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The board is subjected to a tensile force of 200 lcan someone help me with this hw problem

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Answer #1
Concepts and reason

Free body diagram: When all the supports are removed by replacing them with forces that prevents the translation of body in a given direction that diagram is called free body diagram. When their resultant force and couples becomes equal to zero then the body is said to be in equilibrium.

Force in members: If the forces acting in members of truss are considered as tensile and in the end the forces are taken as compressive if the obtained final value is negative.

Stress: When an alloy is loaded with a force, it produces a stress which makes the alloy to deform. Stress is the force per unit area. Normal stress is a stress that occurs when a member is loaded by an axial force and shear force is a stress occurs due to force parallel to the area.

Concept of free-body diagram, equilibrium of forces, and stress must be used to solve the problem. First, draw the free-body diagram and apply equilibrium conditions to find the shear force and normal force in section a – a. Then, using the area and force, find the stresses.

Fundamentals

Basic equations for the calculation of internal forces are as follows:

Moment equilibrium condition,

ΣΜ = 0

Force equilibrium condition along vertical direction,

ΣΕ, = 0

Force equilibrium condition along horizontal direction,

ΣΕ = 0

Consider the relation for normal stress.

Here, is the normal force and is the cross section area of rod.

Consider the relation for shear stress.

Here, is the shear force and is the cross section area of rod.

Draw the free-body diagram.

200 lb

Apply equilibrium of forces along the section a – a.

ΣΕ = 0
-V + 200cos 20° = 0
V = 200cos 20°
V=187.938 1b

Apply equilibrium of forces perpendicular to the section a – a.

ΣF = 0
-Ν + 200sin 20° = 0
N = 200sin 20°
N = 68.404 1b

Calculate the area of cross – section.

A=2x-
sin 20°
= 2x11.695
= 23.39 in

Calculate the normal stress at section a – a.

Here, is the normal force and is the cross section area of rod.

Substitute 68.404 lb
for and 23.39 in?
for .

68.404
0
23.39
= 2.924 lb/in?
1 lb
in?
= 2.924 psi

Calculate the shear stress at section a – a.

Here, is the shear force and is the cross section area of rod.

Substitute 187.938 lb
for and for .

t=
187.938
23.39
=8.0349 ib/in? 10
= 8.0349 psi

Ans: Part A

Average normal stress in wood fibers is 2.924 psi
.

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