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School of Mechanical and ProductionEngineering

NANYANG TECHNOLOGICAL UNIVERSITY

2000/2001

ULTIMATE LOAD OF A BEAM UNDER BENDING

Project P3.8

Submitted byLab group: LA7b

Goh Yan JuanEe Yuan Swee

Dorothy Goh Mei ChiatGoh Yong Chuan

Lily Kor Kwang Sing

Content

Abstract.………………………………………………………………...

Introduction……………………………………………………………..

Scope……………………………………………………………………

Objectives………………………………………………………………

Theory…………………………………………………………………..

Apparatus……………………………………………………………….

Experimental Procedure………………………………………………...

Calculation……………………………………………………………...

Discussion………………………………………………………………

New design……………………………………………………………...

Analysis…………………………………………………………………

Suggestion………………………………………………………………

Question………………………………………………………………...

Conclusion……………………………………………………………...

References

Appendix

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M312Project P 3.8 Ultimate load of a beam under bending

Abstract

This report describes an experiment to determine the strength of four metal beams. Thebeams were subjected to constant loading force at the center of the beam. Not muchdeformation was found on the solid cross-section and C-channel (thick). However, the c-channel (thin) experiences buckling and the hollow cross-section cracks. As such, we wouldneed to reinforce the C-channel (thin) by adding reinforcement to strengthen it.

Introduction

Bending of the beams is a situation that could be seen everywhere. A slendermember of the beam subjected to transverse loading would be termed as a beam underbending. At any cross-section, the transverse loads would generate shear and bendingmoments so as to maintain equilibrium. The bending causes a change in the curvature of thebeam and thus induces tensile and compressive stresses in the cross-section of the beam.Maximum stresses are achieved at the areas furthest from the neutral axis, and the part wherethe strain is zero.

The bending moment causes the maximum stress in the beam to be equivalent to theyield stress. This bending moment is known as the First Yield, Mfy. However, this is not theultimate moment that the structural beam can sustain as there is still a large portion of elasticcore that is capable of supporting higher loads (moments). This is called the ultimate momentor the moment for full plasticity, MP. With this ultimate moment or ultimate load as theparameter, a more cost-effective and efficient design can be established. The shape factor,ratio of the ultimate moment to the moment for the first yield, gives an idea of the designefficiency that can be obtained. Assuming elastic perfectly plastic material, the limit load ofthe beams with any cross-section can be easily calculated.

Scope

This project focuses on the analysis of stresses in beam bending. This was done by loadingvarious beams, namely 2 C-channels (thick and thin), a box beam and 2 rectangular cross-section beams. Theoretical analysis and experiments were carried out to determine thestrength and stiffness of the beams, which were simply supported at 2 ends with a centralload. A C-beam was chosen and a reinforcement scheme was done to prevent buckling and toincrease the strength of the beam.

Objectives

The aims of this investigation are:

• To verify the shape factor and the limit load for beam of rectangular cross-sectioningunder pure bending


• To determine experimentally and theoretically the shape factor and (plastic) limit load ofa simply supported beam.

• To study the mode of failure of a thin walled (1 mm wall thickness) C-section and topropose and experimentally verify an appropriate reinforcement scheme to prevent orreduce buckling.

Theory:

In the simplified engineering theory of bending, in order to establish relation among theapplied bending moment, the cross-sectional properties of a member, the internal stresses anddeformations, we make the following assumptions:

1. The beams are reduced from internally statically indeterminate problems todeterminate ones.

2. The deformations causing strains are related to stresses through the appropriate stress-strain relations

3. The equilibrium requirements of external and internal forces are met.4. Plane section before loading remains plane section after loading.

To determine the shape factor of the beam, we have the following equation:

Shape factor = M ult/ M yd (1.1)

Consider the figure below,

To find ultimate moment M ult , we have to locate the new neutral axis because the section isunsymmetrical about the centroid of the figure.

Let y = the distance from the surface of the flange to the neutral. The area above and belowthe neutral axis is equal.

