For the assembly and loading of Prob. 1.7, determine ( a )...

For the assembly and loading of Prob. $1.7,$ determine ( $a$ ) the average shearing stress in the pin at $B,(b)$ the average bearing stress at $B$ in member $B D,(c)$ the average bearing stress at $B$ in member $A B C,$ knowing that this member has a $10 \times 50$ -mm uniform rectangular cross section.

Key Concepts

Shear Stress

Shear stress is the internal stress that develops in a material when forces are applied parallel or tangential to its cross-sectional area. It is calculated by dividing the applied shear force by the area over which the force is acting. This concept is critical for determining the load a pin or other structural element can withstand before yielding or failing in shear.

Bearing Stress

Bearing stress refers to the contact pressure between two components in a mechanical assembly, where one part exerts force on another over a finite contact area. It is determined by dividing the transmitted force by the effective contact area. Understanding bearing stress is essential for ensuring that the interface between connected parts does not experience excessive local deformation or damage.

Pin Connections in Structural Assemblies

Pin connections are a common method for joining two or more members in a structure, where a pin is used to transfer loads between the components. Analyzing the stresses in pin connections, including both shear and bearing stresses, is fundamental in mechanical design to ensure that the joint performs reliably under applied loads.

Cross-sectional Analysis

Cross-sectional analysis involves calculating the geometric properties—such as area and shape—of a structural member's cross section to determine its ability to resist applied loads. In stress calculations, the cross-sectional area is used to compute average stresses, making this analysis essential for designing components that can safely support the expected loads.

Transcript

00:01 Hi friends in the given diagram we have to find shear stress at connecting points a, b and c given p having 5 kilo newton force.

00:30 The elements are circular cross -section with diameter 18mm.

00:43 Now let us here the force is acting at a, b and c.

00:49 At a, the forces are ax, a by e.

01:05 At b x b bx b by at c c x c by this is the free body diagram now applying the condition of equilibrium at point b so at point b moment of couple must be zero that is minus 5 into point 5 minus 15 into minus 15 into minus 30 4 minus 5 into 5 into 5 .5 plus a y into 5 .5 must be 0 so you will get a y having the value 30 kilo newton.

02:30 Similarly, submission of force fy must be 0.

02:38 By, minus 5, minus 15, minus 30, minus 5 plus a .y must be 0.

02:50 So, by, you will get 25 kilo newton.

02:57 Now applying condition of equilibrium at point c.

03:09 Moment of couple at point c must be 0.

03:14 So you can.

03:14 Can write minus 25 into 4 plus bx into 3 is equal to 0 hence bx you will get 33 .3 kilo now summation of f y must be 0 that is minus b y plus c y is equal to 0 so c y you will get 25 kilo newton.

03:52 Now summation of fx must be 0.

03:57 Hence you will get minus bx plus cx is equal to 0.

04:07 So cx you will get 33 .3 kilo newton...

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