labsheet-cc205.pdf

8/14/2019 labsheet-cc205.pdf

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Shear and bending

moment

CC205

Mechanics of Structure

Deflection and

slope test

Tensile test

aboratory heet

Provided by: Norbaini bt Mohamed


2/15

CIVIL ENGINEERING DEPARTMENTPOLYTECHNIC OF SULTAN AZLAN SHAH

STRUCTURE LABORATORY

CC 205

TITLE SHEAR FORCE AND BENDING MOMENT IN THE BEAM

EXPERIMENT 1A Shear Force and Bending Moment Variation with an increasing Point Load

EXPERIMENT 1B Shear Force and Bending Moment Variation for Various Loading Conditions

DATE OFPRACTICAL

DUE DATE

LECTURER NAME

NAME OFGROUPMEMBERS

NUM NAME REG. NO1

2

3

4

5

6

7

8

9

10

LECTURERS

COMMENT

NUM MARKING CRITERIAL GIVEN MARK

1 Practical Work Title (5%)2 Objective (5%)3 Theory (5%)4 Apparatus / Equipment (5%)5 Procedures (5%)6 Data (5%)7 Analysis / Discussion (5%)8 Comments / Conclusion (5%)

9 Tidy (5%)10 Reference (5%)

Report (50%)

1 Attend for practical (25%)2 Leadership / cooperation (25%)

Practical (50%)

TOTAL MARK (100 %)


3/15

RA L RB

TITLE : SHEAR FORCE AND BENDING MOMENT IN THE BEAM

EXPERIMENT 1A: Shear Force and Bending Moment Variation with an increasing Point Load

OBJECTIVE :This experiment examines how shear force and bending moment varies with an increasing

point load

THEORY :We know that if a body or object of any sort is stationary, then the forces on it balance, as follows:

(1) Vertical equilibrium (total force up = total force down)

(2) Horizontal equilibrium (total force right = total force left)

(3) Moment equilibrium (total clockwise moment = total anticlockwise moment).

We will use the statement:

The Shear force at the cut is equal to the algebraic sum of the forces acting to the left or right

of the cut.

The Bending Moment at the cut is equal to the algebraic sum of the moments caused by theforces acting to the left or right of the cut.

APPARATUS :

a) Computer set

b) Frame structure

c) Digital Force display

PROCEDURE:

(i) Check the Digital Force Display meter reads zero with no load.

(ii) Place a hanger with a 100 g mass to the left of the cut (40 mm away).

w cut

a 40mm

Figure 1a : Force diagram

(iii) Record the Digital Force Display reading in a table as in Table 1a.(iv) Repeat using masses of 200 g, 300 g, 400 g and 500 g.


4/15

RESULT:

Mass (g) Load (N) Experimental

shear force

(N)

Theoretical

shear force

(N)

Experimental

bending

moment (N)

Theoretical

bending

moment

(N)

0 0.98

100 1.96

200 2.94

300 3.92

400 4.90

500 0.98

Table 1b : Results for Experiment

DISCUSSION:

(i) Determine values and directions of vertical reaction, horizontal reaction and moment

(ii) Calculate shear force and bending moment in a beam(iii)Plot graph which compares your experimental results to those you calculated using theory

a) Load v shear force graph

b) Load v bending moment graph

CONCLUSION:

Comment on the shape of the graph. What does it tell us about how shear force varies due to an

increased load? Does the equation we used accurately predict the behavior of the beam?


5/15

RA RB

EXPERIMENT 1B : Shear Force and Bending Moment Variation for Various Loading Conditions

OBJECTIVES: This experiment examines how shear force varies at the cut position of the beam for

various loading conditions.

APPARATUS :

a) Computer set

b) Frame structure

c) Digital Force display

THEORY :We know that if a body or object of any sort is stationary, then the forces on it balance, as follows:

(1) Vertical equilibrium (total force up = total force down)

(2) Horizontal equilibrium (total force right = total force left)

(3) Moment equilibrium (total clockwise moment = total anticlockwise moment).

