# Engineering: Calculation Of Stress, Strain And Moments Of Force (Essay Sample)

Calculation of stress, strain and moments of force.

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ENGINEERING

Student's Name

Course Code

Class

Institution

Date

Engineering

Task 1

MA=0

10 x RB=10x3+8x8

10 RB=94

RB=9.4 kN

MB=0

10 x RA=10x7+8x2

10 RA=86

RA=8.6 kN

V=0

RA+ RB=10+8=18kN

9.4+8.6=10+8=18kN

Distribution of Shearing Force

Starting at A, F = 8.6 kN which is also equal the shear force acting on the section AB. The force remains constant between AB and CD. The beam encounters a downward force of 10kN (load C) resulting into a shear force of -1.4kN to the left of section CD. The force remains constant to the right of section CD and encounters another downward of 8kN which eventually meets an upward reaction at B of 9.4kN. The shear force at B is 0kN.

Distribution of moments

MA = 0kNMsection AB = 2 x RA=2 x 8.6 kN=17.2kNmMsection CD = (RA x 6)-10x3=21.6kNmMB = -(VA x 10) +(F1 x 10) -(F2 x 7) -(F3 x 2) = -(0.8 x 10) +(9.4 x 10) -(10 x 7) -(8 x 2) = 0kN

Task 2

Section a

Given the parameters, the suitable beam size will be a 356 x 171 x 51 universal beam. From the BS4 table, check the under the column second moment of area for the value 14,136mm4. From the BS4 table, a value closer to that is 14,140mm4. Move to the radius of gyration column and under the x-x axis, find the value 14.8. Finally, check for a value of 355.0 under the column depth of section from the BS4 table. Since all the above values are on a given row, locate the corresponding size on the beam by checking its size under the first column of the BS4 table.

Section b

* Stress at the outer edge of the x-axis

Stressσ= Force FArea A

=2x106200 x 200

=50MN/m2

* Slenderness ratio

Slenderness ratio λ=effective length of the column lradius of gyration of the section rxx=50.2=25

* Selection of a suitable size that satisfies the calculation

Considering the parameters identified above, the appropriate size of the column that will satisfy the calculations is a column of serial size 356 x 406 x 634mm.

Task 3

Section a

τmax=TD2π (D4 - d4)32

T = P2 π n

Where,

τ = shear stress (Pa)

T = Torque (Nm)

D = shaft outside diameter (m)

d = shaft in

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