Add to collection(s) Add to saved
  1. No category
Uploaded by Namomo Eria

MECHANICAL ENG SCIENCE

advertisement 
LECTURE NOTES.
STRESS AND STRAIN.
Stress
It’s the resistance per unit area offered by a body against deformation.
Direct stress.
It’s the force acting per unit cross-sectional area.
π‘™π‘œπ‘Žπ‘‘
𝐹
π‘ π‘‘π‘Ÿπ‘’π‘ π‘  = π‘Žπ‘Ÿπ‘’π‘Ž = 𝐴
The derived unit of stress is 𝑁/π‘š2 (Pascal).
Load.
It’s the external force acting on the body.
Types of stresses
1. Normal stress. (Ο­)
It’s the stress that acts in a direction normal to the area.
(a) Tensile stress
It’s the stress induced in a body when subjected to two equal and opposite pulls.
Tensile strain is the ratio of increase in length to original length.
(b) Comprehensive stress
It’s the stress induced in a body when subjected to two equal and opposite pushes.
Comprehensive strain is the ratio of the decrease in length to original length.
2. Shear stress, (τ)
It’s the stress induced in a body when subjected to two equal and opposite forces which are
acting tangentially across the resisting section.
Strain
It’s the extension per unit of original length of a bar.
1|Page
Strain is produced by;
(i)
Application of a load.
(ii)
Change of temperature unaccompanied by load.
π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› =
π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘™π‘’π‘›π‘”π‘‘β„Ž π‘₯
=
π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘™π‘’π‘›π‘”π‘‘β„Ž
𝑙
MODULUS OF ELEASTICITY / YOUNG’S MODULUS, (E)
It’s the ratio of tensile stress to corresponding tensile strain.
Modulus of rigidity/ shear modulus.
It’s the ratio of shear stress to shear strain within the elastic limit.
π‘ β„Žπ‘’π‘Žπ‘Ÿ π‘ π‘‘π‘Ÿπ‘’π‘ 
𝜏
𝐺 = π‘ β„Žπ‘’π‘Žπ‘Ÿ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› = πœ‘
Factor of safety
It’s the ratio of ultimate tensile stress to working/ permissible stress.
π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ π‘œπ‘“ π‘ π‘Žπ‘“π‘’π‘‘π‘¦ =
π‘’π‘™π‘‘π‘–π‘šπ‘Žπ‘‘π‘’ π‘ π‘‘π‘Ÿπ‘’π‘ π‘ 
π‘π‘’π‘Ÿπ‘šπ‘–π‘ π‘ π‘–π‘π‘™π‘’ π‘ π‘‘π‘Ÿπ‘’π‘ π‘ 
OTHER TERMS.
Longitudinal strain/ linear strain
It’s the ratio of axial deformation to original length of the body.
π‘™π‘œπ‘›π‘”π‘–π‘‘π‘’π‘‘π‘–π‘›π‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› =
π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘™π‘’π‘›π‘”π‘‘β„Ž
π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘™π‘’π‘›π‘”π‘‘β„Ž
Lateral strain
It’s the strain at right angles to the direction of applied load.
Poisson’s ratio
2|Page
It’s the ratio of lateral strain to longitudinal strain.
µ=
π‘™π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›
π‘™π‘œπ‘›π‘”π‘–π‘‘π‘’π‘‘π‘–π‘›π‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›
COMPOUND/ COMPOSITE BAR.
It’s a bar made-up of two or more bars of equal lengths but of different material rigidly fixed with
each other behaving as a unit for extension or compression when subjected to an axial tensile or
compressive loads.
For composite bars;
1. The total external load on the composite bar is equal to the sum of the loads carried by each
different material.
2. The extension or compression in each bar is equal.
Consider a composite bar made up of two different materials.
Let:
𝑝 = π‘‘π‘œπ‘‘π‘Žπ‘™ π‘™π‘œπ‘Žπ‘‘ π‘œπ‘› π‘‘β„Žπ‘’ π‘π‘œπ‘šπ‘π‘œπ‘ π‘–π‘‘π‘’ π‘π‘Žπ‘Ÿ.
