CALCULATION OF PHYSICS

SOLUTION

 

1.

 

The beam in question is 410UB59.7. Thus it is a beam of length of 5 m as mentioned in the figure of the question. The weight of the beam is 585.06N(=59.7).

Total force acting on the beam in vertically downward direction=.

Maximum Vertical Shear =27925.3N.

The reaction forces, R1=R2= (25585.06/2)N=13962.65N.

Area of the cross-section=7640=7.64

Shear Stress=== Kpa=3.65Mpa.

Maximum bending moment= (13962.65) Nm=34906.625Nm in the direction as shown in the figure above.

Bending stress=Mpa

Deflection===0.09125mm

The Forces F shown in the above diagram would be equal to the reaction forces(we have cut the beam hypothetically to show the internal forces and the moment).

2.The beam in question is 400 Oregon. Thus, it is a beam of length of 5 m as mentioned in the figure of the question.

Mass of the timber=550 Kg=110 kg

The weight of the beam is 1078N(=110).

Total force acting on the beam in vertically downward direction=.

Maximum Vertical Shear =26078 N.

The reaction forces, R1=R2= (26078/2)N=13039 N.

Area of the cross-section= (400=0.04

Shear Stress=== 651950 pa=0.652 Mpa.

Maximum bending moment= ( ) Nm=32597.5 Nm in the direction as shown in the figure above.

Bending stress=Mpa

Deflection===0.472 mm

3. The beam in question is 180UB18.1. Thus, it is a beam of length of 6 m as mentioned in the figure of the question.

Mass of the timber=18.1 Kg=108.6 kg

The weight of the beam is 1064.28N(=108.6).

Total force acting on the beam in vertically downward direction=.

Maximum Vertical Shear =16064.28 N.

The reaction forces, R1=R2= (16064.28/2) N=8032.14 N.

Area of the cross-section= 2.3

Shear Stress=== 6984.34 Kpa=6.98Mpa.

Maximum bending moment= ( ) Nm=24096 Nm in the direction as shown in the figure above.

Bending stress=Mpa

Deflection===0.2094 mm

Reference:

1. Ferinand P. Beer,E. RusselJhonston  and Elliot R. Eisenberg,Vector Mechanics for Engineers Statics

2. Raymond A. Serway and Jerry S.Faughn, College Physics

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