Chapter 1
Simple Stress
1.1 Introduction
The Mechanics of Deformable Bodies is the continuation of study of forces applied upon a body which
was started in Statics of rigid bodies. The difference lies on the assumption of the types of bodies being
analyzed, rigid body for Statics and deformable body for Strength of Materials. Other considerations are
the cross-sectional areas and dimensions of the body including its thickness, length and width which are
very useful in the measurement of strength, rigidity, deformation, deflection, etc.
The subject deals with the relation between the external forces applied and its internal effects in
the body. In the process of analysis, the body is not anymore assumed to be rigid but will eventually
undergo change in shape. Deformation is quite important however small it may be. The difference
between the two subjects is further explained by an example considering Figure 1-1. In statics, the
problem is to determine the amount of force P applied at the end of the member. The basic assumption
is that the member is rigid enough to resist and permit any load carried by the member. In mechanics of
deformable bodies, the body is analyzed deeper. The member is subject to further investigation to
assure that it will never break or fracture. The theories and principles that involve the fundamental ideas
of strength and rigidity will be studied in the subject.
Figure 1-1
1.2 Analysis of Internal Forces
The analysis of rigid body is primarily done through attaining the equilibrium of the system. It is
concerned with the calculation of external reactions by the forces applied on the body and internal
loadings as in the case of forces exerted by the member of the truss. In the mechanics of deformable
bodies, the analysis is extended to determine the forces internal to the body itself. In addition, we
investigate the orientation and internal distribution of the forces.
In Figure 1-2a, consider the section at C acted upon by external forces F 1, F2, and F3. The internal
force R is resolved into axial force P and shear force V. In Figure 1-2b, the internal moment C R is resolved
into torque T and bending moment M. Generally, the location of the axes is found at the centroid which
is the reference point of the section.
Figure 1-2a
Figure 1-2b
The components acting through C are described as follows:
N
Axial force. This force measures the amount of pull or push over the section. A pulling
force will tend to elongate the body whereas the pushing force tends to compress the
member. It is usually denoted by N.
V
Shear force. This force determines the resistance of one portion of the body to sliding
past another section. It is often designated by V.
T
Torque. The component measures the resistance to rotation of the body and is
denoted by T.
M
Bending Moment.
The component determines the resistance of the member to
bending and is given the symbol M.
The subject basically aims to ensure that the any structure being analyzed will be safe against
maximum internal effects that may be induced by any types of loadings. It is the branch of mechanics
that deals with the study of forces and its internal effects in the forms of stress and strain in a solid body
caused by the application of external loadings.
Illustrative Problems:
1. For the cantilever beam loaded as shown, determine the resultant loadings ( bending moment and
shear force ) acting on the cross section at C. The beam is subjected to a uniformly varying load from
zero at B to 360 N/ m at A on a span of 9 m.
360 N/ m
Solution: Pass a section through C and consider to the right of the cutting section;
240 N/ m
Applying the equations of equilibrium;
Σ FV = 0
VC – ½ (240 N/m) ( 6 m ) = 0
VC = 720 N Ans
Σ MC = M C
MC = ½ (240 N/m) ( 6 m ) ( 2 m )
MC = 1440 Nm Ans
2. The beam is supports a concentrated load of 10 kN and an external moment of 60 kNm. Determine
the internal shear force and bending moment at point C in the beam.
Solution: Compute the support reactions at B using the equations of equilibrium. Consider the whole
figure;
Σ MA = 0
RB ( 2 m ) – 60 kNm + 10 kN ( 4 m ) = 0
RB = 10 kN ( ↓ )
Pass a section through C and consider to the right of the cutting section;
Σ FV = 0
VC – RB – 10 kN = 0
VC = 20 kN Ans
Σ MC = M C
MC = 10 kN ( 3 m ) + RB ( 1 m )
MC = 40 kNm Ans
Exercises:
1. The boom DF of the crane and the column DE have a uniform weight of 40 lb/ ft. A load of 300 lb is
applied at hoist attached to F. Determine the resultant internal loadings in the crane on cross-sections
through points A, B, and C.
Answer: VB = 740 lb, MB = 5720 lbft, NB = 0
2. The beam carries the loadings as shown in the figure. There is an internal hinge located at point B.
Determine the internal normal force, shear and bending moment at D.
