MEC32/A3

MECHANICS OF
DEFORMABLE BODIES

SANTIAGO, CHARL JOSEPH B.

2012107769
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STRENGTH OF MATERIALS

MECHANICS OF DEFORMABLE BODIES

- study of the relationship between externally applied loads and their internal effects on
rigid bodies
RIGID BODY
- bodies which neither change in shape and size after the application of forces
FREE BODY DIAGRAM
- Sketch of the isolated body showing all the forces on it.

THREE MAJOR DIVISIONS OF MECHANICS

1. Mechanics of Rigid Bodies – Engineering Mechanics
2. Mechanics of Deformable Bodies – Strength of Materials
3. Mechanics of Fluids - Hydraulics

Strength
- The strength of a material is its ability to withstand an applied stress without failure
- A material's strength is dependent on its microstructure.
- TWO CATEGORIES OF STRENGTH:
o Yield Strength
 the material begins deformation that
cannot be reversed upon removal of
the loading
 the stress level at which a material
begins to deform plastically
o Ultimate Strength
 the maximum stress
 the maxima of the stress-strain curve.
 the point at which necking will start
o Fracture Strength
 stress calculated immediately before
the fracture

- By setting up the equilibrium conditions, the inner forces of a member subjected to an

external load situation can be determined
 So far neither the material nor the type of cross section applied for the member
are being taken into account.
 Both material and type of cross section obviously have an impact on the
behavior of the member subjected to load.
- To design the member therefore a closer look on how the internal forces act along its
cross section needs to be taken.
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DIRECT OR NORMAL STRESS

- When a force is transmitted through a body, the body tends to change its shape or
deform. The body is said to be strained.

𝑨𝒑𝒑𝒍𝒊𝒆𝒅 𝑭𝒐𝒓𝒄𝒆 (𝑭)

𝑫𝒊𝒓𝒆𝒄𝒕 𝑺𝒕𝒓𝒆𝒔𝒔 =
𝑪𝒓𝒐𝒔𝒔 𝑺𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝑨𝒓𝒆𝒂𝒍 (𝑨)
Units: Usually N/m2 (Pa), N/mm2, MN/m2, GN/m2 or N/cm2
Note: 1 N/mm2 = 1 MN/m2 = 1 MPa

- may be tensile ( σt )or compressive (σc ) and result from forces acting perpendicular to
the plane of the cross-section

Tension

Compression

The normal stress acting along a section of a member only depends

on the external load applied (e.g. a normal force F) and the geometry
of its cross section A (true for statically determinant systems).
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EXAMPLE 103:
Determine the largest weight W which can be supported by the 2
wires shown. The stresses in wires AB and AC are not to exceed
100MPa and 150MPa, respectively. The cross-sectional areas of the
2 wires are 400mm2 for wire AB and 200mm2 for wire AC.
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EXAMPLE 104:
For the truss shown, calculate the stresses in members DF, CE and
BD. The cross-sectional area of each member is 1200mm2. Indicate
tension (T) or compression (C).
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EXAMPLE 110:
A steel tube is rigidly attached between an aluminum rod and
bronze as shown. Axial loads are applied at the positions indicated.
Find the maximum value of P that will not exceed a stress in
aluminum of 80MPa, in steel of 150MPa or in bronze of 100MPa.
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EXAMPLE 109:
Part of the landing gear for a light plane is shown in the figure.
Determine the compressive stress in the strut AB caused by a
landing reaction R=20KN. Strut AB is inclined at 53.1o with BC.
Neglect weights of the members.
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EXAMPLE 105:
For the truss shown, determine the cross-sectional areas of bars BE,
BF and CF so that the stresses will not exceed 100MN/m2 in tension
or 80MN/m2 in compression. A reduced stress in compression is
specified to avoid the danger of buckling.
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BEARING STRESS
- contact pressure between the separate bodies
- differs from compressive stress, as it is an internal stress caused by compressive forces

load W
shear stress  =
area resisting shear A
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EXAMPLE 123:
In the figure shown, assume that a 20mm diameter rivet joins the
plates which are each 100mm wide.
(a) If the allowable stresses are 140MN/m2 for bearing in the
plate material and 80MN/m2 for shearing of the rivet,
determine the minimum thickness of each plate.
(b) Under the conditions specified in part (a), what is the largest
average tensile stress in the plates?
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EXAMPLE 126:
In the clevis shown in the figure, determine the minimum bolt
diameter and the minimum thickness of each yoke that will support
a load P=55kN without exceeding a shearing stress of 70MN/m2 and
a bearing stress of 140MN/m2.
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SHEARING STRESS
- produced by equal and opposite parallel forces not in line
- The forces tend to make one part of the material slide over the other part
- tangential to the area over which it acts
- a measure of the internal resistance of a material to an externally applied shear load
load W
shear stress  =
area resisting shear A
Area
Resisting
Shear

