A metallic bar measuring 300 mm x 100 mm x 40 mm is subjected to tensile forces of 5 kN, 4 kN, and 6 kN along the x, y, and z axes respectively. The document calculates the changes in length, breadth, depth, and volume of the bar due to these tensile forces. It uses the given values of Young's modulus (2x105 N/mm2) and Poisson's ratio (0.25) to determine the stresses in each direction and resulting strains. The calculations show the lengthening, narrowing, and shallowing of the bar under the triaxial tensile loading.

2. Q1. A metallic bar 300 mm x 100 mm x 40 mm is subjected to a force of 5 kN
(tensile), 4 kN (tensile) and 6 kN(tensile) along x, y and z directions respectively.
Determine the change in the length, breadth, depth and volume of the block. Take
E =2 xl05 N/mm2 and Poisson's ratio = 0.25.
Determine:
change in the Length of the block=
change in the Breadth of the block=
change in the depth of the block=
change in the volume of the block=

3. Length of the bar=L= 300 mm,
Width of the bar=b= 100 mm and
Depth of the bar = d = 40 mm
Volume of the bar=
Load in the direction of x = 5 kN = 5000 N
Load in the direction of y = 4 kN = 6000 N
Load in the direction of z = 6 kN = 4000 N
Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
Stress in the x-direction=
𝜎𝑥 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑏.𝑑
Stress in the y-direction=
𝜎 𝑦 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐿.𝑏
Stress in the z-direction=
𝜎𝑧 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐿.𝑑

4. Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
𝜎𝑥 =
𝜎 𝑦 =
𝜎𝑧 =
Volumetric strain for triaxial loading=
𝜺 𝒗 =
𝟏
𝑬
(𝟏-2 𝝂) (𝝈 𝒙 + 𝝈 𝒚 + 𝝈 𝒛 )

5. Length of the bar=L= 300 mm,
Width of the bar=b= 100 mm and
Depth of the bar = d = 40 mm
Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
𝜎𝑥 =
𝜎 𝑦 =
𝜎𝑧 =
Volumetric strain for triaxial loading=𝛿V
𝜀 𝑥 =
𝛿𝐿
𝐿
=
1
𝐸
(𝜎𝑥 - 𝜈 (𝜎 𝑦 + 𝜎𝑧) )
Change in length= 𝛿𝐿=

6. Length of the bar=L= 300 mm,
Width of the bar=b= 100 mm and
Depth of the bar = d = 40 mm
Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
𝜀 𝑦 =
𝛿d
d
=
1
𝐸
(𝜎 𝑦 - 𝜈 (𝜎𝑥 + 𝜎𝑧) )
𝜀 𝑧 =
𝛿𝑏
𝑏
=
1
𝐸
(𝜎𝑧 - 𝜈 (𝜎𝑥 + 𝜎 𝑦) )
Change in depth= 𝛿d=
Change in breadth= 𝛿b=

7. Q2. A metallic bar 250 mm x 100 mm x 50 mm is loaded as shown in Fig. Find the
change in volume. Take E =2 x 105N/mm2 and Poisson's ratio = 0.25. Also find the
change that should be made in the 4 MN load, in order that there should be no
change in the volume of the bar
Determine:
change in the volume of the bar= 𝛿V
change that should be made in the 4 MN load
so that 𝛿V = 0

8. Length of the bar=L= 250 mm,
Width of the bar=b= 100 mm and
Depth of the bar = d = 50 mm
Load in the direction of x = 400kN = 400* 103 N
Load in the direction of y = - 4 MN = - 4* 106 N
Load in the direction of z = 2 MN = 2* 106 N
Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
Stress in the x-direction=
𝜎𝑥 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑏.𝑑
Stress in the y-direction=
𝜎 𝑦 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐿.𝑏
Stress in the z-direction=
𝜎𝑧 =
𝐿𝑜𝑎𝑑 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝐿.𝑑