To find y, we equate the area above and below the neutral axis.

x× y = (1 – y) × x + h × t × 2

To find T1,

T1T2

T3

C

Fig. 1

y

x

h


T1 = yield stress * ( h – y) * x

T2 =yield stress *h* t

where h, y, t and x are in metresyield stress = 300MPa and t is the thickness of the beam.

Taking moments abt force C.M ult = T1(L1 + L) + T2(L2 + L)

To find M yd , we use

M yd = (yield stress * I )/ y where

I = moment of inertia of section,Yield stress= 300 MPay = centriod of section = (Σ Ay )/ΣA

With all parameter known, we substitute them into eqn 1.1 to find shape factor.(Note: for other shapes the same procedure applies)

Apparatus

1) Machine i. X-Y recorder


Experimental Procedure

• 1st piece : solid cross-section1. Put the piece on the four-point bending rig.2. Place the upper rig on the solid cross-section. (enabling the pure bending of the piece to

occur)3. Bring the jet to the center of the upper rig by using the hand-brake.4. Place the linear displacement transducer on the stopper.5. Use the resistor to calibrate the Y-axis by shorting from the amplifier.6. Take measurement (calibration) from the X-Y recorder.7. Press the hand-pump, allowing it to force the piece downwards.8. Constantly check with the graph drawn on the X-Y recorder to see if the line has become

a straight line.9. Then we turn the control valve so as to allow the jet to return to its original position.10. From the graph, we would have seen that the pen on the X-Y recorder has not gone back

to its original position.11. The metal piece has deformed.12. Repeat steps 2 to 9 for C-channel (thick), C-channel (thin) and hollow cross-section.


Calculation for rectangular cross-section

From the theory,

MPay 300=δ

Force 1 = yHB δ×× 5.0

= ( 50.9 × 10-3 ) × ( 0.5 × 12.7 × 10-3 ) × ( 300 × 106 )

= ( 50.9 × 10-3 ) × ( 6.35 × 10-3 ) × ( 300 × 106 ) = 96964.5N

B = 50.9 mm

H = 12.7 × 10-3 mm

D = 6.35 mm

Take moment about Force 2 = 0 (anti-clockwise +)

Pure Bending Moment = Force 1 × D = 96964.5 × 6.35 × 10-3

= 615.7 Nm

200 mm 200 mm

0.5 P

0.5 P

Force 1

Force 2

yδ

yδ

0.5 P

0.5 PP


0.5 P = M / 0.2 P = ( 615.7 / 0.2 ) × 2 = 6157 N

I = bd3/12 = (50.9× 10-3) × (12.7× 10-3)/ 12 = 8.689 × 10-9 m4

Myd = yδ × I / (12.7 × 10-3 / 2) = 300 × 106 × 8.689 × 10-9 / 6.35 × 10-3

= 410.5 Nm

Shape factor = Mult / Myd = 615.7 / 410.5 = 1.5

(See Appendix for picture)

Calculation for C-channel (1mm all about)

For the ultimate moment, we have to locate the new axis due to the axis is not symmetricabout the centroid.

78.5 mm

25.5 mm

yδ

yδ

T1

C y

T2 T3


78.5× y = (1 – y) × 78.5 + 24.5 × 1 × 22 × 78.5y = 78.5 + 24.5 × 2

y = 127.5 / (2 × 78.5) y = 0.8121 mm

T1 = (1.879 × 10-3) × (78.5 × 10-3) × (300 × 10-6) = 4425.05 N

T2 = T3 = (24.5 × 10-3) × (1 × 10-3) × (300 × 10-6) = 7350 N

Take Moment about C = 0 (clockwise+)Moment = T1 × (0.8121 / 2 + 0.1879 / 2) × 10-3

+ 2 × T2 × (0.8121 / 2 + 0.1879 + 24.5 / 2 ) × 10-3

= 4425.05 × (0.5 × 10-3) + 2 × 7350 × (12.844 × 10-3) = 191.02 Nm0.5 P = M / 0.2 P = (660.15 / 0.2 ) × 2 = 6601.5 N

y = centriod of section = (Σ Ay )/ΣA = (76.5 × 1 × 25 + 25.5 × 1 × 12.75 × 2) / (76.5 × 1 + 25.5 × 1 × 2) = 20.1 mm