Shear Force

Shear Force
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EXAMPLE 123:
As in the figure shown, a hole is to be punched out of a plate having
an ultimate shearing stress of 300MPa.
(a) If the compressive stress in the punch is
limited to 400MPa, determine the
maximum thickness of plate from which the
hole 100mm in diameter can be punched.
(b) If the plate is 10mm thick, compute the
smallest diameter hole which can be
punched.
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EXAMPLE 115:
The end chord of a timber truss is framed into the bottom chord as
shown in the figure. Neglecting friction,
(A) compute dimension b if the allowable shearing stress is 900KPA;
(B) Determine dimension c so that the bearing stress does not
exceed 7MPa
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EXAMPLE 119:
The mass of the homogeneous bar AB shown in the figure is 2000Kg.
The bar is supported by a pin at B and a smooth vertical surface at A.
Determine the diameter of the smallest pin which can be used at B if
its shear stress is limited to 60MPa. The detail of the pin support at B
is identical to that of the pin support at D shown in the figure.
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THIN-WALLED CYLINDER
- A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces,
which resist bursting, developed across longitudinal and transverse sections.
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EXAMPLE 132:
• A cylindrical pressure vessel is fabricated from steel plates which
have a thickness of 20mm. The diameter of the pressure vessel is
500mm and its length is 3m. Determine the maximum internal
pressure which can be applied if the stress in the steel is limited to
140MPa.
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EXAMPLE 134:
A water tank is 8m in diameter and 12m high. If the tank is to be
completely filled, determine the minimum thickness of the tank
plating if the stress is limited to 40MPa.
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EXAMPLE 135:
The strength per meter of the longitudinal joint in the figure is
480kN, whereas for the girth joint it is 200kN. Determine the
maximum diameter of the cylindrical tank if the internal pressure is
1.5MN/m2.
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EXAMPLE 139:
The tank shown in the figure is fabricated from 10mm steel plate.
Determine the maximum longitudinal and circumferential stresses
caused by an internal pressure of 1.2MPa
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STRESS STRAIN ANALYSIS

SIMPLE STRAIN (unit deformation)
- the ratio of the change in length caused by the applied force, to the original length
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑳𝒆𝒏𝒈𝒕𝒉 (𝒅𝒍)
𝑫𝒊𝒓𝒆𝒄𝒕 𝑺𝒕𝒓𝒆𝒔𝒔 (𝜺) =
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝑳𝒆𝒏𝒈𝒕𝒉 (𝑳)
where dl is the deformation & L is the original length, thus is dimensionless
DIRECT OR NORMAL STRAIN
- When loads are applied to a body, some
deformation will occur resulting to a
change in dimension.
- Consider a bar, subjected to axial tensile
loading force, F.

- As strain is a ratio of lengths, it is dimensionless.

