Draw shear force diagrams for the overhanging beam, Mechanical Engineering

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Draw shear force diagrams for the overhanging beam:

Draw shear force & bending moment diagrams for the overhanging beam illustrated in Figure

673_Draw shear force diagrams for the overhanging beam.png

Figure

 Solution

Taking moments around A & equating to zero,

 

RB   × 6 - (6 × 8) - 0.8 × 3 × ( 3 +(3/2))  - (5 × 2) = 0

RB = 11.47 kN and RA = 5 + (0.8 × 3) + 6 - 11.47 = 25 - 11.5 = 1.93 kN

Shear Force (beginning from the left end A)

SF at A, FA = + 1.93 kN

SF just left of D, FD = + 1.93 kN

SF just right of D, FD = + 1.93 - 5 = - 3.07 kN

SF at E, FE = - 3.07 kN

SF just left of B, FB = - 3.07 - (0.8 × 3) = - 5.47 kN

 SF just right of B, FB = - 5.47 + 11.47 = + 6 kN

SF just left of C, FC = + 6 kN = Load at E

Bending Moment

BM at C, MC  = 0

BM at B, MB = - 6 × 2 = - 12 kN m

 BM at E, M E =+ (11.46 × 3) - (6 × 5) -( 0.8 × 3 × (3/2) )=+ 0.81 kN m

BM at D, MD = + 1.93 × 2 = + 3.86 kN m

BM at A, MA = 0

Maximum Bending Moment

Maximum positive bending moment take place at D and maximum negative bending moment take place at B.

Mmax (positive) = + 3.86 kN m

Mmax (negative) = - 12 kN m

Point of Contraflexure

As the bending moment changes sign among E and B, let a section XX among E and B at a distance x from C.

BM at section XX,

Mx = - 6x + 11.47 (x - 2) - 0.8 (x - 2) ((x - 2)/2)

= - 6x + 11.47x - 22.94 - 0.4 (x - 2)2

= - 0.4x2 + 7.07x - 24.54

By equating this equation to zero, we achieved the point of contraflexure,

- 0.4x2 + 7.07x - 24.54 = 0

x2 - 17.67x + 61.35 = 0

Solving out by trial and error,

x = 4    Value of (x2 - 17.675x + 61.35) = 6.65

x = 4.9 Value of (x2 - 17.675x + 61.35) = - 1.2475

x = 4.8  Value of (x2 - 17.675x + 61.35) = - 0.45

x = 4.7 Value of (x2 - 17.675x + 61.35) = 0.3675

 x = 4.75 Value of (x2 - 17.675x + 61.35) = - 0.04375        

x = 4.745 Value of (x2 - 17.675x + 61.35) = - 0.00285

Point of contraflexure is at distance of 4.745 m from the end of C.


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