I total = Σ (Io + Ad2) = (76.5 × 13) / 12 + 76.5 × 1 × (25 – 20.1)2

+ [(1 × 25.53) / 12] × 2 + 2 × 25.5 × 1 × (20.1 – 12.75) 2

= 7.3619 × 10-9 m4

Myd = yδ × I / (20.1 × 10-3) = 300 × 106 × 7.3619 × 10-9 / 20.1 × 10-3

= 109.88 Nm



Calculation for C – channel (3mm all about)

For the ultimate moment, we have to locate the new axis due to the axis is not symmetricabout the centroid.


76.2 mm

y

25.4 mm

76.2 × y = (3 – y) × 76.2 + 22.4 × 3 × 22 × 76.2y = 76.4 × 3 + 134.4 y = 363.4 / (2 × 76.4) = 2.38 mm

T1 = (0.62 × 10-3) × (76.2 × 10-3) × (300 × 106) = 14173.2 N

T2 = T3 = (3 × 10-3) × (22.4 × 10-3) × (300 × 106) = 20160 N

Take moment about C = 0 (clockwise+)Moment = T1 × (2.38 / 2 + 0.62 / 2) × 10-3

+ 2 × T2 × (2.38 / 2 + 0.62 + 22.4 / 2) × 10-3

= 14173.2 × (1.5 × 10-3) + 2 × 20160 × (13.01 × 10-3) = 545.8 Nm

0.5 P = M / 0.2 P = (545.8 / 0.2) × 2 = 5458.23 N

y = centriod of section = (Σ Ay )/ΣA = (70.2 × 3 × 23.9 + 25.4 × 3 × 12.7 × 2) / (70.2 × 3 + 25.4 × 3 × 2) = 19.198 mm

I total = Σ (Io + Ad2) = (70.2 × 33) / 12 + 70.2 × 3 × (23.9 – 19.198)2

+ [(3 × 25.43) / 12] × 2 + 2 × 25.4 × 3 × (19.198 – 12.7) 2

= 1.944 × 10-8 m4

Myd = yδ × I / (19.198 × 10-3) = 300 × 106 × 1.944 × 10-8 / 19.198 × 10-3

= 303.82 Nm

T3T2

T1

C




Calculation for Hollow shaft (1mm all about)

76.5 mm

26 mm

C2 = C3 = (1 × 10-3) × (12 × 10-3) × (300 × 106) = 3600 N

T1 = T2 = (1 × 10-3) × (12 × 10-3) × (300 × 106) = 3600 N

T3 = (1 × 10-3) × (76.5 × 10-3) × (300 × 106) = 22950 N

Take moment about C1 = 0 (clockwise+)M = T1 × (0.5 + 12 + 6) × 10-3 × 2 + T3 × (0.5 + 0.5 + 24) × 10-3

– C2 × (0.5 + 6) × 10-3 × 2 = 3600 × 18.5 × 10-3 × 2 + 22950 × 25× 10-3 – 3600 × 6.5 × 10-3 × 2 = 660.15 Nm

0.5 P = M / 0.2 P = (660.15 / 0.2) × 2 = 6601.5 N

y = centriod of section = (Σ Ay )/ΣA = (76.5 × 1 × 25 + 25.5 × 1 × 12.75 × 2) / (76.5 × 1 + 25.5 × 1 × 2) = 20.1 mm

I total = Σ (Io + Ad2) = [(74.5 × 13) / 12 + 74.5 × 1 × (12.5) 2] × 2 + [(1 × 263) / 12] × 2

T1

C3C2

C1

T3

T2


= 2.622 × 10-8 m4

Myd = yδ × I / (13 × 10-3) = 300 × 106 × 2.622 × 10-8 / 13 × 10-3

= 605.146 Nm



Point Load (Solid cross-section)

Yield stress = 300mpaF= (50.9 x 0.001)(1/2 x 12.7 x 0.001)(300 x 1000000) = (50.9 x 0.001)(6.35 x 12.7 x 0.001)(300 x 1000000) = 96964.5N