- Similarly, for compression by amount, dl: Compressive strain = - dl/L
- Note: Strain is positive for an increase in dimension and negative for a reduction in
dimension.
DEFORMATIONS OF MEMBERS UNDER AXIAL LOADING
Consider a homogeneous rod BC of length L and uniform cross section of area A
subjected to a centric axial load P.
If the resulting axial stress does not exceed the proportional limit of the material,
𝑷
𝝈=
𝑨
we may apply Hooke's law and write 𝝈 = 𝑬𝝐
𝝈 𝑷
from which it follows that 𝝐= =
𝑬 𝑨𝑬
Recalling that the strain 𝝐 was defined as 𝝐 = 𝜹/𝑳
we have 𝜹 = 𝝐𝑳 and substituting for 𝝐
𝑷𝑳
𝜹=
𝑨𝑬
The equation may be used only if the rod is homogeneous, has a uniform cross section
of area A, and is loaded at its ends.
If the rod is loaded at other points, or if it consists of several portions of various cross
sections and possibly of different materials, we must divide it into component parts
that satisfy individually the required conditions for the application of the formula.
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EXAMPLE 203:
During a stress-strain test, the unit deformation at a stress of 35MPa
was observed to be 167x10-6m/m and at a stress of 140MPa it was
667x10-6m/m. If the proportional limit was 200MPa, what is the
modulus of elasticity? What is the strain corresponding to a stress of
80MPa?
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EXAMPLE 205:
A steel rod having a cross-sectional area of 300mm2 and a length of
150m is suspended vertically from one end. It supports a load of
20KN at the lower end. If the unit mass of steel is 7850kg/m3 and
E=200x103MPa, find the total elongation of the rod.
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EXAMPLE 206:
A steel wire 10m long hanging vertically supports a tensile load of
2000N. Neglecting the weight of the wire, determine the required
diameter if the stress is not to exceed 140MPa and the total
elongation is not to exceed 5mm. Assume E=200GPa.
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EXAMPLE 208:
An aluminum bar having a cross-sectional area of 160mm2 carries
the axial loads at the positions shown in the figure. If E=70GPa,
compute the total deformation of the bar. Assume that the bar is
suitably braced to prevent buckling.
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EXAMPLE 213:
The rigid bar AB, attached to two vertical rods as shown in the
figure, is horizontal before the load is applied. If the load P=50KN,
determine is vertical movement.
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EXAMPLE 2.17:
Two solid cylindrical rods are joined at B and loaded as shown. Rod
AB is made of steel (E=200GPa) and rod BC of brass (E=105GPa).
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
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EXAMPLE 211:
The rigid bars shown in the figure are separated by a roller at C and
pinned at A and D. a steel rod at B helps support the load of 50KN.
Compute he vertical displacement of the roller at C.
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EXAMPLE 214:
The rigid bars AB and CD shown in the figure are supported by pins
at A and C and the 2 rods. Determine the maximum force P which
can be applied as shown if its vertical movement is limited to 5mm.
Neglect the weights of all members.
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POISSON’S RATIO ()

- When a bar is subjected to a tensile loading there is an increase in length of the bar in
the direction of the applied load, but there is also a decrease in a lateral dimension
perpendicular to the load.
- The ratio of the sidewise deformation (or strain) to the longitudinal deformation (or
strain)
- For most steel, it lies in the range of 0.25 to 0.3, and 0.20 for concrete.

𝛆𝒚 𝛆𝒛
= − =−
𝛆𝒙 𝛆𝒙
- where ε𝑥 is strain in the x-direction
ε𝑦 and ε𝑧 are the strains in the perpendicular direction
The negative sign indicates a decrease in the transverse dimension when 𝜀𝑥 is positive

BIAXIAL DEFORMATION
- If element is subjected simultaneously by tensile stresses, 𝜎𝑥 &𝜎𝑦 , in 𝑥 & 𝑦 directions,
 the strain in the 𝑥 direction is 𝝈𝒙 /𝑬
 the strain in the 𝑦 direction is 𝝈𝒚 /𝑬
 Simultaneously, the stress in the 𝑦 direction will produce a lateral contraction
on the 𝑥 direction of the amount −𝝂𝜺𝒚 or −𝝂𝝈𝒚 /𝑬 .
 The resulting strain in the 𝑥 direction will be
𝝈 𝝂𝝈𝒚 (𝜺𝒙 +𝝂𝜺𝒚 )𝑬
𝜺𝒙 = 𝒙 − − 𝝈𝒙 =
𝑬 𝑬 𝟏−𝝂𝟐
𝝈𝒚 𝝂𝝈𝒙 (𝜺𝒚 +𝝂𝜺𝒙 )𝑬
𝜺𝒚 = −− 𝝈𝒚 =
𝑬 𝑬 𝟏−𝝂𝟐

