UNIVERSITY OF PERPETUAL - HELP
CALAMBA CAMPUS
COLLEGE OF ENGINEERING
MECHANICAL ENGINEERING DEPARTMENT
MODULE 1
FENOL, MELCHOR M.
BSME-4
ENGR.GERALD GAZA
INSTRUCTOR
A machine is a tool containing one or more parts that uses energy to perform an intended
action.
Machines are usually powered by mechanical, chemical, thermal, or electrical means,
and are
often motorized. Historically, a power tool also required moving parts to classify as a
machine.
Mechanism (engineering) - rigid bodies connected by joints in order to accomplish a
desired force and/or motion transmission.
Slider mechanism
Tensile stress (or tension) is the stress state leading to expansion; that is, the length of a
material ends to increase in the tensile direction.
Compressive stress is the stress on materials that leads to a smaller volume.
Shear stress arises from the force vector component parallel to the cross section.
Torsion is the
twisting of an
object due to an
applied torque.
Bending (also known as flexure) characterizes the behavior of a slender structural
element subjected to an external load applied perpendicularly to a longitudinal axis of the
element.
ADAPTIVE DESIGN existing designs
DEVELOPMENT DESIGN a design that needs
evolution or growth
NEW DESIGN - not existing before
1.A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN.
Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.
P=A
P=A
where:
P=400kN=400000N
=120MPa
A=41D241(1002)
A=41(D210000)
thus,
400000=120[41(D210000)]
400000=30D2300000
D2=30400000+300000
D=11935mm answer
2.Weight of bar = 800 kg
Maximum allowable stress for bronze = 90 MPa
Maximum allowable stress for steel = 120 MPa
Required: Smallest area of bronze and steel cables
By symmetry:
Pbr=Pst=21(7848)
Pbr=3924N
Pst=3924N
For bronze cable:
Pbr=brAbr
3924=90Abr
Abr=436mm2 answer
For steel cable:
Pst=stAst
3924=120Ast
Ast=327mm2 answer
3.Given:
Diameter of cable = 0.6 inch
Weight of bar = 6000 lb
Required: Stress in the cable
MC=0
5T+10334T=5(6000)
T=295713lb
T=A
295713=41(062)
=1045872psi answer
4.Given:
Axial load P = 3000 lb
Cross-sectional area of the rod = 0.5 in2
Required: Stress in steel, aluminum, and bronze sections
For steel:
stAst=Pst
st(05)=12
st=24ksi answer
For aluminum:
alAal=Pal
al(05)=12
al=24ksi answer
For bronze:
brAbr=Pbr
br(05)=9
br=18ksi answer
5.Given:
Maximum allowable stress for steel = 140 MPa
Maximum allowable stress for aluminum = 90 MPa
Maximum allowable stress for bronze = 100 MPa
Required: Maximum safe value of axial load P
For bronze:
brAbr=2P
100(200)=2P
P=10000N
For aluminum:
alAal=P
90(400)=P
P=36000N
For Steel:
stAst=5P
P=14000N
For safe P, use P=10000N=10kN answer
6.Given:
Required diameter of hole = 20 mm
Thickness of plate = 25 mm
Shear strength of plate = 350 MN/m2
Required: Force required to punch a 20-mm-diameter hole
The resisting area is the
shaded area along the
perimeter and the shear
force V is equal to the
punching force P.
V=A
P=350[(20)(25)]
P=5497787N
P=5498kN answer
7.Given:
Force P = 400 kN
Shear strength of the bolt =
300 MPa
The figure below:
Required: Diameter of the
smallest bolt
The bolt is subject to double shear.
V=A
400(1000)=300[2(41d2)]
d=2913mm answer
8.In Fig. 1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The
allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet.
Determine (a) the minimum thickness of each plate; and (b) the largest average tensile stress in the
plates.
Part (a):
From shearing of
rivet:
P=Arivets
P=60[41(202)]
P=6000textN
From bearing of
plate material:
P=bAb
6000=120(20t)
t=785mm answer
Part (b): Largest average tensile stress in the
P=A
6000=[785(11020)]
=2667MPa answer
plate:
9.Given:
Allowable shearing stress in the pin at B = 4000 psi
Allowable axial stress in the control rod at C = 5000 psi
Diameter of the pin = 0.25 inch
Diameter of control rod = 0.5 inch
Pin at B is at single shear
Required: The maximum force P that can be applied by the operator
MB=0
6P=2Tsin10 Equation (1)
FH=0
BH=Tcos10
From Equation
(1), T=3Psin10
BH=3Psin10cos10
BH=3cot10P
FV=0
BV=Tsin10+P
From Equation (1), Tsin10=3P
BV=3P+P
BV=4P
R2B=B2H+B2V
R2B=(3cot10P)2+(4P)2
R2B=30547P2
RB=1748P
P=RB1748 Equation (2)
Based on tension of rod (equation 1):
P=31Tsin10
P=31[500041(05)2]sin10
P=5683lb
Based on shear of rivet (equation 2):
P=17484000[41(025)2]
P=1123lb
Safe load P=1123lb answer
10.Given:
Diameter of
each rivet = 3/4
inch
Maximum
allowable shear
stress of rivet =
14 ksi
Maximum
allowable
bearing stress
of plate = 18
ksi
The figure below:
Required: The maximum safe value of P that can be applied
Based on shearing of rivets:
P=A
P=14[4(41)(43)2]
P=2474kips
Based on bearing of plates:
P=bAb
P=18[4(43)(87)]
P=4725kips
Safe load P=2474kips answer
