Question

Determine the average normal stress developed in rod AB if the load has a mass of 66 kg . The diameter of rod AB is 8 mm.

4 8 mm

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Concepts and reason

Beam: A beam may be defined as a structural element that is used to bear load in transverse direction or horizontal direction.

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 force in rod AB. Then, using the diameter and force in rod, find the stress.

Fundamentals

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

Moment equilibrium condition,

ΣΜ-0

Force equilibrium condition along vertical direction,

Σ , =0

Force equilibrium condition along horizontal direction,

ΣΕ-0

Consider the relation for normal stress in rod.

F

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

Consider the relation for area of circular rod.

ла
A
4

Here, is the diameter of circular rod.

Consider the relation for normal stress in circular rod.

F
о
па?
4
4F
о
2
ла

Here, F
is the normal force and is the diameter of circular rod.

Draw the free-body diagram.

АС
3
FAB
4
(66x9.81) N
t

Apply equilibrium of forces along the vertical direction.

Σε-0
3
FAC
(66x9.81) 0
66x9.81x5
F
3
Fc 1079.1 N
AC

Apply equilibrium of forces along the horizontal direction.

ΣΗ-0
«)-ка-о
FAC
0
AВ
1079.1(0.8)-F
FR863.28 N
АВ

Calculate stress in rod AB.

4F
о 3
ла?
АВ
АВ

Here, АВ
is the normal force in member AB and АВ
is the diameter of circular rod AB.

Substitute 863.28 N
for АВ
and 8 mm
for АВ
.

4x863.28
лх8
3453.12
201.062
1 MPa
1 N/mm2
17.1744 N/mm2
=17.1744 MPa

Ans:

Normal stress in rod AB is 17.1744 MPa
.

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