Explanation Given information: The modulus of elasticity of the material is E = 12 GPa . Calculation: Show the free-body diagram of the beam AD as in Figure 1. Write the singularity equation for load intensity as follows; d V d x = − w ( x ) = − 5 〈 x − 1 〉 0 Integrate the equation to find the shear force. V = A y − 4 〈 x − 0.5 〉 0 − 5 〈 x − 1 〉 1 By definition, the change in bending moment with respect to change in distance is shear force. It is expressed as follows: d M d x = V = A y − 4 〈 x − 0.5 〉 0 − 5 〈 x − 1 〉 1 Integrate the equation to find the bending moment. M = A y x − 4 〈 x − 0.5 〉 1 − 5 2 〈 x − 1 〉 2 (1) Write the second order differential equation as follows; d 2 y d x 2 = M ( x ) E I Here, the moment at the corresponding section is M ( x ) , the modulus of elasticity of the material is E , and the moment of inertia of the section is I . Substitute ( A y x − 4 〈 x − 0.5 〉 1 − 5 2 〈 x − 1 〉 2 ) for M ( x ) . d 2 y d x 2 = A y x − 4 〈 x − 0.5 〉 1 − 5 2 〈 x − 1 〉 2 E I E I d 2 y d x 2 = A y x − 4 〈 x − 0.5 〉 1 − 5 2 〈 x − 1 〉 2 Integrate the equation with respect to x ; E I d y d x = A y x 2 2 − 2 〈 x − 0.5 〉 2 − 5 6 〈 x − 1 〉 3 + C 1 (2) Integrate the Equation (2) with respect to x . E I y = A y x 3 6 − 2 3 〈 x − 0.5 〉 3 − 5 24 〈 x − 1 〉 4 + C 1 x + C 2 (3) Boundary condition 1: At the point D ; x = 2 m ; M = 0 . Substitute 2 m for x and 0 for M in Equation (1). 0 = A y ( 2 ) − 4 〈 2 − 0.5 〉 1 − 5 2 〈 2 − 1 〉 2 0 = 2 A y − 6 − 2.5 2 A y = 8.5 A y = 4.25 kN Boundary condition 2: At the point A ; x = 0 ; y = 0 . Substitute 4.25 kN for A y , 0 for x and 0 for y in Equation (3). E I ( 0 ) = 4.25 ( 0 ) 3 6 − 2 3 〈 0 − 0.5 〉 3 − 5 24 〈 0 − 1 〉 4 + C 1 ( 0 ) + C 2 0 = 0 − 0 − 0 + 0 + C 2 C 2 = 0 Boundary condition 3: At the point D ; x = 2 m ; y = 0 . Substitute 4.25 kN for A y , 2 m for x , 0 for y , and 0 for C 2 in Equation (3). E I ( 0 ) = 4.25 ( 2 ) 3 6 − 2 3 〈 2 − 0.5 〉 3 − 5 24 〈 2 − 1 〉 4 + C 1 ( 2 ) + 0 0 = 5.66667 − 2.25 − 0.20833 + 2 C 1 C 1 = − 1.60417 kN-m 2 Determine the moment of inertia ( I ) of the rectangular cross section using the equation. I = b d 3 12 Here, the width of the beam is b and the depth of the beam is d . Substitute 50 mm for b and 150 mm for d . I = 50 × 150 3 12 = 14 , 062 , 500 mm 4 × ( 1 m 1,000 mm ) 4 = 1.40625 × 10 − 5 m 4 At point A ; x = 0 ; d y d x = θ A . Substitute 12 GPa for E , 1.40625 × 10 − 5 m 4 for I , θ A for d y d x , 4.25 kN for A y , 0 for x , and − 1.60417 kN-m 2 for C 1 in Equation (2). 12 GPa × 10 6 kN/m 2 1 GPa × 1.40625 × 10 − 5 × θ A = 4.25 × 0 2 2 − 2 〈 0 − 0.5 〉 2 − 5 6 〈 0 − 1 〉 3 − 1.60417 12 × 10 6 × 1.40625 × 10 − 5 × θ A = 0 − 0 − 0 − 1.60417 θ A = − 9.51 × 10 − 3 rad At point B ; x = 0

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Chapter 9.3, Problem 59P

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Find the magnitude and location of the largest downward deflection of the beam.

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Chapter 9.3, Problem 58P

Chapter 9.3, Problem 60P

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Given answers to be: i) 14.65 kN; 6.16 kN; 8.46 kN ii) 8.63 kN; 9.88 kN iii) Bearing 6315 for B1 & B2, or Bearing 6215 for B1

(b) A steel 'hot rolled structural hollow section' column of length 5.75 m, has the cross-section shown in Figure Q.5(b) and supports a load of 750 kN. During service, it is subjected to axial compression loading where one end of the column is effectively restrained in position and direction (fixed) and the other is effectively held in position but not in direction (pinned). i) Given that the steel has a design strength of 275 MN/m², determine the load factor for the structural member based upon the BS5950 design approach using Datasheet Q.5(b). [11] ii) Determine the axial load that can be supported by the column using the Rankine-Gordon formula, given that the yield strength of the material is 280 MN/m² and the constant *a* is 1/30000. [6] 300 600 2-300 mm wide x 5 mm thick plates. Figure Q.5(b) L=5.75m Pinned Fixed

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Find the magnitude and location of the largest downward deflection of the beam.

Solution Summary: The author shows the free-body diagram of the beam AD as in Figure 1. Write the singularity equation for load intensity.

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Find the magnitude and location of the largest downward deflection of the beam.

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