We will use the statement:

The Shear force at the cut is equal to the algebraic sum of the forces acting to the left or right

of the cut.

The Bending Moment at the cut is equal to the algebraic sum of the moments caused by the

forces acting to the left or right of the cut.

PROCEDUR:

(i) Check the Digital Force Display meter reads zero with no load.

(ii) Carefully load the beam with the hangers in the positions shown in Figure 1, using the loads

indicated in table 2a.

cut

140mm

W1

W1=3.92 N (400 g)

Figure 1 : Force diagram

(iii) Record the Digital Force Display reading as in a figure 2 and figure 3.

(iv) Repeat the procedure with the beam loaded as in figure


6/15

RA RB

RA RB

(v) Convert the mass into a load (in Nrefer table 1a)

cut140mm260mm

W1 W2

W1=1.96 N (200 g)W2=3.92 N (400 g)


cut240mm400mm

W1 W2

W1=4.91 N (500 g)W2=3.92 N (400 g)


RESULT:

Figure W1

(N)

W2

(N)

Force

(N)

RA (N) RB (N) Experiment

al shear

force (N)

Theoretical

shear force

(N)

Experimental

bending

moment (N)

Theoretical

bending

moment

(N)

Table 2a : Results for Experiment

DISCUSSION:

(i) Calculated the support reactions (RA and RB)

(ii) Calculated the theoretical shear force and bending moment at the cut.

CONCLUSION:

Comment on how the results of the experiments compare with those calculated using the theory.


7/15


8/15

TITLE : DEFLECTION AND SLOPE IN THE BEAM

EXPERIMENT 2a : Simply Supported Beam with a Point Load

OBJECTIVE : To compare the value of deflection and gradient (slope) from the test and theory for the

simply supported beam

THEORY :Macaulays methods

The application of a double integration method to a beam subjected to a discontinuous load leads to a

number of bending equations and their constants. The derivation of the deflection curve by this method is tedious

to say the least. We therefore use a step function which is more commonly known as Macaulys method. The

method of solution requires only one equation for the entire beam and thus only two constants of integration. The

step function is a function of x of the form f(x)=[x-a]nsuch that for x < a, f(x)=0 and for x >a, f(x)=(x-a)

n. The

important feature to note is that if the quantity inside the square brackets becomes negative we omit it from any

subsequent analysis. Care must be taken to retain the identity of the square bracket term under integration. Formathematical continuity where we have a distributed load we continue it to x=1 and superimpose additional

loadings which cancel out the extra load we added to the problem in order to obtain a solution. Three common

step functions for Bending Moment are shown below;

Mo

x = a x = a , M = Mo[x-a]0

W

x = a x = a , M = W [x-a]1

W

x = a x = a , M =

2

APPARATUS :

a) Two knife edge Supports

b) Dial gauge

c) Steel beam (25 mm x 6 mm)

d) One Load hangere) Weights


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Support A Support B

PROCEDURE:

(v) Placed both of supported A and B with a distance of 700mm as in the figure 1.

(vi) Placed a load hanger at the middle of test beam and put carefully on the support (make sure

the steel beam on both of support is balanced).

weight

350mm 350mm


(vii) Placed the dial gauge at the middle of the beam. Ensure that there are no weights in the

hangers.

(viii) Set the dial gauge to the zero.

(ix) Apply the loads in increase of 10 N up to a maximum of 20 N and in each case note the

reading of the dial indicator. Tabulate your results as the table 2a. Remember,

Experiment Deflection, = Dial gauge reading x 0.01 mm

Experiment Slope, =

x 0.01 radian

RESULT:

Load (N) Dial

gaugereading

Experimental

Deflection(mm)

Theoretical

Deflection(mm)

Experiment

Slope(radian)

Theoretical

slope(radian)

0

10

20

30

40

50

Table 2a : Results for Experiment

DISCUSSION:

Calculated deflection and slope theory using a macaulays method. Given Youngs Modulus for mild steel

E = 207 x 103N/mm

2and second moment area is I = bd

3/ 12 , where b = 25 mm and d = 6 mm.

