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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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-
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Home work • Problems 2.19, 2.27, 2.28
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 – 14
Spring 2015
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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.
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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.
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Self Study • Example 2.02, 2.03, Superposition method,
example 2.05
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Self study
• Example 2.06
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Problem 2.60
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 – 15
Spring 2015
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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)
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Generalized Hooke’s Law
• What is v?
• What is ε?
• What is γ?
• What are E and G?
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
Home work • Problem 2.51 (answer given)
• Problem 2.53 (answer given)
• Problem 2.76 (answer given)
• Problem 2.79 (0.048 in)
Spring 2015 By Hafiz Kabeer Raza MM222 Strength of Materials
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