Fakultas Teknik Jurusan Teknik Sipil Universitas Brawijaya...
Transcript of Fakultas Teknik Jurusan Teknik Sipil Universitas Brawijaya...
Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to Direct Stresses in Two Mutually Perpendicular
Directions
Another graphical method for the determination of normal tangential and resultant stress is known as Mohr’s Circle
When the two mutually perpendicular principal stresses are unequal and alike
Consider a rectangular body subjected to two mutually perpendicular principal
tensile stresses and an oblique section, on which we are required to find out the
stresses.
Let,
p1 = major tensile stress
p2 = minor tensile stress
Θ = angle which the oblique section makes with the minor tensile stress
• Take some suitable point O and draw a horizontal line OX
• Cut off OA and OB equal to the stress p1 and p2 respectively tosome suitable scale on the same side of O (because the twostresses are alike)
• Bisect BA at C. Now with C as center, and radius equal to CB or CAdraw a circle
Draw the circle of the stresses
• Now through C, draw a line CP making an angle 2θ with CX meeting the circle atP
• Through P, draw PQ perpendicular to OA. Now join OP
• Now OQ and PQ will give the required normal and tangential stresses to thescale. OP will give the required resultant stress to the scale. The angle POA iscalled the angle of obliquity
Proof:
From the geometry of the figure, we find that:
2
21 ppCPCABC
22
2
2
21212
212
ppppp
pppBCOBOC
Normal Stress
2cos22
2cos2
2121
21
ppppp
CPpp
CQOCOQp
n
n
Tangential Stress 2sin2
2sin 21 ppCPPQpt
NOTE:• Since A and B are the ends of the horizontal diameter, therefore maximum
normal stress will be equal to p1 and minimum principal stress will be p2
• On the plane having maximum or minimum principal stresses, there will be notangential stress
• Shear stress on mutually perpendicular planes are numerically equal• The maximum shear stress will be equal to radius of the Mohr’s Circle and will
act on planes inclined at 45° to the principal planes. Mathematically:
• The angle of obliquity will be maximum, when OP is tangential to the Mohr’sCircle
2max 21 pp
pt
Consider a rectangular body subjected to two mutually perpendicular unlike
principal tensile stresses and an oblique section, on which we are required to find
out the stresses.
Let,
p1 = major principal tensile stress
p2 = minor principal compressive stressΘ = angle which the oblique section makes with the minor principal stress
• Take some suitable point O and draw a horizontal line X’OX
• Cut off OA and OB equal to the stresses p1 and p2 to some suitablescale on the opposite sides of O (because the two stresses areunlike)
• Bisect BA at C. Now with C as center, and radius equal to CB or CAdraw a circle
Draw the circle of the stresses
• Now through C, draw a line CP making an angle 2θ with CXmeeting the circle at P
• Through P, draw PQ perpendicular to OA. Now join OP
• Now OQ and PQ will give the required normal and tangentialstresses to the scale. OP will give the required resultant stress tothe scale.
Proof:
From the geometry of the figure, we find that:
2
21 ppCPCABC
22
2
2
21221
221
ppppp
ppp
OBBCOC
A point in a strained material the normal tensile stresses are 60 N/mm2 and30 N/mm2. Determine by Mohr’s Circle, the resultant intensity of stress on aplane inclined at 40° to the axis of the minor stress. Also check the answeranalytically
Example :
Solution :
Given:Major tensile stress p1 = 60 N/mm2Minor tensile stress p2 = 30 N/mm2Angle which the plane makes with the axis of minor stress = 40°
• Take some suitable point O and draw a horizontal line OX• Cut off OA equal to 60 and OB equal to 30 to some suitable scale on the
same sides of O• Bisect BA at C. Now with C as center, and radius equal to CB or CA draw a
circle• Now through C, draw a line CP making an angle 2x40 = 80° with CX, meeting
the circle at P• Joint OP. By measurement, we find that the resultant stress,
pR = OP = 49,8 kg/cm2
Analytical Check:
Let,
pn = normal stress on the plane
pt = tangential stress on the plane
pR = resultant stress on the plane
Using the relation,
2
2121
6,4780cos2
3060
2
3060
2cos22
mmNp
ppppp
n
n
2
21
8,1480sin2
3060
2sin2
mmNp
ppp
t
t
22222
8,498,146,47cm
kgppp tnR
A point in a strained material is subjected to a tensile stress of 800kg/cm2 and compressive stress of 500 kg/cm2. Draw the Mohr’s stresscircle and find out the resultant stress at a plane making 64° with thetensile stress. Also find out the normal and tangential stress on theplane.
Example :
Solution :
Given:Major stress p1 = 800 kg/cm2Minor stress p2 = -500 kg/cm2Angle which the plane makes with the tensile stress = 64°
• Take some suitable point O and draw a horizontal line X’OX• Cut off OA equal to 800 and OB equal to 500 to some suitable scale on the
opposite sides of O• Bisect BA at C. Now with C as center, and radius equal to CB or CA draw a circle• Through C, draw a line CP making an angle 2x64 = 128° with CX, meeting the circle
at P• Through P. draw PQ perpendicular to OA. Join OP,• By measurement, we find that:
Normal stress pn = OQ = -250 kg/cm2Tangential stress pt = PQ = 510 kg/cm2Resultant stress pR = OP = 570 kg/cm2
Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to a Direct Stress in One Plane Accompanied by A Simple
Shear Stress
Consider a rectangular body ABCD subjected to a tensile stress in one plane
accompanied by shear stress, and an oblique section, on which we are required to
find out the stresses.