TRIAXIAL DEFORMATION
• If an element is subjected simultaneously by three mutually perpendicular normal
stresses 𝜎𝑥 , 𝜎𝑦 , and 𝜎𝑧 , which are accompanied by strains 𝜀𝑥 , 𝜀𝑦 , and 𝜀𝑧 , respectively,
𝟏
𝜺𝒙 = [𝝈 − 𝝂(𝝈𝒚 + 𝝈𝒛 )]
𝑬 𝒙
𝟏
𝜺𝒚 = [𝝈𝒚 − 𝝂(𝝈𝒙 + 𝝈𝒛 )]
𝑬
𝟏
𝜺𝒛 = [𝝈𝒛 − 𝝂(𝝈𝒙 + 𝝈𝒚 )]
𝑬
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EXAMPLE 223:
A rectangular aluminum block is 100mm long in the X direction,
75mm wide in the Y direction and 50mm thick in the Z direction. It is
subjected to a triaxial loading consisting of a uniformly distributed
tensile force of 200KN in the X direction and uniformly distributed
compressive forces of 160KN in the Y direction and 220KN in the Z
direction. If v=1/3 and E=70GPa, determine a single distributed
loading in the X direction that would produce the same Z
deformation as the original loading.
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EXAMPLE 224:
A welded steel cylindrical drum made of 10mm plate has an internal
diameter of 1.20m. By how much will the diameter be changed by
an internal pressure of 1.5MPa? Assume Poisson’s ratio is 0.30 and
E=200GPa.
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EXAMPLE 226:
A welded steel cylindrical drum made of 10mm plate has an internal
diameter of 1.20m. By how much will the diameter be changed by
an internal pressure of 1.5MPa? Assume Poisson’s ratio is 0.30 and
E=200GPa.
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STATICALLY INDETERMINATE MEMBERS

INDETERMINATE MEMBERS
- the reactive forces or the internal resisting forces over a cross section exceed the
number of independent equations of equilibrium
- These cases require the use of additional relations that depend on the elastic
deformations in the members
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EXAMPLE 223:
A reinforced concrete column 250mm in diameter is designed to
carry an axial compressive load of 400KN. Using allowable stresses
of Sc=6MPa and Ss=120MPa, determine the required area of
reinforced steel. Assume that Ec=14GPa and Es=200GPa.
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EXAMPLE 235:
A rigid block of mass M is supported by three symmetrically spaced
rods as shown in the figure. Each copper rod has an area of 900mm2;
E=120GPa; and the allowable stress is 70MPa. The steel rod has an
area of 1200mm2; E=200GPa; and the allowable stress is 140MPa.
Determine the largest mass M which can be supported.
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EXAMPLE 237:
The lower ends of the three bars in the figure are at the same level
before the rigid homogeneous 18Mg block is attached. Each steel
bar has an area of 600mm2 and E=200GPa. For the bronze bar, the
area is 900mm2 and E=83GPa. Find the stress developed in each bar.
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EXAMPLE 244:
The bar shown in the figure is firmly attached to unyielding
supports. Find the stress caused in each material by applying an axial
load P=200KN.
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EXAMPLE 246:
A rod is composed of three segments shown in the figure and carries
the axial loads P1=120KN and P2=50KN. Determine the stress in
each material if the left wall yields 0.60mm.
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EXAMPLE 255:
Three rods, each with an area of 300mm2, jointly support the load
of 10KN, as shown in the figure. Assume there was no slack or stress
in the rods before the load was applied, find the stress in each rod.
Here, Es=200GPa and Eb=83GPa
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THERMAL STRESS
• Temperature changes cause the body to expand or contract.
• The amount 𝛿𝑇 , is given by
𝜹𝑻 = 𝜶𝑳(𝑻𝒇 − 𝑻𝒊 ) = 𝜶𝑳𝚫𝑻
where 𝛼 is the coefficient of thermal expansion in m/m°C,
𝐿 is the length in meter
𝑇𝑓 𝑎𝑛𝑑 𝑇𝑖 are the initial and final temperatures, respectively in °C.
For steel, 𝛼 = 11.25𝑥10−6 m/m°C

DEFORMATION DUE TO TEMPERATURE CHANGES

- For a homogeneous rod mounted between unyielding supports as shown, the thermal
stress is computed as
𝜹𝑻 = 𝜶𝑳(𝑻𝒇 − 𝑻𝒊 ) = 𝜶𝑳𝚫𝑻

DEFORMATION DUE TO EQUIVALENT AXIAL STRESS

- If the wall yields a distance of as shown, the following calculations will be made:

𝑃𝐿 𝜎𝐿
𝛿𝑃 = =
𝐴𝐸 𝐸
𝛿𝑃 = 𝛿𝑇
𝜎𝐿
𝛼𝐿Δ𝑇 =
𝐸
𝝈 = 𝑬𝜶𝚫𝑻 where is the thermal stress in MPa,
is the modulus of elasticity of the rod in MPa.