𝑙 = π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘π‘œπ‘šπ‘π‘œπ‘ π‘–π‘‘π‘’ π‘π‘Žπ‘Ÿ π‘Žπ‘›π‘‘ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘π‘Žπ‘Ÿ π‘œπ‘“ π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘ π‘šπ‘Žπ‘‘π‘’π‘Ÿπ‘–π‘Žπ‘™.
𝐴1 = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Ÿπ‘œπ‘ π‘  − π‘ π‘’π‘π‘‘π‘–π‘œπ‘› π‘π‘Žπ‘Ÿ 1.
𝐴2 = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Ÿπ‘œπ‘ π‘  − π‘ π‘’π‘π‘‘π‘–π‘œπ‘› π‘π‘Žπ‘Ÿ 2.
𝑃1 = π‘™π‘œπ‘Žπ‘‘ π‘ β„Žπ‘Žπ‘Ÿπ‘’π‘‘ 𝑏𝑦 π‘π‘Žπ‘Ÿ 1.
𝑃2 = πΏπ‘œπ‘Žπ‘‘ π‘ β„Žπ‘Žπ‘Ÿπ‘’π‘‘ 𝑏𝑦 π‘π‘Žπ‘Ÿ 2.
𝐸1 = π‘¦π‘œπ‘’π‘›π‘”π‘  π‘šπ‘œπ‘‘π‘’π‘™π‘’π‘  π‘œπ‘“ π‘π‘Žπ‘Ÿ 1
𝐸2 = π‘¦π‘œπ‘’π‘›π‘”π‘  π‘šπ‘œπ‘‘π‘’π‘™π‘’π‘  π‘œπ‘“ π‘π‘Žπ‘Ÿ 2
𝜎1 = π‘†π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛𝑑𝑒𝑐𝑒𝑑 𝑖𝑛 π‘π‘Žπ‘Ÿ 1
𝜎2 = π‘†π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛𝑑𝑒𝑐𝑒𝑑 𝑖𝑛 π‘π‘Žπ‘Ÿ 2
Now the total load on the composite bar is the sum of the load carried by the two bars.
3|Page
𝑃 = 𝑃1 + 𝑃2 … … … … … … … … … … … … … … … . . (𝑖)
Stress in bar 1
𝜎1 =
𝑃1
𝐴1
AND
𝑃1 = 𝜎1 𝐴1
Stress in bar 2.
𝜎2 =
𝑃2
𝐴2
And
𝑃2 = 𝜎2 𝐴2
Substituting in equation (i)
𝑃 = 𝜎1 𝐴1 + 𝜎2 𝐴2 … … … … … … … … … … … … … … … … … (𝑖𝑖)
Since the ends of the bars are rigidly connected, each bar will change in length by the same amount.
Now
π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝑖𝑛 π‘π‘Žπ‘Ÿ 1 =
𝜎1
𝐸1
π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝑖𝑛 π‘π‘Žπ‘Ÿ 2 =
𝜎2
𝐸2
AND
π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝑖𝑛 π‘π‘Žπ‘Ÿ 1 = π‘†π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝑖𝑛 π‘π‘Žπ‘Ÿ 2
𝜎1
𝐸1
𝜎
= 𝐸2……………………………………………………………….(iii)
2
From equation (ii) and (iii) the stress in each bar can be calculated.
THERMAL STRESSES.
4|Page
These are stresses induced in a body due to change in temperature.
Thermal stresses are setup in the body when the temperature of the body is raised or lowered and
the body is not allowed to expand or contract freely, no stresses will be set up in the body.
Consider that is heated to a certain temperature,
𝑙 = π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘™π‘’π‘›π‘”π‘‘β„Ž
𝑇 = π‘Ÿπ‘Žπ‘–π‘ π‘’ π‘–π‘›π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’
𝐸 = π‘¦π‘œπ‘’π‘›π‘”π‘  π‘šπ‘œπ‘‘π‘’π‘™π‘’π‘ 
𝛼 = π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘›
𝑑𝑙 = 𝑒π‘₯π‘‘π‘’π‘›π‘ π‘–π‘œπ‘› π‘œπ‘“ π‘‘β„Žπ‘’ π‘Ÿπ‘œπ‘‘ 𝑑𝑒𝑒 π‘‘π‘œ π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘Ÿπ‘–π‘ π‘’.
