beams giacomo bo introduction general principles beams analysis … · 2020. 4. 16. · analysis of...

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Design of Beams

Beams

Giacomo Boffi

http://intranet.dica.polimi.it/people/boffi-giacomo

Dipartimento di Ingegneria Civile Ambientale e TerritorialePolitecnico di Milano

October 19, 2018

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Design of Beams

Introduction

Definition:

I a prevalent dimension (the span),

I loads act perpendicular to the prevalent dimension, so no axial forces butshear and bending moment,

I the cross section dimensions are the depth (parallel to loading) and thewidth.

We restrict our attention to rectilinear beams but curvilinear beams are possibleas well.

Historical remarks:

I Galileo Galilei,

I Coulomb,

I De Saint Venant.

Beams

Giacomo Boffi

Introduction

General Principles

Beams in Buildings

Basic Stress Distributions

Analysis of Beams

Design of Beams

Beams in Buildings

Beams are almost always present in a building.When beams form the principal structural system they are arranged in asystematic manner, usually following a hierarchical arrangement.Decking or planks have limited span so it is typical to support them, at regularintervals with orthogonal beams, forming a two-level system.

It is common (especially in steel structures) to support these deck-supportingbeams on larger, orthogonal beams forming a three-level system.

The loads on the beams are progressively increasing, as well as the lengths of thebeams and this leads in turn to an increase in the size of the beams in a threelevel system, from the secondary beams to the primary ones.

Beams

Giacomo Boffi

Introduction

General Principles

Beams in Buildings


Analysis of Beams

Design of Beams

Beams in Buildings

The stresses in a beam depend on the bending moment value and on shape ofthe cross section.

I The bending moment depends on the value of the loading, on the span ofthe beam and on the support conditions.

I In general, the larger the section the lower the stresses, but also the shapeand its orientation w/r to loading direction is important, in particular anincrease of the depth leads to an important decrease of the maximumstresses.

Beams

Giacomo Boffi

Introduction

General Principles

Beams in Buildings


Analysis of Beams

Design of Beams


At any cross section the resultant of the stresses must equal the shear force andthe bending moment.

I The shear force, that acts in the direction of the depth of the section, isequilibrated by tangential stresses.

I The bending moment, that acts perpendicular to the cross section, isequilibrated by normal stresses.

Beams

Giacomo Boffi

Introduction

General Principles

Beams in Buildings


Analysis of Beams

Design of Beams


Beams

Giacomo Boffi

Introduction

General Principles

Beams in Buildings


Analysis of Beams

Design of Beams


I The moment cause the beam to bow, in our figure top fibers are shortedand bottom fibers are elongated.

I There are fibers whose length is not varied, these fibers cross the section ona line called the neutral axis.

I The amount of length variation depends linearly on the distance y from theneutral axis.

I The stresses, for a linear elastic material, depends on the amount of lengthvariation(a.k.a. deformation) and are hence linearly varying, with positiveand negative extremes on the borders of the cross section.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Bending Stresses

The stresses are proportional from the distance y from the neutral axis, areproportional to the bending moment and are inversely proportional to a propertyof the dimensions and shape of the cross section, that we call moment of inertiaand denote with the symbol J. (NB dimensional analysis)

σ = My/J

Example

Simply supported beam, L = 8 m, p = 60 kNm , Mmax = pL2

8 = 480 kNm.For a rectangular section it is J = bd3/12 and ymax = b/2.

The maximum stress is σmax =PL2

8× d

2bd3/12

and with b = 0.3 m, d = 0.6 m it is

σmax = 26 666 667 Nm2 = 26.67 MPa

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Bending Stresses

For a rectangular section it is

J =bd3

12, σmax =

Md/2

bd3/12=

M

bd2/6= M/W

where W = bd2/6 is the section modulus.It is clear that if the double the amount of material in a beam by doubling thewidth b the maximum stress is halved, if we double the depth d the stress isreduced to one quarter.

Example

Simply supported beam, L = 7.2 m, P = 20 kN applied ad mid span,Mmax = PL/4 = 36 kN m = 36 MN mm.Rectangular section, b = 120 mm and d = 200 mm, J = 80× 106 mm4 andW = 800× 103 mm3, eventually σmax = 45 MPa.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

A little theory

In general we have to perform our computation in two steps

1. find the position of the neutral axis,

2. find the moment of inertia with respect to the neutral axis.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

A little theory

To find the position of the neutral axis

1. draw an axis parallel to the neutral one at a(n unknown) distance yG from it,

2. the distance from the neutral axis, in terms of the distance of the temporary axis,is y − yG ,

3. the stress is σ = (y − yG )k ,

4. the stress resultant is N =∫Aσ dA = k

∫A

(y − yG ) dA = k∫Ay dA− kyGA

5. we have N = 0 hence

yG =SxA

where Sx =

∫A

y dA

Similarly xG = Sy/A with Sy =∫Ax dA.

We have found the position of the centroid of the cross section w/r to two arbitrary axes.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

A little theory

The moment of the stresses about the neutral axes, if now we measure thedistances from it, is M =

∫A yσ dA = k

∫A y2 dA = k Jx by the definition of

moment of inertia, so that we can solve for k = M/Jx and eventually we canwrite

σ =M

Jxy .

The expression above coincides with the expression that was shown earlier w/o anexplanation.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Computing moments of inertia

The moments of inertia of many simple figures are tabulated, with respect totheir own centroids. We can use these known results along with parallel axistheorem.