M= F(1/2 x 12.7 x 0.001) =615.7 N

Length between supports is 0.73m.P/2 = M/0.3515P = (615.7/ 0.3515) x 2 = 3503.3N

Thus the theoretical ultimate load (for point load) is 3503.3N It tallies rather well withexperimental data which is in the region of 3150N

(See graph and picture in appendix)

X-SECTION

THEORETICALULTIMATE

LOAD

EXPERIMENTALULTIMATE

LOAD

RADIUS OFCURVATURE

SHAPEFACTOR

1 Solid cross-section

6157 N 6800 N 1070mm 1.5

2 C- channel(1mm)

1910.02 N 1800N 305mm 1.738

3 C- channel(3mm)

5458.23 N 5400 N 570mm 1.796

4 Hollowrectangle

cross-section

6601.5 N 3500 N - 1.091


Discussion

As seen above, all except the thin C-channel and the hollow rectangle cross-section have theultimate load near the theoretical calculated one.

The C-channel could not meet the value due to the buckling at the side. Once the side startedto buckle, there will be a change in shape of the C-channel and the shape factor of the C-channel will be altered. With this change, the ultimate bending moment will change, thuschanging the rate of deformation with respect to the load applied.

As seen from the figures below, there’s a notch at the right hand side of the hollow beam.This can be explained by the fact that the beam experiences compressive stresses from theroller placed underneath it. Hence it reaches yield stress faster than the rest of the beam andthus deform first, causing the imbalance of force distributed on the beam. Therefore any extraload applied will result in that portion to be more deformed and the rest of the beamremaining horizontal.

NEW DESIGN

From the results shown above, it shows that buckling occurs only on the thin C-channel(1mm wall thickness) where under a sufficiently large lateral load, the side walls tend tobuckle and the top plate will tend to wrinkle under compression load imposed by the steelbar.

Buckling during bendingled to change in shapefactor


Thus, further modification was carried out on the thin C-channel in order to increase thestrength and stiffness of the beam so as to reduce or prevent buckling from occurring.

When the applied lateral load P has a small value, the C-channel remains straight andundergoes only slight compression, which means that the C-channel is still stable and it willreturn will return to its original position when the lateral load is removed.

As the applied lateral load P is gradually increased, we reach a condition of neutralequilibrium in which the C-channel may have a bent shape. The corresponding value of theload is the critical load Pcr. At this load the ideal C-channel may undergo small lateraldeflections which will produce a bent shape that does not disappear when the lateral load isremoved.

At higher values of the load, the C-channel is considered unstable and will collapse bybending or buckling. Thus, we reach a condition of unstable equilibrium and the C-channelwill buckle sideways and wrinkled on the top plated.

As summarized:• If P<Pcr, the C-channel is in stable equilibrium in the straight position.• If P=Pcr, the C-channel is in neutral equilibrium in the slightly bent position.• IF P>Pcr, the C-channel is in unstable equilibrium in the bent position, and hence its

buckles.

The critical load of the C-channel is important. Due to the fact that we can’t use anothermaterial therefore the strength of the material is fixed. Thus, using a stronger material doesnot increase the critical load. However, the load can be increased by using a stiffer material(larger modulus of elasticity E).

After much consideration, we decided to use the following new design because it willincreased the stiffness and provided rigidity to the C-channel by having small aluminumplates (in the form of ‘A’ shape) placed inside the C-channel. The aluminum plate is thenbeing riveted to the side of the C-channel to prevent buckling from the side walls by having:• F1 which will tends to pull the side walls inwards• F2 which will tends to prevent shearing at point 1 and point 2 when subject to bendingand bending from the top plate by having:

• F3, which is the reaction force form from the base of the ‘A’ to act against the lateralloading.

At the same time, this design is the simplest among all and required the shortest fabricationtime. Also, it contributed the least weight with respect to the current C-channel and theminimum cost.