Let,
p = tensile stress on the faces AD and BC
q = shear stress across the faces DA and BC
Θ = angle which the oblique section EF makes with the normal cross section EG
• draw a horizontal line X’X and cut off AB equal to the stress tosome suitable scale, and bisect it at C
• Now erect a perpendicular at B and cut off BE equal to the shearstress q to the scale
• Now with C as center and radius equal to CE, draw a circlemeeting the line X’X at H and G
Draw the circle of the stresses
• On the base CE, draw a line CF at an angle 2θ meeting the circle at F
• From F, draw FD perpendicular to AD. Join AF
• Now AD and DF will give the required normal and tangential stress to the scale.AF will give the required resultant stress to the scale
• Moreover, AG and AH will give the maximum and minimum values of normalstresses to the scale.
Proof:
From the geometry of the figure, we find that normal stress:
CECFCFCFp
p
CFp
p
CFp
CDACADp
n
n
n
sin2sincos2cos2
sin2sincos2cos2
2cos2
2sin)2cos1(2
2sin2cos22
2sin2cos2
cossin2sincos2cos2
qp
p
qpp
p
BECBp
p
CECBCECEp
p
n
n
n
n
And tangential stress:
2cos2sin2
2cos2sin
cossin2coscos2sin
sin2coscos2sin
sin2coscos2sin
2sin
qp
p
BECBp
CECBCECEp
CECFCFCFp
CFp
CFDFp
t
t
t
t
t
t
The maximum value of normal stress
2
2
1
22
1
1
22
2
qpp
p
BECBp
p
CEACCGACAGp
n
n
n
The minimum value of normal stress
ngan tekansikan tegamengindika hanya minus tanda
22
22
22
2
2
2
2
2
2
2
2
2
2
22
2
2
qpp
p
qpp
p
pq
pp
pBECBp
CACECACHAHp
n
n
n
n
n
A point in a strained material is subjected to a compressive stress of 800kg/cm2 and shear stress of 560 kg/cm2. Determine graphically orotherwise the maximum and minimum intensity of direct stresses.
Example :
Solution :
Given: Compressive stress p = 800 kg/cm2 Shear stress q = 560 kg/cm2
• Draw a horizontal line X-X and cut off AB equal to 800 to some suitable scale, andbisect it at C
• Now erect a perpendicular at B, and cut off BE equal to 560 to the scale• Now with C as centre, and radius equal to CE, draw a circle meeting the line X-X’ at
H and G• By measurement, we find that:
Maximum direct stress AG = 1008,2 kg/cm2 (compressive)Minimum direct stress AH = 288,2 kg/cm2 (tensile)
Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to a Direct Stress in Two Mutually Perpendicular
Directions Accompanied by A Simple Shear Stress
Consider a rectangular body ABCD subjected to a tensile stress and shear stress
and an oblique section, on which we are required to find out the stresses.
Let,
p1 = tensile stress on the faces AD and BC
p2 = tensile stress on the faces AB and CD
q = shear stress across the faces DA and BC
Θ = angle which the oblique section EF makes with the normal cross section EG
• Take some point O and draw a horizontal line OX
• Cut off OA and OB equal to the stresses p1 and p2 respectively to some suitablescale on the same side of O ( because the two stresses are alike). Bisect BA at C
• Now erect perpendiculars A and B and cut off AE and BF equal to the shearstress q to the scale.
Draw the circle of the stresses
• Now with C as center and radius equal to CE, draw a circle
meeting the line OX at Q and P
• On the base CE, draw a line CJ at an angle 2θ meeting the
circle at J• From J, draw JK perpendicular to AX. Join OJ
• Now OK and KJ will give the required normal and tangential
stresses to the scale. OJ will give the required resultant
stress to the scale.
• Moreover, OQ and OP will give the maximum and minimum
values of normal stresses to the scale and CG will give the
maximum value of shear stress
Proof:
From the geometry of the figure, we find that normal stress:
2sin2cos22
2sin2cos2
sin2sincos2cos2
sin2sincos2cos2
sin2sincos2cos2
2cos
2121
21
21
21
21
qpppp
p
AECApp
p
CECEpp
p
CJCECJCJpp
p
CJpp
p
CJOCCKOCOKp
n
n
n
n
n
n
And tangential stress:
2cos2sin2
2cos2sin
sin2coscos2sin
sin2coscos2sin
sin2coscos2sin
2sin
21 qpp
p
AECAp
CECEp
CJCECJCJp
CJp
CJKJp
t
t
t
t
t
t
The maximum value of normal stress:
2
2
21211
1
22q
ppppp
CEOCCPOCOPp
n
n
The minimum value of normal stress:
2
2
21212
2
22q
ppppp
CEOCCQOCOQp
n
n
2max 21 nn
t
ppCGp
The maximum shear stress:
A point in a strained material is subjected to a tensile stress of 60N/mm2 and a compressive stress of 40 N/mm2, acting on two mutuallyperpendicular planes and a shear stress of 10 N/mm2 on these planes.Determine principal as well as maximum shear stresses. Also find outthe value of maximum shear stress..
Example :
Solution :
Given:Major stress Tensile stress p1= 60 N/mm2Minor stress Compressive stress p2 = -40 N/mm2 Shear stress q = 10 N/mm2
• Take some point O and draw a horizontal line X’OX
• Cut off OA equal to 60 as OB equal to 40 to some suitable scale on theopposite sides of O. Bisect BA at C
• Now erect a perpendicular at A and cut off AE equal to 10 to the scale
• Now with C as center and radius equal to the CE draw a circle meetingthe line XX’ at Q and P. Also erect a perpendicular at C meeting the circleat G
• By measurement, we find:Major principal stress pn1 = OP = 61 N/mm2Minor principal stress pn2 = OQ = -41 N/mm2Maximum shear stress = pt = CG = 51 N/mm2