Answer: MD = 13.5 kipft
3. For the structure carrying a uniform load of 4 kN/ m, compute the internal axial force, shear force,
and bending moment at point C in the beam.
Answer: VC = 2 kN, MC = 4 kNm, NC = 24 kN
4 kN/ m
4. The cable will fail when subjected to a tension of 8000 N. Determine the largest vertical load P the
frame will support and compute the resultant internal loadings at the cross section through point C.
Answer: P = 2133.33 N, MC = 1600 Nm, NC = 8000 N, VC = 2133.33 N
5. For the beam loaded as shown, calculate the internal axial force, shear force, and bending moment
on the cross section through points C and D. Assume the reactions at the supports A and B are vertical.
Answer: VD = 1.875 kN, MD = 3.94 kNm
6. Determine the resultant internal loadings acting on the cross sections through points C and D. The
support reactions are assumed to be vertical.
Answer: VD = 0 , MD = 54 kip-ft
9 kip/ ft
9 kip/ ft
7. The beam is supported by a hinge at A and roller supported at B. Determine the internal shear force
and bending moment at point C in the beam.
Answer: VC = 50 lb, MC = 750 lbft
200 lb/ ft
8. The post is fixed to the ground and a rectangular pressure 0f 7 lb/ ft 2 acts perpendicular to the face
of the signboard. What resultant internal loadings act on the cross section through point B of the sign
post.
Answer: VBx = 105 lb, MBy = 787.50 lbft
9. A 200-lb bucket is suspended from a cable on the wooden frame shown. Calculate the internal
normal force, shear force, and bending moment on the cross section at E.
Answer: NE = 430.94 lb, VE = 133.33 lb, ME = 400 lbft
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1.3 Simple Stress
The prime objective of an engineer is to choose which material will work efficiently in order to resist the
applied load. In this regard, it is important to identify the properties of material such as strength,
stiffness, rigidity, etc. in order to achieve its maximum productivity with minimum wasted effort. The
first criterion in the design is the measurement of the strength of the member. It is basically measured
in terms of the stress developed in the body.
Simple stresses are expressed as the ratio of the applied force to its resisting area. Stresses are
classified as normal stress, shearing stress, and bearing stress. It is the expression of force divided by the
area. Notable structures that carry these stresses are bridges, trusses, beams, and columns.
1.4 Normal Stress
Normal stress, also called axial stress is the stress caused by forces that act perpendicular to the cross
sectional area of the body. It can be referred to as either tensile stress or compressive stress depending
on the direction of the applied force. If the force will tend to elongate the body, it is called tensile stress
as shown in Figure 1-4. When the load applied will cause the member to contract, it is called
compressive stress.
Figure 1-4
Normal stress is denoted by a Greek letter called sigma, σ, and expressed as:
where : σ = stress ( MPa or psi )
P = normal force ( N or lb )
A = cross sectional area ( mm2 or in2 )
In the analysis of problems involving normal stress, the following assumptions must be taken
into consideration: load must be axial, material is homogeneous, and cross sectional area is uniform or
constant.
Illustrative Problems:
1. The bar ABCD is composed of three segments with varying lengths and cross sectional areas. Axial
loads are applied as shown in the figure. Determine the normal stress developed in each segment.
1.25’
1.60’
1.75’
Solution: Consider to the left of section ①;
Consider to the right of section ③;
Σ FH = 0
Σ FH = 0
PAB – 4000 lb = 0
PCD – 7000 lb = 0
PAB = 4000 lb ( tension )
PCD = 7000 lb ( compression )
Calculating the stress;
Calculating the stress;
2
σAB = 4000 lb / 1.2 in
σCD = 7000 lb / 1.6 in2
σAB = 3333.33 psi Ans
σCD = 4375 psi Ans
Pass a section through ② and analyze the left side of the section;
Σ FH = 0
– PBC – 4000 lb + 9000 lb = 0
PBC = 5000 lb ( compression )
Calculating the stress;
σBC = 5000 lb / 1.8 in2
σAB = 2777.78 psi Ans
2. For the truss loaded as shown, compute the axial stress developed in members BE and BD. Assume
that the cross sectional area for each member is 800 mm 2.
