Structville

A hub for civil engineering related designs, analyses, discussion, information, and knowledge...

Trending Posts:

Search This Blog

Friday, June 30, 2017

Solved Examples on Shear Deformation of one-span Beams Using Virtual Work Method


1.0 INTRODUCTION
In this post, we are are going to present some solved examples on the shear deformation of one-span beams due to externally applied load. The predominant cause of deformation in beams subjected to externally applied load is bending, and that is the one usually considered. However, additional deformation is produced due to shear forces in the form of mutual sliding
of adjacent sections across each other. As a result of non-uniform distribution of shear stresses, the sections previously plane now become curved due to shearing alone.

Using the simple virtual work method and employing Verecshagin's combination rule, we are going to calculate the deflection at the critical points of some beams due to bending and due to shear forces. You will discover why shear deflection are neglected in some cases, but in sections that are significantly deeper, shear forces can be quite influential.

The displacement equation due to shear forces is given below;
Where;
k = factor that accounts for non-uniform distribution of shearing stresses. For rectangular sections, k = 3/2, and for circular sections, k = 3/4
G = Shear modulus of the section
A = Area of the cross-section
Q = Shear force due to externally applied load
Q ̅ = Shear force due to a unit virtual load at the point where the deflection is sought
Similarly, the displacement equation due to bending moment is given below;
Where;
E = Elastic modulus of the section
I = Moment of inertia of the cross-section
M = Bending moment due to externally applied load
M ̅ = Bending moment due to a unit virtual load at the point where the deflection is sought



2.0 SOLVED EXAMPLES
For all examples shown below, the section shown below will be used for all calculations.


2.1 Geometrical Properties
Area (A) = bh = 0.2m × 0.4m = 0.08 m2

Moment of inertia (I) = bh3/12 = (0.2 × 0.43)/12 = 0.0010667 m4

E = 21.7 KN/mm2 = 21.7 × 106 KN/m2

G = E / 2(1 + ν) = (21.7 × 106)/2(1 + 0.2) = 9.0416 × 106

EI = (21.7 × 106) × 0.0010667 = 23147.39 KN.m2

GA = (9.0416 × 106) × 0.08 = 723333.333 KN

2.2 Example 1: Simply supported beam carrying a uniformly distributed load. We are to find the deformations at the mid span

The internal forces diagram due to externally applied load is shown below;

Placing a unit load at the mid-span and plotting the internal stresses diagram;


We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

EIδb1 = 2 [5/12 × 62.5 × 1.25 × 2.5] = 162.760
δb1 = 162.760/23147.39 = 0.0070314 m = 7.0314mm

(b) Deflection due to shear;

GAδs1 = 2 [3/2 × 1/2 × 50 × 0.5 × 2.5] = 93.75
δc1 = 93.75/723333.333 = 0.0001296 m = 0.1296 mm

2.3 Example 2: Simply supported beam carrying a concentrated load at the mid-span. We are to find the deformations at the mid span

The internal forces diagram due to externally applied load is shown below;


Placing a unit load at the mid-span and plotting the internal stresses diagram;


We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

EIδb2 = 2 [1/3 × 125 × 1.25 × 2.5] = 260.417
δb2 = 260.417/23147.39 = 0.01125 m = 11.25 mm

(b) Deflection due to shear;


GAδs2 = 2 [3/2 × 50 × 0.5 × 2.5] = 187.5
δs2 = 187.5/723333.333 = 0.0002592 m = 0.2592 mm

2.4 Example 3: Cantilever beam carrying a uniformly distributed load. We are to find the deformations at the free end



Placing a unit load at the free end and plotting the internal stresses diagram;


We can now obtain the deflection at the mid-span by diagram combination;

(a) Deflection due to bending;

EIδb3 = [1/4 × 250 × 5 × 5] = 1562.5
δb3 = 1562.5/23147.39 = 0.0675 m = 67.502 mm

(b) Deflection due to shear;

GAδs3 = [3/2 × 1/2 × 100 × 1.0 × 5] = 375
δs3 = 375/723333.333 = 0.00051843 m = 0.5184 mm

As you can see in all the examples we considered, there was no case where the additional deflection due to shear was up to 1mm, and for practical purposes, they are very negligible for normal one span beams. Watch out for future posts where we will present cases where shear deflection becomes significant.

Thank you for visiting www.structville.com. We hope you visit us again, and share this post with your friends.

No comments:

Post a Comment