CONCLUSION:

How well does your value of experiment compare with the value of theory.


10/15

TITLE : DEFLECTION AND SLOPE IN THE BEAM

EXPERIMENT 2b: Cantilever Beam with a Point Load

OBJECTIVE :To compare the value of deflection and gradient (slope) from the test and theory for the

cantilever beam

THEORY :Macaulays methods

The application of a double integration method to a beam subjected to a discontinuous load leads to a

number of bending equations and their constants. The derivation of the deflection curve by this method is tedious

to say the least. We therefore use a step function which is more commonly known as Macaulys method. The

method of solution requires only one equation for the entire beam and thus only two constants of integration. The

step function is a function of x of the form f(x)=[x-a]nsuch that for x < a, f(x)=0 and for x >a, f(x)=(x-a)

n. The

important feature to note is that if the quantity inside the square brackets becomes negative we omit it from any

subsequent analysis. Care must be taken to retain the identity of the square bracket term under integration. Formathematical continuity where we have a distributed load we continue it to x=1 and superimpose additional

loadings which cancel out the extra load we added to the problem in order to obtain a solution. Three common

step functions for Bending Moment are shown below;

Mo

x = a x = a , M = Mo[x-a]0

W

x = a x = a , M = W [x-a]1

W

x = a x = a , M =

2

APPARATUS :

a) Built in supports

b) Stirrup

c) Double end hooke

d) Dial gaugee) Steel beam (25 mm x 6 mm)

f) One Load hanger

g) Weights


11/15

Support

PROCEDURE:

(i) Placed the support at left side.

(ii) Put the test beam with a distance of 500mm as in the figure 2. Clip the test beam by tighten

screw at the support.

weight

500mm


(iii) Placed the load hanger and dial gauge at the free end of the beam. Ensure that there are no

weights in the hangers.

(x) Set the dial gauge to the zero.

(xi) Apply the loads in increase of 10 N up to a maximum of 20 N and in each case note the

reading of the dial indicator. Tabulate your results as the table 2b. Remember,

Experiment Deflection, = Dial gauge reading x 0.01 mm

Experiment Slope, =

x 0.01 radian

RESULT:

Load (N) Dial

gauge

reading

Experimental

Deflection

(mm)

Theoretical

Deflection

(mm)

Experiment

Slope

(radian)

Theoretical

slope

(radian)

0

10

20

30

40

50

Table 2b : Results for Experiment

DISCUSSION:

Calculated deflection and slope theory using a macaulays method. Given Youngs Modulus for mild steel

E = 207 x 103N/mm

2and second moment area is I = bd

3/ 12 , where b = 25 mm and d = 6 mm.

CONCLUSION:

How well does your value of experiment compare with the value of theory.


12/15

CIVIL ENGINEERING DEPARTMENTPOLYTECHNIC OF SULTAN AZLAN SHAH

STRUCTURE LABORATORY

CC 205

TITLE TENSILE TEST

EXPERIMENT 3

EXPERIMENT 3

DATE OF

PRACTICAL

DUE DATE

LECTURER NAME

NAME OF

GROUP

MEMBERS

NUM NAME REG. NO

1

2

3

4

5

6

7

8

9

10

LECTURERS

COMMENT

NUM MARKING CRITERIAL GIVEN MARK

1 Practical Work Title (5%)

2 Objective (5%)

3 Theory (5%)

4 Apparatus / Equipment (5%)

5 Procedures (5%)

6 Data (5%)7 Analysis / Discussion (5%)

8 Comments / Conclusion (5%)

9 Tidy (5%)

10 Reference (5%)

Report (50%)

1 Attend for practical (25%)

2 Leadership / cooperation (25%)

Practical (50%)

TOTAL MARK (100 %)


13/15

TITLE : TENSILE TEST

EXPERIMENT : TENSILE TEST

OBJECTIVE :To determine the stress, strain and young modulus and behavior of mild steel bar when

subjected to an axial tensile test.