4 panels at 4 m = 16 m
Solution: Pass a section and consider to the left of the cutting section;
①
C
RAv
30 kN
Consider the whole system and calculate the reaction at the left support;
Σ MH = 0
RAv ( 16 m ) – 70 kN ( 4 m ) – 30 kN ( 12 m ) = 0
RAv = 40 kN (↑)
Consider the left side of section ①;
Σ FV = 0
RAv – 30 kN – PBE ( 3/5 ) = 0
PBE = 16.67 kN ( tension )
Calculate the stress;
σBE = 16.67 x 103 N / 800 mm2
σBE = 20.83 MPa Ans
Σ ME = 0
RAv ( 8 m ) – 30 kN ( 4 m ) + PBD ( 3 m ) = 0
PBD = –66.67 kN ( compression )
Calculate the stress;
σBD = 66.67 x 103 N / 800 mm2
σBD = 83.33 MPa Ans
Exercises:
1. A rod consists of three segments rigidly attached to one another. Axial loads of 4P and P are applied
as shown. Assume that P = 2500 lb and the cross sectional area is 0.75 in 2. Determine the stress
developed in each segment.
Answer: σSt = 13.33 ksi, σBr = 10 ksi
1.75’
2.75’
2.50’
2. The figure shows the two wires supporting the block with weight W. The cross sectional areas of the
wires are 700 mm2 for AB and 450 mm2 for AC. Determine the maximum safe weight W if the stresses
are limited to 100 MPa for AB and 125 MPa for AC.
Answer: W = 72.30 kN
3. Two steel cables support a homogeneous bar AB weighing 2000 lb. The allowable stress for both
steel cables is 16500 psi. Calculate the minimum area required for each cable.
Answer: A = 0.06 in2
4. Determine the weight “W” of the heaviest cylinder that can be supported in the position as shown in
the figure. Allowable stress for the cable is 40 MPa. Neglect any frictional resistances. Assume that the
cross sectional area of cable BC is 120 mm2. Also, calculate the reaction between the uniform cylinder
and the vertical wall.
Answer: W = 5760 N
5. For the truss loaded as shown, determine the normal stress developed in members AB, BC, and CD.
The cross sectional area of each member is 1600 mm 2. State whether the member is tension or
compression.
Answer: σAB = 35.35 MPa, σBC =43.75 MPa
6. The truss is subjected to two concentrated loads as shown in the figure. The allowable stresses are
25 ksi for tension and 12 ksi for compression. Determine the minimum cross sectional areas for
members BC, CE, and BE.
Answer: ABC = 2.67 in2, ACE = 3.75 in2
45 kips
50 kips
7. The uniform beam shown carries a triangular load from zero at A to a maximum “w” at C. It is
supported by two rods AB and CD that have cross sectional areas of 12 mm 2 and 18 mm2, respectively.
Determine the intensity “w” of the triangular load so that the allowable normal stress in each rod is
limited 400 KPa.
Answer: w = 3.60 N/ m
8. The 500-kN force is assumed to act through the centroid of the cross-section. Determine the
location of the centroid and the axial stress developed on the section.
Answer: σ = 8.33 MPa
500 kN
9. A rod has three segments as shown in the figure. The position of applied loads and diameter of each
segment are indicated. Calculate the normal stress developed at points A, B, and C.
Answer: σA = 0.95 ksi, σB = 0.48 ksi
2 in
4 in
2 in
10. The homogeneous bar ABC that weighs 5000 lb is supported by a smooth pin at C and a cable that
runs from A to B around the frictionless pulley at D. If the diameter of the cable is 0.80 inch, calculate its
axial stress.
Answer: σ = 4.90 ksi
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1.5 Shearing Stress
Shearing stress, also called tangential stress, is the stress caused by forces that act along or parallel to
the area resisting the force. Shear stress arises whenever the applied force causes one section of the
body to slide past another body. Several examples are shown in Figures 1.5a, 1.5b, and 1.5c.
In Figure 1.5a, two plates are fastened together by a rivet. The rivet resists shear across its
cross-sectional area and is called single shear. Figure 1.5b shows the resistance of bolt across two crosssectional areas and is referred to as double shear. In Figure 1.5c, a circular slug is to be punched out of
the plate. The resisting area in that case is the same as the milled edge of a coin. It can be noticed that
in each case, the shear occurs over an area parallel to the applied load.