If the rod is free to expand, then extension of the rod is given by;
𝑑𝑙 = 𝛼𝑇. 𝑙
Suppose an external load is applied so that the rod is decreased in length from
(𝑙 + 𝛼𝑇. 𝑙) to 𝑙
Now
π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘–π‘£π‘’ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› =
=
π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ 𝑖𝑛 π‘™π‘’π‘›π‘”π‘‘β„Ž
π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘™π‘’π‘›π‘”π‘‘β„Ž
𝛼𝑇. 𝑙
𝛼𝑇. 𝑙
=
= 𝛼𝑇
𝑙 + 𝛼𝑇. 𝑙
𝑙
But
π‘ π‘‘π‘Ÿπ‘’π‘ π‘  = π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝑋 𝐸
= 𝛼𝑇. 𝐸
And
5|Page
π‘™π‘œπ‘Žπ‘‘ = π‘ π‘‘π‘Ÿπ‘’π‘ π‘  𝑋 π‘Žπ‘Ÿπ‘’π‘Ž
= 𝛼𝑇. 𝐸. 𝐴
Thermal stress is also referred to as temperature stress and thermal strain is also referred to as
temperature strain.
THERMAL STRESSES IN COMPOSITE BARS
Consider a composite bar consisting of two members, a bar of brass and another of steel. Let the
composite bar be heated through some temperature.
If the members are free to expand, then no stress will be induced in the members. The two members
are rigidly fixed hence the composite bar will expand as a whole by the same amount.
Since the coefficient of linear expansion of brass is more than that of steel hence the expansion of
brass will be more than that of steel but both members are not free to expand so the expansion of
the composite bar as a whole will be less than that of brass but more than that of steel hence the
stresses induced in brass will be compressive and tensile in steel.
Now
π‘™π‘œπ‘Žπ‘‘ π‘œπ‘› π‘π‘Ÿπ‘Žπ‘ π‘  = π‘ π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛 π‘π‘Ÿπ‘Žπ‘ π‘  𝑋 π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Ÿπ‘Žπ‘ π‘ 
= πœŽπ‘ 𝐴𝑏
π‘Žπ‘›π‘‘ π‘™π‘œπ‘Žπ‘‘ π‘œπ‘› 𝑠𝑑𝑒𝑒𝑙 = π‘ π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛 𝑠𝑑𝑒𝑒𝑙 𝑋 π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑠𝑑𝑒𝑒𝑙
= πœŽπ‘  𝐴𝑠
π‘™π‘œπ‘Žπ‘‘ π‘œπ‘› π‘π‘Ÿπ‘Žπ‘ π‘  = π‘™π‘œπ‘Žπ‘‘ π‘œπ‘› 𝑠𝑑𝑒𝑒𝑙
πœŽπ‘ 𝐴𝑏 = πœŽπ‘  𝐴𝑠
Actual expansion of steel = actual expansion of brass
π‘Žπ‘π‘‘π‘’π‘Žπ‘™ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› π‘œπ‘“ 𝑠𝑑𝑒𝑒𝑙
= π‘“π‘Ÿπ‘’π‘’ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› π‘œπ‘“ 𝑠𝑑𝑒𝑒𝑙 + 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› 𝑑𝑒𝑒 π‘‘π‘œ 𝑑𝑒𝑛𝑠𝑖𝑙𝑒 π‘ π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛 𝑠𝑑𝑒𝑒𝑙
𝜎
𝛼𝑠 𝑇. 𝑙 + 𝐸𝑠 . 𝑙…………………………………………………………………(ii)
𝑠
6|Page
And
actual
expansion
in
copper= π‘“π‘Ÿπ‘’π‘’ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› 𝑖𝑛 π‘π‘œπ‘π‘π‘’π‘Ÿ −
π‘π‘œπ‘›π‘‘π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› 𝑑𝑒𝑒 π‘‘π‘œ π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘–π‘£π‘’ π‘ π‘‘π‘Ÿπ‘’π‘ π‘  𝑖𝑛𝑑𝑒𝑐𝑒𝑑 𝑖𝑛 π‘π‘Ÿπ‘Žπ‘ π‘ 
= 𝛼𝑏 𝑇. 𝑙 −
πœŽπ‘
. 𝑙 … … … … … … … … … … … … … … … … … … … … … . (𝑖𝑖𝑖)
𝐸𝑏
Equating eqn (ii) and (iii) gives.