𝝈𝑳
𝜶𝑳𝚫𝑻 = 𝒙 + where represents the thermal stress
𝑬

As the temperature rises above the normal, the rod will be in compression,
and if the temperature drops below the normal, the rod is in tension.
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EXAMPLE 261:
A steel rod with a cross-sectional area of 150mm2 is stretched
between two fixed points. The tensile load at 200C is 5000N. What
will be stress at –200C? At what temperature will the stress be zero?
Assume =11.7m/m0C and E=200GPa.
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EXAMPLE 262:
A steel rod is stretched between two rigid walls and carries a tensile
load of 5000N at 200C. If the allowable stress is not to exceed
130MPa at –200C, what is the minimum diameter of the rod?
Assume =11.7m/m0C and E=200GPa.
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EXAMPLE 263:
Steel railroad rails 10m long are laid with a clearance of 3mm at a
temperature of 150C. At what temperature will the rails just touch?
What stress would be included in the rails at that temperature if
there were no initial clearance? Assume =11.7m/m0C and
E=200GPa.
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EXAMPLE 264:
At a temperature of 900C, a steel tire 10mm thick and 75mm wide
that is to be shrunk onto a locomotive driving wheel 1.8m in
diameter just fits over the wheel, which is at a temperature of 200C.
Determine the contact pressure between the tire and wheel after
the assembly cools to 200C. Neglect the deformation of the wheel
caused by the pressure of the tire. Assume =11.7m/m0C and
E=200GPa.
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TORSION
• Consider a bar to be rigidly attached at one
end and twisted at the other end by a torque
or twisting moment T equivalent to F × d,
which is applied perpendicular to the axis of
the bar, as shown in the figure. Such a bar is
said to be in torsion.
TORSIONAL SHEARING STRESS, 
- For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing
stress τ at a distance ρ from the center of the shaft is
𝑻𝑷 𝑻𝜸
𝝉= 𝒂𝒏𝒅 𝝉𝒎𝒂𝒙 =
𝑱 𝑱
where J is the polar moment of inertia of the section
r is the outer radius
SOLID CYLINDRICAL SHAFT: HOLLOW CYLINDRICAL SHAFT:
𝝅 𝟒 𝝅
𝑱= 𝑫 𝑱= (𝑫𝟒 − 𝒅𝟒 )
𝟑𝟐 𝟑𝟐
𝟏𝟔𝑻 𝟏𝟔𝑻𝑫
𝝉𝒎𝒂𝒙 = 𝝉𝒎𝒂𝒙 =
𝝅𝑫𝟑 𝝅(𝑫𝟒 − 𝒅𝟒 )

ANGLE OF TWIST
- The angle θ through which the bar length L will twist is
𝑻𝑳
𝜽= 𝒊𝒏 𝒓𝒂𝒅𝒊𝒂𝒏𝒔
𝑱𝑮
where T is the torque in N·mm,
L is the length of shaft in mm,
G is shear modulus in MPa,
J is the polar moment of inertia in mm4,
D and d are diameter in mm,
r is the radius in mm
POWER TRANSMITTED BY THE SHAFT
- A shaft rotating with a constant angular velocity ω (in radians per second) is being acted
by a twisting moment T. The power transmitted by the shaft is
𝑷 = 𝑻𝝎 = 𝟐𝝅𝑻𝒇
where T is the torque in N·m,
f is the number of revolutions per second,
P is the power in watts
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EXAMPLE 304:
What is the minimum diameter of a solid steel shaft that will not
twist through more than 3o in a 6m length when subjected to a
torque of 14KN-m? What maximum shearing stress is developed?
Use G=83GN/m2.
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EXAMPLE 306:
Determine the length of the shortest 2mm diameter bronze wire
which can be twisted through two complete turns without
exceeding a shearing stress of 70MPa. Use G=35GPa.
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EXAMPLE 310:
Determine the maximum torque that can be applied to a hollow
circular steel shaft of 100mm outside diameter and 70mm inside
diameter without exceeding a shearing stress of 60x106N/m2 or a
twist of 0.5deg/m. Use G=83x109N/m2.
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EXAMPLE 305:
A solid steel shaft 5m long is stressed to 60MPa when twisted
through 40. Using G=83GPa, compute the shaft diameter. What
power can be transmitted by the shaft at 20r/s?
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EXAMPLE 311:
A stepped steel shaft consists of a hollow shaft 2m long, with an
outside diameter of 100mm and an inside diameter of 70mm, rigidly
attached to a solid shaft 1.5m long, and 70mm in diameter.
Determine the maximum torque which can be applied without
exceeding a shearing stress of 70MPa or a twist of 2.50 in the 3.5m
length. Use G=83GPa.
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EXAMPLE 314:
A solid steel shaft is loaded as shown. Using G=83GPa, determine
the required diameter of the shaft if the shearing stress is limited to
60MPa and the angle of rotation at the free end is not to exceed 4 0.
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SHEAR & MOMENT