EQUIPMENT/ APPARATUS:

a) Universal testing machineb) Steel reinforcing bars / specimenc) Cut-off machined) Measurement apparatus such as ruler, divider, etc.

THEORY :One of the simplest tests for determining mechanical properties of a material is the tensile test.

In this test, a load is applied along the longitudinal axis of a circular test specimen (Figure 1). The applied

load and the resulting elongation of the member are measured. The resulting stress-strain curve ordiagram gives a direct indication of the material properties(Figure 2).

BD

A

E

F

Stress (f)

Strain ()Figure 2

AProportional limit

BElastic limit

CPlastic limit / upper yield limit

DYield limit / Lower yield limit

EUltimate stress

FBreaking point

C

ELASTIC

PLASTIC

P P

Figure 1

L L


14/15

This curve is typical of class of steel known as mild steel. The relationship between stress and

strain is linear up toproportional limit; the material is said to follow Hookes Lawuntil pointA, namely

proportional limit. After this point, Hookes Lawwouldnt happen although the material has an elastic

characteristic. The Bpoint called Elastic Limit. The elastic limitof the material is a stress that liesbetween theproportional limitand the upper yield point (point C).Up to this stress, the specimen can be

unloaded without permanent deformation; the unloading will be along the linear portion of diagram,

the same path followed during loading. This part of the stress-strain diagram is called the elastic range.

However, the elastic limit (Point B)is seldom determined, since it is very close to theproportional limit

(Point A)and therefore rather difficult to detect.

A peak value, the upper yield point (point C), is quickly reached after that, followed by leveling

off at the lower yield point (Point D). At this stage of loading, the test specimen continues to elongate as

long as the load is not removed, even though the load cannot be increased. This constant stress region is

called theplastic range.

When a further load is applied to the specimen, the curve will rise continuously but became

flatter until it reaches a maximum stress referred as the ultimate stress (Point E).Throughout the test,

while the specimen is elongating, its cross-sectional area will decrease. At the ultimate stress (Point E),

the cross-sectional area begins to decrease in a localized region of the specimen, instead of over the

specimens entire gage length. This phenomenon is caused by slip planes formed within the material,

and the actual strains produced are caused by shear stress. As a result, a constriction or neck gradually

tends to form in this region.

Since the cross-sectional area within this region is continually decreasing, the smaller area can only carry

an ever-decreasing load. Hence, the stress-strain diagram tend to curve downward until the specimenbreaks at thefracture stress (Point F).

PROSEDURE:

1. Cut the specimen 40 mm long using the cut-off machine

2. Install the electronic extensometers into the specimens (the extensometer has to be

fitted centrally on the specimens)

3. Place centrally the specimen on the grips of the testing machine

4. Lock the specimens by tightening the upper and lower clamps

5. Insert the cable into the electronic extensometer

6. Switch ON the pump and move up the first piston, move the second piston toward the

motor and move up the last piston.

7. Once the specimen failed, switch OFFthe pump and move the piston back to the

normal position

8. The deformation was transferred automatics to computer set

9. Record the stress and strain value into the table.


15/15

RESULT:

a- Specimen diameter = ___________ mm

b- Reduction of area, (A) = ____________mm2

c- Specimen Gauge Length (L) = ____________mm

Load Cell Reading

(KN)

Stroke Reading

Change in Length, L

(mm)

Stress, ()

(N/mm2)

Strain, ()

( mm/mm)

0 0.98

100 1.96

200 2.94

300 3.92

400 4.90

500 0.98

Table 3 : Results for Experiment

DISCUSSION:

1. From the table plot the graph of stress () verses strain ().

2. From the graph mark the ;

i. Upper yield point/Plastic limit

ii. Lower yield point/Yield limit

iii. Point of maximum load/Ultimate stress and

iv. Breaking point

3. Determine the slope of the graph and the point where the graph starts to be non linear.

Elastic Modulus, (E) =

Youngs modulus () = Slope

= Stress ()

Strain ()

CONCLUSION:

Comment on the shape of the graph. What does it tell us about how shear force varies due to an

increased load? Does the equation we used accurately predict the behavior of the beam?

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