Skip to main content

Combined effect of bending and torsion

IIf a body is subjected to Shear Force, Bending moment and Twisting moments, then the combined stress in the shaft will be calculated by using of principle of superposition.

1.Shear stress due to direct shear:


The average Shear Stress due to direct shear force q = F/A

Where F= Shear Force at the section

A= Area of the cross section, 
But the shear stress distribution in the section will get by using the equaction
qd = FQ/Ib
Where Q= Ay
Where, A = Area between the extreme face of beam and the plane at which the shear stress is q
y = Distance of the centroid of area from N.A
F = Shear force at the cross-section.
I = Moment of Inertia of the beam cross section about N.A.
b = Width of the fiber at the plane at which shear stress is q

2.Shear stress due to Torsion:

For solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are:


where:
R is the outer radius of the shaft.
τ is the maximum shear stress at the outer surface.
φ is the angle of twist in radians.
T is the torque (N·m ).
ℓ is the length of the object the torque is being applied to or over.
G  is the shear modulus or modulus of rigidity
J is the polar moment of inertia for a round shaft or concentric tube only.
The shear stress at a point within a shaft is:


qS = TR/J
Note: The combined shear stress in body due to direct shear and torsion is                   q = qd + qs

3.Stresses due to Bending Moment:

Stresses caused by the bending moment are known as flexural or bending stresses. These stresses are calculated by using bending equation,
M/I = f/y = E/R
Where
M = Bending Moment
I = Moment of inertia about Neutral axis (N.A.)
f = Bending stress
y = Distance of the fiber from N.A.
R = Radius of Curvature
E = Young's Modulus

2.Combined effect of bending and torsion


In some applications the shaft are simultaneously subjected to bending moment M and Torque T. 

The Bending moment comes on the shaft due to gravity or Inertia loads. So the stresses are set up due to bending moment and Torque.

Equivalent twisting moment :


A twisting moment which, if acting alone, would produce in a circular shaft a shear stress of the same magnitude as the shear stress produced by a given twisting moment and a given bending moment acting simultaneously.

Equivalent bending moment :
A bending moment which, acting alone, would produce in a circular shaft a normal stress of the same magnitude as the maximum normal stress produced by a given bending moment and a given twisting moment acting simultaneously.

Comments

Popular posts from this blog

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø           ...

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us co...

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method  and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces.  Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns.  Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method   1. The points of inflection are located at the mid-height of e...