𝛼𝑠 𝑇 +
πœŽπ‘ 
πœŽπ‘
= 𝛼𝑏 𝑇 −
𝐸𝑠
𝐸𝑏
Trial questions.
THERMAL STRAIN
Thermal strain is caused by the change in the temperature of a material.
The change in temperature t in a bar of length l will cause it to extend by an amount;
π‘₯ = 𝛼𝑑𝑙
Where α is the coefficient of thermal expansion of the material.
Thermal strain=
π‘₯
𝑙
= 𝛼𝑑
Note
Values of coefficient of thermal expansion of;
(i)
Carbon steel is 12𝑋10−6 /0 𝐢
(ii)
Austenitic steel is 18𝑋10−6 /0 𝐢
(iii)
Aluminum is 24𝑋10−6 /0 𝐢
(iv)
Copper is 17𝑋10−6 /0 𝐢
(v)
Cast iron is 10𝑋10−6 /0 𝐢
(vi)
Brass is 16𝑋10−6 /0 𝐢
7|Page
NB:
Both the thermal strain and the elastic strain may exist together, therefore the total stain is the sum
of the two.
𝓔 = πœΆπ’• +
Ο­
𝑬
And.Extension
= 𝓔𝒍
Ο­
= (πœΆπ’• + ) 𝒍
𝑬
Poisson’s ratio: Lateral strain (𝐯)
It’s the ratio of lateral strain to longitudinal strain.
𝑣=
π‘™π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›
π‘™π‘œπ‘›π‘”π‘–π‘‘π‘’π‘‘π‘–π‘›π‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›
Volumetric strain (change in volume)
When a bar is pulled, its length increases and its transverse dimensions decrease, and there is an
increase in volume.
Let the length of the bar be L and the cross-section to be square side B.
If Ο­ is the stress produced, then
π‘™π‘œπ‘›π‘”π‘–π‘‘π‘’π‘‘π‘–π‘›π‘Žπ‘™ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› 𝓔 =
8|Page
Ο­
𝑬
Strain energy: Resilience. (U)
This is the energy stored per unit volume in a strained bar.
π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ =
1
𝐹π‘₯
2
Where F is the force and x is the extension.
In terms of maximum stress,Ο­ where A is the cross-sectional area and l being the length;
𝐹 = ϭ𝐴
And
π‘₯=
ϭ𝑙
𝐸
THUS
π‘ˆ=
=
1
ϭ𝑙
𝑋(ϭ𝐴)𝑋
2
𝐸
=
=
1
𝐹π‘₯
2
Ο­2
𝑋 𝐴𝑙
2𝐸
Ο­2
π‘‹π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘Žπ‘Ÿ.
2𝐸
APPLICATION OF STRAIN ENERGY TO IMPACT AND SUDENLY APPLIED LOADS
1. IMPACT LOADS.
If a load is suddenly applied to a bar, as an impact, the bar stretches and behaves as a spring
oscillating about a mean position.
Assume a load of weight W falls through a height h on the collar at the end of a vertical
bar of length l, let x be the maximum instantaneous extension of the bar and Ο­ the
corresponding stress.
Then;
9|Page
π‘₯=
ϭ𝑙
𝐸
At the point of maximum extension;
π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘œπ‘“ π‘™π‘œπ‘Žπ‘‘ = π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘› π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ 𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿ
π‘Š(β„Ž + π‘₯) =
π‘Š (β„Ž +
Ο­2
𝑋𝐴𝑙
2𝐸
Ο­
Ο­2
𝑙) =
𝑋𝐴𝑙
𝐸
2𝐸
Assumptions made;
ο‚·
All connections except the bar are completely rigid.
ο‚·
The limit of proportionality of stress is not exceeded.
