F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER
2ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila
Contact Number/s: 0995-351-7556; 0945-734-3928
ENGINEERING MECHANICS – STATICS (PART 2)
Prepared by: Engr. Francis Arjay Pastores Luz
ARCHES
PROBLEM 1: The parabolic arch is loaded as shown.
Determine the internal forces at C.
a. Determine the shear at C.
b. Determine the Normal Force at C.
c. Determine the moment at C.
FLEXIBLE CABLES
One important type of structural member is the flexible
cable which is used in suspension bridges, transmission
lines, messenger cables for supporting heavy trolley or
telephone lines, and many other applications.
PROBLEM 2: The light cable supports a mass of 12 kg per
meter of horizontal length and is suspended between the
two points on the same level 300 m apart. If the sag is 60 m,
a. find the tension at midlength, Ans. 2250 kg.
b. the maximum tension, Ans. 2881.41 kg.
c. and the total length of the cable. Ans. 328.93 m
PROBLEM 3: A parabolic cable is loaded as shown. If 𝑤 =
50 𝑘𝑁/𝑚,
a. Determine the minimum tension in the cable. Ans. 3055.7
kN
b. Determine the maximum tension in the cable. Ans. 4120.3
kN
c. Determine the total length of the cable. Ans. 87.27 m
Parabolic Cable – a cable will assume a parabolic curve if it
is loaded uniformly along the horizontal plane.
Length of the parabolic cable, S:
Summation of moment at any point along the cable:
𝑥
𝑇𝑜 (𝑦) = 𝑤𝑥 ( )
2
𝑤𝑥 2
𝑑𝑦 𝑤𝑥
𝑦=
;
=
2𝑇𝑜
𝑑𝑥
𝑇𝑜
The length of the cable is:
𝑠
𝑙
∫ 𝑑𝑠 = ∫ √1 + (
0
0
𝑙
𝑑𝑦 2
𝑤𝑥 2
) ∙ 𝑑𝑥 = ∫ √1 + ( ) ∙ 𝑑𝑥
𝑑𝑥
𝑇𝑜
0
express the radical as a convergent series and then
integrate it term by term. Using the binomial expansion,
𝑛(𝑛 − 1) 2 𝑛(𝑛 − 1)(𝑛 − 2) 3
(1 + 𝑥 )𝑛 = 1 + 𝑛𝑥 +
𝑥 +
𝑥 +. ..
2!
3!
Where n = ½
𝟐 𝒉 𝟐 𝟐 𝒉 𝟒
𝒔 = 𝒍 [𝟏 + ( ) − ( ) + ⋯ ]
𝟑 𝒍
𝟓 𝒍
ℎ
1
This series is convergent for values of 𝑙 < 2 , which holds for
most practical cases.
PROBLEM 4: The suspension bridge in the figure is
constructed using two stiffening trusses that are pin
connected at their ends C and supported by a pin at A and
rocker at B. The cable has a parabolic shape.
a. Determine the minimum tension in the cable.
b. Det. The magnitude of the equivalent uniform loading,
w.
c. Det. The max. tension in the cable.
CATENARY CABLE
Catenary Cable - A cable is considered as a catenary when
the loading is distributed along and throughout the cable.
The term “catenary” comes from the Latin word meaning
“chain”. It is a graph of the equation y = cosh (x)
F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER
2ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila
Contact Number/s: 0995-351-7556; 0945-734-3928
CENTROID AND MOMENT OF INERTIA
Second Moment of Area. Also known as moment of
inertia of plane area, area moment of inertia, or second
area moment, is a geometrical property of an area which
reflects how its points are distributed with regard to an
arbitrary axis.
Tension at the supports:
𝐓𝟏 = 𝐰𝐲𝟏 ;
2
A= πr
IX= πr4/4
IY= πr4/4
𝐓𝟐 = 𝐰𝐲𝟐
Minimum Tension, To :
𝐓𝐨 = 𝐰𝐜
Distance between supports (span):
𝐋 = 𝟐𝐱
𝐬𝟏 + 𝐲𝟏
𝐱 = 𝐜 𝐥𝐧 (
)
𝐜
Relationship between S, y and c:
(𝐲𝟏 )𝟐 = (𝐬𝟏 )𝟐 + (𝐜)𝟐
Length of catenary cable:
𝐓𝐨
𝐰𝐱
𝐬 = (𝐬𝐢𝐧𝐡
)
𝐰
𝐓𝐨
Equation of a catenary cable:
𝐓𝐨
𝐰𝐱
𝐲 = (𝐜𝐨𝐬𝐡
− 𝟏)
𝐰
𝐓𝐨
PROBLEM 5: A cable weighing 0.4 kg/m and 800 m long is
to be suspended with a sag of 80 m. Determine the
maximum tension and the span.
PROBLEM 6: Find the total length L of cable which will have
the configuration shown when suspended from points A
and B. Ans. 46.20 m
centoid, x=
centoid, y=
A=
IX=
IY=
4r/3π
4r/3π
2
πr /4
0.055r4
4
0.055r
centoid, x=
centoid, y=
A=
IX=
IY=
b/2
h/2
bh
bh3/12
3
b h/12
centoid, x=
centoid, y=
A=
IX=
(a+b)/3
h/3
bh/2
3
bh /36
2
A= πr
IX= πr4/4
IY= πr4/4
centoid, x=
centoid, y=
A=
IX=
IY=
4r/3π
4r/3π
πr2/4
0.055r4
0.055r4
centoid, x=
centoid, y=
A=
IX=
IY=
b/2
h/2
bh
bh3/12
b3h/12
centoid, x=
centoid, y=
A=
IX=
(a+b)/3
h/3
bh/2
bh3/36
Parallel Axis Theorem:
Used to determine the moment of inertia of a rigid body
about any axis.
𝐼𝑜 = 𝐼𝑐.𝑔. + 𝐴𝑑 2
PROBLEM 8: Determine the centroid of the following
figure measured from point O.
CABLES WITH CONCENTRATED LOADS
PROBLEM 7: The cable supports three 400-N loads as
shown. If ℎ𝑐 = 16𝑚, determine the tension in segment DE
and BC.
F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER
2ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila
Contact Number/s: 0995-351-7556; 0945-734-3928
PROBLEM 9: Refer to figure in problem 1. Determine the
moment of inertia of the section with respect to its
centroidal x and y axis.
PROBLEM 10: A built-up steel section made from plates
is shown in the figure.
a. Determine the location of its neutral axis from the base.
b. Determine the moment of inertia of the section with
respect to its neutral axis.
c. Determine the Section Modulus, 𝑆𝑥 .
Product of Inertia
It is the measure of symmetry of a section. The product of
inertia of a section about its axis of symmetry is zero.
𝑃𝑥𝑦 = ∫ 𝑥𝑦 𝑑𝐴
Transfer Formula,
𝑃𝑜𝑧 = 𝑃𝑥̅ 𝑦̅ + 𝐴(𝑥)(𝑦)
Mohr’s Circle
1. A radius in the Mohr’s Circle represents axis in the
section.
2. The coordinates that the radius touches along the
circumference of the circle are the Moment of Inertia and
Product of Inertia of the axis it represents.
3. The angle between two radii in the circle is twice the
angle between the axes they represent.
PROBLEM 11: From the Z-Section shown, determine the
following:
a. Moment of Inertia about x-axis, 𝐼𝑥
b. Moment of Inertia about y-axis, 𝐼𝑦
c. Product of Inertia, 𝑃𝑥𝑦
d. Minimum moment of inertia, 𝐼𝑚𝑖𝑛.
e. Maximum moment of inertia, 𝐼𝑚𝑎𝑥.