Figure 1.5a
Figure 1.5b
Figure 1.5c
Shearing stress is denoted by a Greek letter called tau, τ, and expressed as:
where : τ = shear stress ( MPa or psi )
V = shear force ( N or lb )
A = shear area ( mm2 or in2 )
Any problems related to shearing stress will assume that uniform shear stress will be developed
when the resultant shearing force passes through the centroid of the cross section being sheared. It is
also noted that the shear stress produced across any section is practically never distributed uniformly
thus must be calculated as average shearing stress. But in any case, the distribution of stress becomes
approximately uniform when the distance between the application of load and the depth of the shear
area is small.
1.6 Bearing Stress
Bearing stress is the contact pressure between separate bodies. When two different bodies are pressed
against each other, forces are developed on the contact area. The pressure caused by these loads is
called bearing stress. Some examples of bearing stress are the forces on bearing plates, pressure
between the pier and soil, and the contact pressure between the rivet and the side of the plate at the
rivet connection.
Consider the two plates fastened together by a rivet as shown in Figure 1.6. It shows ( a ) a rivet
in a lap joint, ( b ) the bearing stress is not constant, and ( c ) the stress caused by applied force P b is
assumed to be uniform in the shaded area.
(a)
(b)
(c)
Figure 1.6
Bearing stress is denoted by Greek letter sigma ( with subscript of lower case letter b ), σb, and
written as:
where : σb = bearing stress ( MPa or psi )
Pb = bearing load ( N or lb )
Ab = projected area ( mm2 or in2 )
The bearing stress caused by the rivet is not constant but varies from zero at the sides of the
hole to maximum at the rivet. The assumption is that the bearing stress is uniformly distributed over
the projected area. The applied load is likewise assumed to be equally resisted by each rivet. The
projected area is computed as the product of the thickness of the plate and the diameter of the rivet.
Illustrative Problems:
1. The beam is supported by a smooth pin at A and cable at B. Determine the average shear stress in
the 16-mm diameter pin at A and the 32-mm diameter pin at B.
Solution: Consider the whole figure and compute the reactions by equations of equilibrium;
Σ MA = 0
30 kN ( 2 m ) – FBC ( 4/5 ) ( 6 m ) = 0
FBC = 12.5 kN
Calculate the pin reaction at A;
Σ FH = 0
– RAH + FBC ( 3/5 ) = 0
RBH = 7.5 kN ( ← )
Σ FV = 0
RAV – 30 kN + FBC ( 4/5 ) = 0
RAV = 20 kN ( ↑ )
RA2 = RBv2 + RBH2
RA = 21.36 kN
Calculate the shear stress;
τ=V/A
VA = RA = 21.36 kN
τA = V A / A
τ B = VB / A
3
2
τA = 21.36 x 10 N / 2 (πrr )
τB = 12.5 x 103 N / (πrr2 )
τA = 73.84 x 103 N / 2 ( πr x 82 ) mm2
τB = 12.5 x 103 N / ( πr x 162 ) mm2
τA = 53.12 MPa Ans
τB = 15.54 MPa Ans
2. The member shown is supported by a smooth pin at B and roller at C. The detail of the pin at B is
double shear. The pin diameter is 16 mm. Determine the shearing stress developed in the pin.
50 kN
Solution: Calculate the vertical component of the pin reaction at B;
Σ MC = 0
– RBv ( 250 mm ) + 50 kN ( Cos 350 ) ( 200 mm ) + 50 kN ( Sin 350 ) ( 250 mm ) = 0
RBv = 61.44 kN (↓)
Calculate the horizontal component of the pin reaction at B;
Σ FH = 0
– RBH + 50 kN ( Cos 350 ) = 0
RBH = 40.96 kN ( ← )
RB2 = RBv2 + RBH2
RB = 73.84 kN
Calculate the shear stress;
τ=V/A
VB = RB = 73.84 kN
τ B = VB / A
τB = 73.84 x 103 N / 2 (πrr2 )
τB = 73.84 x 103 N / 2 ( πr x 82 ) mm2
τB = 183.62 MPa Ans
3. Assume that a 20-mm diameter rivet joins the plates that are each 110 mm wide. The allowable
stresses are 100 MPa for bearing in the plate material and 55 MPa for shearing of rivet. Calculate the
largest tensile stress developed in the plate.
