direction of the desired dicplacement EI = flexural rigidity 11. State the Principle of Virtual work. It states that the work done on a structure by external loads is equal to the internal energy stored in a structure (Ue = Ui) Work of external loads = work of internal loads 12. What is the strain energy stored in a rod of length l and axial rigidity AE to an axial force P? Strain energy stored P2 L U= -------2AE 13. Define Virtual work. The term virtual work means the work done by a real force acting through a virtual displacement or a virtual force acting through a real displacement. The virtual work is not a real quantity but an imaginary one. 14. Explain the procedure involved in the deflection of pin jointed plane frames. 1. Virtual forces k: Remove all the real loads from the truss. Place a unit load on the truss at the joint and in the direction of the desired displacement. Use the method of joints or the method of sections and calculate the internal forces k in each member of the truss. 2. Real forces F: These forces arre caused only by the real loads acting on the truss. Use the method of sections or the method of joints to determine the forces F in each member. 3. Virtual work equation: Apply the equation of virtual work, to determine the desired displacement. 15. In the truss shown in fig. no load acts. The member AB gets 4mm too short. The cross sectional area of each member is A = 300 mm2 and E = 200 GPa. Determine the vertical displacement of joint C. DEPARTMENT OF CIVIL ENGINEERING/ FMCET

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FMCET CE6501 – STRUCTURAL ANALYSIS FMCET 5 Solution: Virtual forces, k: Since the vertical displacement of joint C is required, a vertical force of 1 kN is applied at C. The force k in each member is determined as below: By symmetry, RA = RB = ½ Joint A: ∑V = 0 gives KAC cos 36º 52’ + ½ = 0 KAC = 0.625 kN (comp) ∑H = 0 gives kAB + kAC cos 53º 08’ = 0 kAB = 0.375 kN (tensile) Joint B: ∑V = 0 gives KBC cos 36º 52’ + ½ = 0 KBC = 0.625 kN (comp) Member AB undergoes a deformation, ∆L = 0.004 m ∆ = ∑(k.∆L) (∆C)V = (0.375) (-0.004) + 0 +0 = -0.0015m = -1.5 mm The negative sign indicates that joint C displaced upward, i.e. opposite to the 1 kN vertical load. 16. Using the method of virtual work, determine the vertical displacement of point B of the beam shown in fig. Take E = 2x 105 MPa and I = 825x 107 mm4. Solution: l mMdx EI 0 Virtual moment, m. Remove the external load. Apply a unit vertical load at B. Consider a section XX at a distance x from B. m = -1 * x (Hogging moment) Real moment, M. Using the same x co-ordinate, the internal moment (due to the given loading) M is formulated as, M = -25. *.x/2 M = -12.5x2 ∆= DEPARTMENT OF CIVIL ENGINEERING/ FMCET