Can you please use Castalianos method. thanks. Question # B: For the shown frame in the...
The rigid frame shown below is supported by Pin A and Roller C. [Point B is a rigid joint.] The frame supports a uniformly distributed load of 20 kN/m (downward) in Region BC, and a 250 kN point load (downward) located halfway between Pin A and rigid joint B. The modulus of elasticity of the entire frame is E = 200 GPa and the moment of inertia is I = 500 x 106 mm4. Determine the rotation (slope) at Joint...
A=1200mm2 Problem #4: Determine the horizontal deflection at joint C of the frame shown in the Figure including the effect of axial deformations, by the virtual work method. El- constant, E 70 GPa, l = 554(106) mmt (25 Points) 10 m 15 kN/m -75 kN- 6 m BHinge 6 m Problem #4: Determine the horizontal deflection at joint C of the frame shown in the Figure including the effect of axial deformations, by the virtual work method. El- constant, E...
Portalio Problems-Virtal Work M ork Method Problem 3 For the frame structure below: a) Use the method of Virtual Work to determine the slope and horizontal b) Use SpaceGass to determine the slope and horizontal displacement at c) Compare the results by the two methods and provide a sensible displacement of joint C joint C discussions why they are/are not equal. Take E 200 GPa and /- 200(10) mm 6 kN/m 3 m 2.4 m C 3 kN/m Portalio Problems-Virtal...
Use slope-deflection method to analyze the frame shown below. Segments AB and BD of the frame have moment of inertia I. Segment BC has moment of inertia 2/. Modulus of elasticity E is constant throughout the frame. The frame is supported by fixed-supports at A and D, and by a roller-support at C. Joint B is rigid. A downward point load of 20 kN is applied at mid-span of AB. Uniformly distributed load of intensity 2 kN/m acting downwards is...
Question 5 The members of the truss shown are made of steel and have the cross-sectional areas shown. Use the work energy method to determine the vertical deflection of joint C caused by the application of the 210 kN load. E = 200 GPa 15 m 1200 L5 m 210 kN 1800 Question 5 The members of the truss shown are made of steel and have the cross-sectional areas shown. Use the work energy method to determine the vertical deflection...
9. For the beam loaded and supported as shown in Figure (see Week 4), use the integration method to determine (a) The equation of the elastic curve using the xi and x2 coordinates (b) The slope at A. (c) The deflection at C Take E 200 GPa and1- 4 x 108 mm4 30 kN 20 kNm 4 m 2 m 9. For the beam loaded and supported as shown in Figure (see Week 4), use the integration method to determine...
structure design Name 1Pag Question 1 (35 marks) Use the method of virtual work to determine the vertical displacement of joint C of the truss below The cross-sectional area of each member is A 300 mm2 and E- 200 GPa Note 1 GPa -1.0 x 10 kN/m 1 mm2-10 x 10-*m2 3 kN 2 m 2 m 1.5 m EA Set out wour computations as follows (a) Find the reactions at hinge support E and at roller support A b)...
5. Use the virtual work method using visual integration and only bending deformations to find the indicated deflections/rotations in each structure. Calculate the deflection at point D and the rotation at point B.I-1.46-10'mm and E 200 GPa. (20pts) 80 KN Moment release 60 KN 5 m 5 m 5 m 5 m 6. Use the virtual work method include both shear and bending deformations to determine the vertical deflection and rotation at point B. E-29,000 ksi; G 11000 ksi 1...
Question 3 (30 points): Determine the smallest moment of inertia I required for the members of the frame shown, so that the horizontal deflection at joint C does not exceed 1 inch. Use the virtual work method. E 29000 ksi EI - Constant. 7k Hinge 20 ft 10 ft10 ft Question 3 (30 points): Determine the smallest moment of inertia I required for the members of the frame shown, so that the horizontal deflection at joint C does not exceed...
For the cantilever steel beam shown below, find the deflection at point B. Q5 Assume that: L 3 m, Mo 70 kN-m, E 200 GPa, I 129 x 106 mm4 and w= 15 kN/m. Mo Figure 5 [20 marks]