Solution: From the shearing of the rivet;
τ=V/A
55 N/ mm2 = P / ( πr x 102 )
P = 17278.76 N
From the bearing of the plate;
σb = Pb / Ab
where Ab = t x d
2
100 N/ mm = 17278.76 N / ( t x 20 mm )
t = 8.64 mm
Calculate the largest average tensile stress in the plate;
σ=P/A
σ = 17278.76 N / ( 100 mm – 20 mm ) ( 8.64 mm )
σ = 25 MPa Ans
Exercises:
1. How much force is required to punch a 25-mm diameter hole in the plate that is 30 mm thick? The
punching shearing stress is limited to 340 MPa.
Answer: P = 801 kN
2. Two blocks of wood, 25 mm x 75 mm, are glued together along the joint inclined at 15 0. Determine
the average normal stress and shear stress developed in the wood fiber along section a-a.
Answer: σ = 15.20 kPa, τ = 56.7 kPa
3. The uniform 3-Mg bar is supported by a smooth wall at A and pin connected at B ( double shear ).
Determine the diameter of the smallest pin that can be used without exceeding the allowable shearing
stress in the rivet is 70 MPa.
Answer: D = 16.9 mm
4. The allowable bearing stress for the material under support A is 2 MPa. Determine the size of
square bearing plates at A’ required to support the load. Dimension the plates to the nearest multiple of
10 mm.
Answer: s = 110 mm
35 kN/ m
90 kN
5. The lap joint shown is fastened by 4-0.75” diameter rivets. The allowable shearing stress in the rivet
is 15 ksi and the bearing stress in the plates is limited to 20 ksi. Assuming that the applied loads are
uniformly distributed among the number of rivets, calculate the maximum safe load P that can be
applied.
Answer: P = 26.51 kips
6. The A-frame is loaded by two concentrated forces as shown in the figure. The pins at B and C have a
diameter of 0.50”. The two pins are both subjected to single shear. Determine the shear stress
developed in each pin.
Answer: τB = 3026 psi
7. The bell crank shown which attains the equilibrium under the applied forces is supported by smooth
pin at D. Determine the shear stress developed in the 20-mm diameter pin at D that is in double shear.
Also, calculate the required diameter of the connecting rod AB if the tensile stress is limited to 90 MPa.
Answer: τD = 84.33 MPa
8. The Howe truss is loaded by three concentrated forces. The rivet connection at joint B is shown in
the figure. If the allowable shearing stress in the rivet and bearing stress in the plate are 60 MPa and
145 MPa, respectively, how many 16-mm diameter rivets are required to fasten members BC and BE to
the gusset plate.
Answer: nBC = 8, nBE = 7
9. A force P is applied on the foot pedal as shown in the figure. The allowable stress for the 6-mm
diameter pin at B is 30 MPa. Assume the pin to be in single shear. The 5-mm diameter cable attached at
C has a an axial stress limited to 150 MPa . Compute the maximum value of P to satisfy the given
conditions.
10. The bar AB is supported by a hinge support at B that is in double shear and smooth frictionless
surface at A. Determine the shear stress in the pin at B when the diameter of the pin is ¾ inch. Neglect
the weight of the bar.
Answer: τB = 1.24 ksi
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1.7 Thin-Walled Pressure Vessels
Thin-walled pressure vessels are tanks or vessels that carry either a gas or fluid under an internal
pressure “ρ “ ( Greek letter rho ). It is subjected to tensile forces developed along the thickness which
resist the bursting forces across longitudinal and transverse sections. There are two types of stresses
developed in the pressure vessel: tangential stress and longitudinal stress.
Consider the tank shown in Figure 1.7a subjected to an internal pressure, ρ. The length of the
tank is denoted as “ L “ and the wall thickness is “ t “.Pass a cutting section and consider the right half of
the tank.
Figure 1.7a
Σ FH = 0
– 2T+ F = 0
T=F/2
but F = ρ x AF
where AF = D x L
F = bursting force ( N, lb )
F=ρxDxL
T = (ρ x D x L ) / 2
From the normal stress developed along the thickness of the tank;
σ = P / At
where At = t x L
but P = T
σt = (ρ x D x L ) / 2 (t x L )
therefore σt = ρD / 2tD / 2t
where: σt = tangential stress ( MPa or psi )
ρ = internal pressure (MPa or psi)
D = diameter of the cylinder ( mm or in )
t = thickness ( mm or in )
It is noted that the bursting force acting normal to the cutting plane is resisted by the equal
forces P located on the surface of the cylinder wall. The stress developed along the surface is usually
called tangential stress. It is called tangential since it is caused by forces that act tangent to the
thickness of the cylindrical vessel. The stress is sometimes referred to as circumferential stress, girth
stress, or hoop stress.