Analysis

The thin walled C-section beam (with reinforcement) exhibited lower ultimate loadcharacteristics when compared to the regular thin walled C-section beam under pure bending.Inspection carried out after failure showed a crack propagating from a riveted joint situated atthe mid-span of the beam (see figure). The occurrences of the crack at that location whenthere are also similar joints along the beam can be attributed to the fact that bending momentis the greatest at the mid-span. The particular hole drilled in the beam to accommodate therivet serves as a stress concentration point/location. This resulted in the beam at thatparticular location to reach yield stress very much earlier than expected even though the loadapplied to the beam is still relatively light. Thus the anticipated higher strength offered by thereinforcement did not materialize as it was found upon failure that it has lower strengthcompared to that of the unmodified one.


Suggestion

The beam is made of aluminum. This makes the option of welding the reinforcement to thebeam non-feasible, as the workshop facilities do not have MIG and TIG welding equipmentin their inventory. In addition, aluminum is rather hard to weld because of its high heatconductivity. Welding the reinforcement into place would have been the best mode ofattaching the reinforcement compared to riveting and other means of fastening. This isbecause welding would allow the applied load to be uniformly distributed to the reinforcingstructure throughout he entirety of the welding joints but the same cannot be said of riveting,where the load and stresses are concentrated in the few rivets and holes that hold thereinforcement into place.

Questions

a) If the beam is not in pure bending, shear stresses will be present in the beam too. This is inaddition to the bending moment that is already present. Thus the curvature of the beamcaused by bending alone and in tandem with shear forces would be different. Also theultimate load that it can sustain would be lower.

For full plasticity condition, yield stress must be reached or exceeded throughout the crosssection of the beam.

And when the bending moment for full plasticity is reached along a particular section of thebeam, a plastic hinge is formed. The material gives way and it behaves like a hinge thatallows large rotations to occur at constant plastic moment.

b) Plastic hinge occurs at 1 section. This buckling is local and it is termed lateral torsionalbuckling. For conventional buckling, the load applied acts axially along its length. For ourexperiment our load is lateral. Global buckling refers to the beam undergoing bendingthroughout its length which is in contrast to local buckling and it occurs at a restricted andwell defined location. To control buckling we can either reduce the length of the beam,increase its width and/or thickness or add structural reinforcement (webs in the axialdirection) to enhance its rigidity so as to prevent buckling.


Conclusion

Among the beams of 4 different cross-sections tested, results showed that the solid crosssection is the strongest and the hollow cross section the weakest. In considering the beststructure we would have to take into account of the respective weights of the beams relativeto that of their weights. As such the solid cross section may have the highest strength but thisis more than offset by its higher weight. We can compare this using the strength to weightratio.

The best combination would be to be a design that is able to provide high strength but not atthe expense of weight.... ie it should be lightweight. One way to achieve this is to reinforceone of the thin C-channels by adding ribs or webs so as to support the underlying structure.However we have been unable to achieve this experimentally to get the higher strengthrequired because our reinforced specimen failed prematurely. We have come to theconclusion that it might have been an oversight not to take into account the weakening of thebeam due to the drilling of holes in it. In hindsight we should have added reinforcementplates where the holes are drilled.

All in all, our experimental results tallied well with theoretical calculations with the soleexception of the reinforced c-channel.

References

(1) Popov, E.P., Engineering Mechanics of Solids, 2nd Edition, SI Version, Prentice-Hall, New Jersey, 1990.

(2) Gere, J.M. & Timonshenko S.P., Mechanics of Materials, SI Edition, Chapman &Hall, London, 1991.

(3) Riley, W.F. & Zachary, L. Introduction to Mechanics of Materials, John Wiley &Sons, New York,1989

(4) Beer, W.F. & Johnston, E.R., Mechanics of Materials, 2nd Edition, SI Units,McGraw-Hill, 1


AppendixRectangular cross-section

Before bending

After Bending


C-channel (1mm all about)

Before bending

After Bending


C – channel (3mm all about)

Before bending

After Bending

Hollow shaft (1mm all about)


Before bending

After Bending

Point Load (Solid cross-section)


After bending


Time at which thecrack start

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