Mechanical Engineering: Shear Force Calculation and Analysis Report

This report provides a comprehensive analysis of shear forces, focusing on calculations and the impact of pin arrangements. It begins by calculating shear stress using the formula, considering both the applied force and the cross-sectional area of the pins. The report then examines the shear stress in a joint with multiple bolts and calculates the maximum shear stress before failure. The analysis considers the effects of both vertical and horizontal forces on the pins. The conclusion highlights how pin arrangements determine the total and average shear stress, the yield strength, and the allowable shear stress before failure. It also emphasizes the role of pin diameter and capacity in controlling shear properties and failure mechanisms. Finally, it discusses the distribution of shear across the pins and the equilibrium of vertical and horizontal shear forces, linking these concepts to the diagonal tensile web crack in an RC beam. This report provides a thorough understanding of shear force analysis in mechanical engineering.

Results and Discussion
For this example, shown below, we are to calculate the total shear force due to failure on the
pins.
The total shear load on a bolt is usually equal to the vector sum of the individual loads.
Calculating the following;
the brass pin diameter is 6.36 mm,
Single pin has a force of 9.782 kN,
And rectangular 6 pin total force equals 27.561 kN,
Pk load is 27561N
Pk ext is 2.72 mm

Shear stress of pins is calculated using the formula:
In which, S= shear stress, Q= the moment about the neutral axis of the area above the point you
want to calculate the shear stress, I = moment of inertia, b = width of the desired section.
But also can be found the simple formula;
In this case pins can be acted upon by forces in both vertical and horizontal direction.
Shear stress = applied force / area
= 4F/( πD2)
= 27561/ pi *0.00636^2
= 2.1688567 *10^5 N/m^2
where A is the cross-sectional area of the pins. If the pins are of equal area the equation reduces
to for a joint carrying a load P and consisting of n bolts.
The application of a moment to the joint produces shearing stresses which are assumed to be
proportional to the distances r of 17 the bolt centers from the center of the bolt areas.
And the maximum shear stress before failure is calculated as;

= 16 *27561/3*pi*6.36^2
=3.4702*10^3 N/m^2
This is because when a load is applied to a group of bolts in such fashion that its line of action
does not pass through the centroid of the group, it is usually convenient to superimpose the effect
of a force of equal magnitude and direction, passing through the center id, upon the effect of a
moment about the center which is equal to the product of the actual force and its perpendicular
distance from the center.
Conclusion
It can be concluded that pin arrangement determines the total and average shear stress. And the
pin arrangement also determines yield strength and allowable shear stress before failure stress.
The shear properties and failure mechanisms on this task is controlled by two factors namely; the
capacity content and the diameter of the pins used. Pins usually inhibit crack growth at the
failure lines and transfer weights even after cracking has occurred.
And therefore it is shown that the shear is distributed over the total pins and both in vertical and
horizontal direction. In addition, it can be deduced that total shear force is the average shear
stress in all pins shared uniformly. And also it can be said that at equilibrium, vertical shear of
equal magnitude to that of horizontal shear is present at every level of fiber. The resultant of
these two perpendicular stresses acts diagonally and directly contributes to the diagonal tensile
web crack of an RC beam.