Consider now the free body diagram in the transverse section of the tank as shown in Figure
1.7b. The total bursting force acting at the rear tank must be equated to the total forces along the
circumference of the vessel.
Figure 1.7b
Σ FH = 0
–P+F=0
P=F
but F = ρ x AF
where AF = πr x D2 / 4
F = bursting force ( N, lb )
2
P = ρ x πr x D / 4
From the normal stress developed along the thickness of the tank;
σ = P / At
where At = t x πr x D
2
σl = (ρ x πr x D / 4) / (t x πr x D)
therefore σl = ρD / 2tD / 4t
where: σl = longitudinal stress ( MPa or psi )
ρ = internal pressure (MPa or psi)
D = diameter of the cylinder ( mm or in )
t = thickness ( mm or in )
From the illustration, the busting force acting over the end of the cylinder is counteracted by the
resultant force P acting over the transverse section. The area of the transverse section is equal to the
thickness multiplied by the mean circumference. Since the thickness is quite small as compared to the
diameter, the area is closely approximated by πrDt. The stress is called longitudinal stress because it is
the stress caused by forces acting parallel to the longitudinal axis of the cylinder.
Illustrative Problems:
1. The wall thickness of a 1.5-m diameter spherical tank is 8 mm. Calculate the allowable internal
pressure if the stress will not exceed 55 MPa.
Solution: Pass a cutting section through the diameter;
Σ FV = 0
–P+F=0
P=F
but F = ρ x AF
where AF = πr x D2 / 4
P = ρ x πr x D2 / 4
From the normal stress developed along the thickness of the tank;
σ = P / At
where At = t x πr x D
σ = (ρ x πr x D2 / 4) / (t x πr x D)
therefore σ = ρD / 4t
55 N/ mm2 = ρ ( 1500 mm ) / 4 ( 8 mm )
ρ = 1.17 MPa Ans
2. A cylindrical pressure vessel is fabricated from steel plate with a thickness of 20 mm. The diameter
of the cylinder is 450 mm and its length is 2 m. The tangential stress and longitudinal stress are limited
to 65 MPa and 150 MPa, respectively. Determine the maximum internal pressure that can be applied.
Solution: From the tangential stress formula;
σt = ρD / 2t
65 N/ mm2 = ρ ( 450 mm ) / 2 ( 20 mm )
ρ = 5.78 MPa
From the longitudinal stress formula;
σl = ρD / 4t
150 N/ mm2 = ρ ( 450 mm ) / 4 ( 20 mm )
ρ = 26.67 MPa
therefore use ρ = 5.78 MPa Ans
Exercises:
1. A cylindrical pressure vessel is 500 mm in diameter and has a wall thickness of 22 mm. If it is
subjected to an internal pressure of 5 MPa, determine the tangential and longitudinal stresses
developed in the cylinder.
Answer: σt = 56.82 MPa, σl = 28.41 MPa
2. The strength per meter of the longitudinal joint is 500 kN and 220 kN for the girth joint. If the
internal pressure is 2 MPa, determine the maximum diameter of the thin-walled cylinder.
Answer: D = 440 mm
3. Determine the minimum thickness for a cylinder to contain a gas at a pressure of 1300 psi. The
diameter of the cylindrical vessel is 2.5 ft with an allowable stress of 10000 psi.
Answer: t = 1.95 mm
4. A water tank is made from 0.5 inch steel plate and has a diameter of 20 ft. Calculate the maximum
height to which the tank may be filled if the girth stress is not to exceed 5000 psi. The water has a
specific weight of 62.4 lb/ ft3.
Answer: h = 48.10 ft
5. The tank shown is fabricated using a steel plate 1/6 of an inch thick. Determine the maximum
tangential and longitudinal stresses developed caused by an internal pressure of 130 psi.
Answer: σt = 16.40 ksi, σl = 5.12 ksi
Figure 1.7.5
6. The tank shown in Figure 1.7.5 is fabricated from a steel plate. What minimum thickness of plate
may be used if the allowable stress is 15 ksi and the internal pressure is 125 psi.
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