9. Value of E = 2 X 105 N/mm2
Poisson's ratio = 0.25
𝜎𝑥 =
𝜎 𝑦 =
𝜎𝑧 =
Volumetric strain for triaxial loading=
𝜺 𝒗 =
𝟏
𝑬
(𝟏-2 𝝂) (𝝈 𝒙 + 𝝈 𝒚 + 𝝈 𝒛 )

10. To make change in Volume as zero=
𝛿V = 0
𝟏
𝑬
(𝟏-2 𝝂) (𝝈 𝒙 + 𝝈 𝒚 + 𝝈 𝒛 ) *V = 0
(𝝈 𝒙 + 𝝈 𝒚 + 𝝈 𝒛 ) = 0
The stresses 𝝈 𝒙 and 𝝈 𝒛 are not to be changed. Only the stress corresponding to
the load 4 MN (i.e., stress in y direction 𝝈 𝒚 ) is to be changed.
𝝈 𝒚 = − (𝝈 𝒙+ 𝝈 𝒛 )

15. Prismatic bar in tension showing the stresses acting on cross
section mn: (a) bar with axial forces P, (b) three-dimensional
view of the cut bar showing the normal stresses, and (c) two-
dimensional view
❑ For axially loaded members,
the only stresses we
considered were the normal
stresses acting on cross
sections.
❑ There are no shear stresses
acting on the cut section,
because it is perpendicular to
the longitudinal axis of the bar.

16. Stress Elements
❑The most useful way of representing the
stresses in the bar of Fig is to isolate a
small element of material, such as the
element labeled “C” in Fig, and then
show the stresses acting on all faces of
this element. An element of this kind is
called a stress element.
❑The stress element at point “C” is a small
rectangular block with its right-hand face
lying in cross section mn.
Stress element at point C of the axially loaded bar shown in Fig.
(a) three-dimensional view of the element, and (b) two-
dimensional view of the element

17. stresses on Inclined sections when Bar
subjected to Uniaxial loading
Prismatic bar in tension showing the stresses acting on an
inclined section pq: (a) bar with axial forces P, (b) three-
dimensional view of the cut bar showing the stresses, and (c)
two-dimensional view
Procedure to find stresses on inclined plane:
1. Find the area of the cut section
2. Represent the stresses on the element
3. Represent forces on the element
4. Draw the free body diagram of forces
5. Take the force balance in a direction
normal to cutting plane
6. Take the force balance in a direction
tangential to cutting plane

18. Prismatic bar in tension showing the stresses
acting on an inclined section pq

20. Free body diagram of forces on the element:

24. Graph of normal stress 𝜎u and shear stress 𝜏u versus angle u of
the inclined section

25. Element A:
The maximum normal
stress
occurs at
𝜽 = 0o and is
These maximum shear
stresses occurs at
𝜽 = 45o have the
magnitude:

26. Principle of complementary Shear stress:
A set of shear stresses acting across a
plane will always be accompanied by a
set of balancing shear stresses of
similar intensity across the plane and
acting normal to it.

27. Stresses on Inclined sections when Bar subjected to Pure
torsion:
Stresses acting on a stress element cut from a bar in torsion

28. Analysis of stresses on inclined planes: (a) element in pure shear,
(b) stresses acting on a triangular stress element

30. Analysis of stresses on inclined planes:(c) forces acting on the
triangular stress element (free-body diagram)
=>

35. Graph of normal stresses 𝜎ϴ and shear stresses
𝜏ϴ versus angle ϴ of the inclined plane
We see that for ϴ= 0, which is the right-hand face of the stress element the
graph gives 𝜎ϴ= 0 and 𝜏 max = 𝜏
We see that for ϴ= 45o, which is the right-hand face of the stress element the
graph gives 𝜎ϴ= 𝜏 and 𝜏 max = 0

36. Stress elements oriented at 0o and 45° for pure
shear

39. Linear strain of the diagonal AC can approximately be taken as