ο‚·
There is no loss of energy at impact (mass of the bar is negligible)
ο‚·
The modulus of elasticity, E is the same for impulsive loading
2. SUDDENLY APPLIED LOADS.
If the load is placed in contact with the collar without impact and suddenly let go, h=0 and equating
potential energy to strain energy gives;
Ο­2
π‘Šπ‘₯ =
𝑋𝐴𝑙
2𝐸
(π‘Š
Ο­
Ο­2
𝑙) =
𝑋𝐴𝑙
𝐸
2𝐸
Ο­=2
π‘Š
𝐴
Hence the maximum stress produced by the suddenly applied load is twice that due to the same
load gradually applied.
The maximum instantaneous extension is also twice that of gradually applied load.
THIN WALLED PRESSURE VESSELS.
10 | P a g e
1. Hoop stress in a cylinder.
Consider a cylinder containing a fluid under pressure being subjected to a uniform radial pressure
normal to the walls.
There will be tensile or hoop stress setup in the circumferential direction.
There is also a uniform downward pressure p acting on the detrimental surface section.
This is balanced by the upward force due to the loop stress along the two edges.
Axial stress in a cylinder.
For the cylinder or closed pipe, there is axial stress in addition to hoop stress.
If the pressure p is acting on the cross-section, the axial force is given by;
= π‘πœ‹π‘Ÿ 2
This force is balanced by the force due to axial stress and it’s given by;
= π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ 𝑋 π‘‘β„Žπ‘–π‘π‘˜π‘›π‘’π‘ π‘ 
= 2πœ‹π‘Ÿπ‘‘
Hence;
π‘πœ‹π‘Ÿ 2 = Ο­π‘Ž 2πœ‹π‘Ÿπ‘‘
Ο­π‘Ž =
And Ο­β„Ž =
π‘π‘Ÿ
2𝑑
π‘π‘Ÿ
𝑑
Hence
1
Ο­π‘Ž = Ο­β„Ž
2
The axial stress is half the hoop stress.
11 | P a g e
TANGENTIAL STRESSES
If a thin spherical shell is subject to internal pressure p, a tensile stress is setup in the shell wall
due to the tendency of the shell to expand under pressure.
Consider the forces acting on the half shell;
1. The diametral force due to pressure
= π‘πœ‹π‘Ÿ 2
2. The resisting force due to tangential stress
ϭ𝑑 . 2πœ‹π‘Ÿπ‘‘
Equating the two gives;
ϭ𝑑 =
π‘π‘Ÿ
2𝑑
ROTATING RIMS
If F is the inertia force;
Then
𝐹 = π‘šπœ”2 π‘Ÿ
Where m is the mass of the element.
Now, if 𝜌 is the density of the material and a is the cross-sectional area, then;
π‘š = 𝜌. 𝐴𝐡. π‘Ž
= πœŒπ‘Ÿπœƒπ‘Ž
12 | P a g e
Hence;
𝐹 = πœŒπ‘Ÿπœƒπ‘Ž. πœ”2 π‘Ÿ
= πœŒπœƒπ‘Žπœ”2 π‘Ÿ 2
The element is maintained in equilibrium by the inertia force F and by tangential forces T at A and
B exerted by the material of the ring at these points.
From the force diagram;
π‘‡πœƒ = 𝐹 = πœŒπœƒπ‘Žπœ”2 π‘Ÿ 2
Thus;
𝑇 = πœŒπ‘Žπœ”2 π‘Ÿ 2
If
Ο­ is the stress set up due to T, then;
𝑇 = Ο­π‘Ž
= πœŒπ‘Žπœ”2 π‘Ÿ 2
Thus;
Ο­ = πœŒπœ”2 π‘Ÿ 2 = 𝑝𝑣 2
Where v is the linear speed of the rim and r is the mean radius.
13 | P a g e
Related documents
Differentiated Table For Types of Energy[1]
Differentiated Table For Types of Energy[1]
aaMaterial Science Homework 2
aaMaterial Science Homework 2
Multiple-Choice Test Chapter 01.06 Propagation of Errors
Multiple-Choice Test Chapter 01.06 Propagation of Errors
3 b Roles Analysis 1 (1)
3 b Roles Analysis 1 (1)
Status
Status
An MS Word format deposit sheet for fungal strains.
An MS Word format deposit sheet for fungal strains.
Download
advertisement