MM222 Lec 13-15

Hafiz Kabeer Raza Research Associate

Faculty of Materials Science and Engineering, GIK Institute Contact: Office G13, Faculty Lobby

[email protected], [email protected], 03344025392

MM222

Strength of Materials

Lecture – 13

Spring 2015

Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials

Relative end displacement • The previous examples had one

end attached to a fixed support. In each case, therefore, the deformation δ of the rod was equal to the displacement of its free end

• What will be the δ of the bar of both ends are moving?

• δ𝐵/𝐴 = δ𝐵 − δ𝐴 =𝑃𝐿

𝐴𝐸

• Example: compute, how much point B will move downward? If the applied load P was 10 kN. For L = 1.0 m, A = 2.5 mm2, and E = 200 GPa δB/A = 20 mm δA = 10 mm So δB = 30 mm

• Solve the same problem for δB = 10 mm upward and this time compute the magnitude of P.

• Conclusion: whatever the configuration and design of the structure is the deformations of the members are consistent same as compared to each other


Relative end displacement • In all these related problems, 1st, compute the overall deformation by

considering that opposite ends are not moving

• This was the case; For one of the components, Both ends were moving, Rest were moving from one end only (ignoring the signs of the deformations, weather it is compressive or tensile)

In this case

δ𝐵/𝐴 = δ𝐵 − δ𝐴 =𝑃𝐿

𝐴𝐸

Another thing; making the free body diagram of the member undergoing relative deformation, will result • If one component, for example, is under tension, the others will be under compression

• And opposite is true for vice versa

• But all the forces are in same direction, and these forces will be balanced by reactions at the supports

• A new case of relative end displacement (Sample problem 2.2) When there is no support/reaction Now the forces will be opposite to each other Here you will use a negative sign for one of the deformations….. OR

δ𝐴/𝐵 = δ𝐵 + δ𝐴 =𝑃𝐿

𝐴𝐸


Sample Problem 2.2

• In order to calculate the stress, We need to find out the force on EF

• Assume a force P on EF and make free body diagram.

• Compute the relative displacement in the rod and make a relation with the deformations of individual components

• Put the formulae of individual displacements and solve for P

• 0.025 = δ1 + δ2

• δ2 = 7.03 E-7 P

• δ1 = 6.41 E-7 P

• P = 18.6 ksi


Quizzes • Alternate Thursdays, It is 5th week now

• Each quiz will held; Venue: LH4 FMSE Time: 2:00 PM

Academic Week #

Quiz # Date (Thursday)

Remarks

3 1 25-02-2015 Done with date change

5 2 12-03-2015 Date and time updated

7 3 26-03-2015 -do-

9 4 16-04-2015 -do-

11 5 30-04-2015 -do-

13 6 14-05-2015 -do-

15 7 28-05-2015 -do-


Home work • Problems 2.19, 2.27, 2.28




MM222


Lecture – 14

Spring 2015


Static Indeterminacy • Structures for which internal forces and reactions

cannot be determined from statics alone are said

to be statically indeterminate.

0 RL

• Deformations due to actual loads and redundant

reactions are determined separately and then

added or superposed.

• Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

• A structure will be statically indeterminate

whenever it is held by more supports than are

required to maintain its equilibrium.


Example 2.04 Determine the reactions at A and B for the steel

bar and loading shown, assuming a close fit at

both supports before the loads are applied.

• Solve for the reaction at A due to applied loads

and the reaction found at B.

• Require that the displacements due to the loads

and due to the redundant reaction be compatible,

i.e., require that their sum be zero.

• Solve for the displacement at B due to the

redundant reaction at B.

SOLUTION:

• Consider the reaction at B as redundant, release

the bar from that support, and solve for the

displacement at B due to the applied loads.


Example 2.04 SOLUTION:

• Solve for the displacement at B due to the applied

loads with the redundant constraint released,

EEA

LP

LLLL

AAAA

PPPP

i ii

ii9

L

4321

2643

2621

34

3321

10125.1

m 150.0

m10250m10400

N10900N106000

• Solve for the displacement at B due to the redundant

constraint,

i

B

ii

iiR

B

E

R

EA

LPδ

LL

AA

RPP

3

21

262

261

21

1095.1

m 300.0

m10250m10400


Example 2.04 • Require that the displacements due to the loads and due to

the redundant reaction be compatible,

kN 577N10577

01095.110125.1

0

3

39

B

B

RL

R

E

R

E

• Find the reaction at A due to the loads and the reaction at B

kN323

kN577kN600kN 3000

A

Ay

R

RF

kN577

kN323

B

A

R

R


Self Study • Example 2.02, 2.03, Superposition method,

example 2.05


Thermal expansion and thermal stresses • A temperature change results in a change in length or

thermal strain. There is no stress associated with the

thermal strain unless the elongation is restrained by

the supports.

coef.expansion thermal

AE

PLLT PT

• Treat the additional support as redundant and apply

the principle of superposition.

0

0

AE

PLLT

PT

• The thermal deformation and the deformation from

the redundant support must be compatible.

TEA

P

TAEP


Self study

• Example 2.06


Problem 2.60




MM222


Lecture – 15

Spring 2015


Static Indeterminacy • If a material is not homogeneous in its cross

section, the load distribution will not be

homogeneous

• Example of a steel bar reinforced concrete

column


Problem 2.35

• The force will not be homogeneously distributed

• No. of steel rods = 4, rest is concrete

• Suppose steel is ‘1’ and concrete is ‘2’

• Statically indeterminate

• P = 4P1+P2

• Equating the two δ

• Solve for P1 and P2

• Calculate the stresses

• ζ1 = -15.82 ksi, ζ2 = -1.96 ksi


Inhomogeneous distribution of loads • Suppose two materials;

one is hard and the other is soft are joined together

• May be two boxes, may be two pipes, may be one rod and the other pipe, may be one or multiple rods and the other solid block


Home work • Problem 2.33 (-67.54 MPa, 0.29 mm)

• Problem 2.39 (answer given in the book)

• Problem 2.40 (RC = 9.74 kips, RA = 2.26 kips,

ζlower part = 3.10 ksi, ζupper part = 1.84 ksi)

• Problem 2.41 (answer given in the book)


Generalized Hooke’s Law

• What is v?

• What is ε?

• What is γ?

• What are E and G?


Problem 2.68

• Using the equations of Hooke’s Law for the particular dimension

• Use the definition of strain to find the change in length

• δAB = 10.2 μm, δBC = 2.4 μm, δAC = 8.91 μm


Problem 2.75

• The rubber is under shear stress (double shear)

• P = 25000/2 N, A = 150*100-mm

• η = 0.83 MPa,

• γ = 1.5/30 = 0.05

• G = η/γ = 16.67 MPa


Home work • Problem 2.51 (answer given)

• Problem 2.53 (answer given)

• Problem 2.76 (answer given)

• Problem 2.79 (0.048 in)


Problem 2.46

• ΣMF = 0 600*8 = 2FDE + 4FBC

2400 = FDE + 2FBC

• Compute δ for each

• Compute the relation between δ of BC and DE while F is not removed

• This will result 2.5 FDE = FBC

• Use the equation to solve

• FDE = 400 lb, FBC = 1000 lb

• Use the slope method to compute deflection of A

• 2.2E-3 in

Documents

MM222 Lec 13-15