Theorem

The moment of inertia J of an area A, with respect to a axis parallel to an axispassing for its centroid is equal to the sum of the centroidal moment of inertia J ′

plus the moment of transport ∆2A

J = J ′ + ∆2A

where ∆ is the distance between the two axes.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams


30mm

36

mm

6mm

6mm

A = (6× 30 + 6× 30)mm2 = 360mm2

Sx = (180× 15 + 180× 33)mm3 = (180× 48)mm3

yG =180× 48

180× 2mm = 24mm

J1

mm4 =303 × 6

12+180×(15−24)2 = (13500+14580) = 28080

J2

mm4 =30× 63

12+180×(33−15)2 = (540+14580) = 15120

J = J1 + J2 = 43200mm4.

Wt = J/12mm = 3600mm3

Wb = J/24mm = 1800mm3

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams


16mm

20mm

25

mm

35

mm

A = (20× 35− 16× 25)mm2 = 300mm2

Sx = 10mm, yG = 17.5mm

J1

mm4 =353 × 20

12= 71458.33

J2

mm4 =25× 163

12= 20833.33

J = J1 − J2 = 50625.00mm4.

W = J/17.5mm = 2892.86mm3

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Rectangle

G

y

a

b

Sides a, b

Centroid xG = a/2, yG = b/2

Area A = ab,

Moment of Inertia J =ab3

12

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Triangle

For a right triangle.

y

G

a

b

Sides a, b

Centroid xG = a/3, yG = b/3

Area A = ab/2,

Moment of Inertia J =ab3

36

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Oval

When a = b = D = 2R the oval is a circle.

x

y

a

b

Axes a, b

Centre of Mass xG = yG = 0

Area A =πab

4= πR2,

Moment of Inertia J = πab3

64= π

R4

4

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Lateral Buckling

As was the case with trusses, one side of a beam is subjected to compression andit’s subjected to the possibility of a lateral sway.Possible remedies:

I lateral bracing

I stiffening the compressed part in the lateral direction

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Shear Stresses

I On the left, a “beam” composed of three equal, separate sheets, each one behavesthe same,

I on the right, the sheets have been glued and the top sheet is shortened while thebottom one is elongated.

This behavior is possible only if the sheets exchange tangential stresses at theirinterfaces.These tangential stresses are the shear stresses.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Shear Stresses

E.g., for a rectangular beam the resultants of compressive and tensile stresses lieat d/6 from the borders, so their arm is a = 2d/3 and we have

T = C = M/a = 1.5M/d .

Consider now a slice of beam cut between x1 and x2 = x1 + δ and separated intwo halves at the neutral axis, the bottom part is subjected to two forcesT1 = 1.5M1/d and T2 = 1.5M2/d , we have an unbalanced force

U = 1.5M(x1 + δ)−M(x1)

d.

For δ → dx it isdU = 1.5dM/d = 1.5V dx/d ,

the area available for equilibrating the unbalanced force using tangential stressesis dA = b dx so eventually

τ = dU/dA = 1.5V /(bd).

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Shear Stresses

The procedure can be generalized, we know that σ = My/J so the resultant foran area A∗ is

F =M

J

∫A∗

y dA =M

JS∗

where S∗ is the first moment of the area A∗ w/r to the neutral axis.The variation of F over dx , dF = S∗V dx/J, is the unbalanced force toequilibrate over an area dA = b dx so that

τ =V S∗

b J.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Shear Stresses

For the T section we have previously studied, consider a maximum shearV = 4800 N and consider that the flange and the web have been glued together(b = 6 mm).We have to compute S∗ for the flange

S∗ =

∫ 36

3030y dy = 5940 mm3

and compute τ

τ =V S∗

b J=

4800× 5940

6× 43200MPa = 110MPa.

Your glue must resist to a tangential stress of 110MPa.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Deflections

Just a few formulas, E is the Young modulus of the material... for the cantilever,the concentrated load is at the free end, in the other cases the concentrated loadis at mid span — the deflections are measured at the same points.

Cantilever Supported Fixed Ends

P PL3

3EJPL3

48EJPL3

192EJ

w wL4

8EJ5wL4

384EJwL4

384EJ

Note how much the displacements depend on the beam span, note also thatdifferent end conditions influence very much the deflections.

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Bending Stresses

Lateral Buckling

Shear Stresses

Bearing Stresses

Torsion

Shear Center

Deflections

Principal Stresses

Design of Beams

Deflections

A simply supported beam, L = 6m, has a live load w = 3kN m−1. The section isrectangular, b = 0.2m and d = 0.4m, the Young modulus is E = 12GPa.Compute the mid span displacement.

∆ ≈ 4 mm

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Design of Beams

General Principles

Timber

Steel

Reinforced Concrete:Principles

Reinforced Concrete:Procedures

General Principles

I Approaches

I Strength and stiffness

I Cross-section design

I Material properties variation

I Shaping the beam

I Position of supports

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Design of Beams

General Principles

Timber

Steel



Different Design Procedures

I Admissible Stress Design (ASD)

I Load and Resistance Factor Design (LRFD)

Beams

Giacomo Boffi

Introduction

General Principles

Analysis of Beams

Design of Beams

General Principles

Timber

Steel



Steel Beams

The design of a steel beam in most cases coincides with choosing a section in a catalogof commonly available steel members.The section you are going to choose has to satisfy two essential requirements in terms ofstrength and deformability.

I Strength: the section must be strong enough to resist the maximum value of thebending moment,

fyW ≥ Mmax → W ≥ Mmax/fy .

I Deformability: the maximum deflection must be less than a fraction of the beamspan, ∆ ≤ L/k.

The value of k depends on the load cases used to compute the deflection (live vsdead vs combination) and the function that the beam has to perform (support aroof, a floor, a wall, etc).

E.g., for a simply supported beam subjected to a distributed load, supporting a wall

∆ =5wL4

384EJ≤ L

200→ J ≥ 1000wL3

384E.

beams giacomo bo introduction general principles beams analysis … · 2020. 4. 16. · analysis of...

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