BEAMS
- subject to forces/couples that lie in a plane containing longitudinal section of the bar
- According to determinacy, a beam may be determinate or indeterminate

STATICALLY DETERMINATE BEAMS STATICALLY INDETERMINATE BEAMS

- are those beams in which the - number of reactions exerted upon a beam
reactions of the supports may be exceeds the number of equations in static
determined by the use of the equilibrium
equations of static equilibrium - In order to solve the reactions of the beam,
- The beams shown below are the static equations must be supplemented
examples of statically determinate by equations based upon the elastic
beams deformations of the beam.
• The degree of indeterminacy is taken as
the difference between the numbers of
reactions to the number of equations in
static equilibrium that can be applied.
• For propped beam, there are three
reactions R1, R2, and M and only two
equations (ΣM = 0, ΣFv = 0) can be applied,
thus the beam is indeterminate to the first
degree (3 - 2 = 1).
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TYPES OF LOADING
- Loads applied to the beam may consist of a concentrated load (load applied at a point),
uniform load, uniformly varying load, or an applied couple or moment. These loads are
shown in the following figures.

SHEAR & MOMENT DIAGRAM

- Consider a simple beam shown of length L that carries a uniform load of w (N/m)
throughout its length and is held in equilibrium by reactions R 1 and R2.
- Assume that the beam is cut at point C a distance of x from he left support and the
portion of the beam to the right of C be removed.
- The portion removed must then be replaced by vertical shearing force V together with a
couple M to hold the left portion of the bar in equilibrium under the action of R 1 and wx.

The couple M is called the resisting moment or moment

The force V is called the resisting shear or shear.
The sign of V and M are taken to be positive if they have
the senses indicated above.
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INSTRUCTION
Write shear and moment equations for the beams in the following problems. In each problem,
let x be the distance measured from left end of the beam. Also, draw shear and moment
diagrams, specifying values at all change of loading positions and at points of zero shear.
Neglect the mass of the beam in each problem.
EXAMPLE 403:
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EXAMPLE 404:
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EXAMPLE 405:
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STRENGTH OF MATERIALS

SHEAR & MOMENT MOVING LOADS

- The maximum moment occurs at a point of zero shears.
- For beams loaded with concentrated loads, the point of zero shears usually occurs
under a concentrated load and so the maximum moment.
- Beams and girders such as in a bridge or an overhead crane are subject to moving
concentrated loads, which are at fixed distance with each other.
o The problem here is to determine the moment under each load when each load is in
a position to cause a maximum moment.
o The largest value of these moments governs the design of the beam.
SINGLE MOVING LOAD
- For a single moving load, the maximum moment occurs when the load is at the midspan
and the maximum shear occurs when the load is very near the support (usually assumed
to lie over the support).
𝑷𝑳
𝑴𝒎𝒂𝒙 =
𝟒
𝑽𝒎𝒂𝒙 = 𝑷
TWO MOVING LOADS
- For two moving loads, the maximum
shear occurs at the reaction when the larger load is over that support. The maximum
moment is given by
(𝑷𝑳−𝑷𝒔 𝒅)𝟐
𝑴𝒎𝒂𝒙 =
𝟒𝑷𝑳
where Ps is the smaller load,
Pb is the bigger load,
P is the total load (P = Ps + Pb).

THREE OR MORE MOVING LOADS

- In general, the bending moment under a particular load is a maximum when the center
of the beam is midway between that load and the resultant of all the loads then on the
span.
- With this rule, we compute the maximum moment under each load,
o use the biggest of the moments for the design
o Usually, the biggest of these moments occurs under the biggest load.
- In determining the largest moment and shear, it is sometimes necessary to check the
condition when the bigger loads are on the span and the rest of the smaller loads are
outside.
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EXAMPLE 453:
A truck with axle loads of 40 kN and 60 kN on a wheel base of 5 m
rolls across a 10-m span. Compute the maximum bending moment
and the maximum shearing force.
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STRENGTH OF MATERIALS

EXAMPLE 454:
Repeat Prob. 453 using axle loads of 30 kN and 50 kN on a wheel
base of 4 m crossing an 8-m span.