mohrs circle bits
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Mohrs Circle BitsTRANSCRIPT
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MOHRS CIRCLE:
1. Two unequal like principal stresses are acting at a point across two perpendicular planes. The
resultant stress acting on a plane inclined at an angle with the major principal plane is to be
determined by Mohrs circle of stress as shown in Fig. 8.19. Two principal stresses in Fig. 8.19 acting
on the point are given by
(a) AO and OB
(b)AC and BC
(c)AB and AC
(d)AD and AB
2. In Fig. 8.19, the radius of the
Mohrs circle is
(a) sum of two principal stress
b) difference of two principal
stress
(c) half of difference of principal
stresses
d) half of sum of two principal stress
83. In Fig. 8.19, the normal stress is given by
(a )AE b)ED c) CE d) AD
84. In Fig. 8.19, the tangential stress is given by
(a )AE b)ED c) CE d) AD
85. In Fig. 8.19, the resultant stress is given by
(a )AE b)ED c) CE d) AD
86. In Fig. 8.19, obliquity is represented by the
(a) angle ECB (B) angle EOD c) angle EAD d) angle AED
87. In Fig. 8.19, the angle of oblique plane is represented by
(a) angle ECB (b) angle EOD (c) angle EAD (d) angle AED.
88. In Fig. 8.19, the maximum tangential stress is equal to
(a) radius of the Mohr's circle (b) diameter of Mohr*s circle (c) circumference of Mohr*s circle
(d)half of radius of Mohr's circle.
89.ln which of the following cases, Mohrs circle is used to determine the stresses on an oblique
plane ?
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(a) two unequal like principal stresses
(b) two unequal unlike principal stresses
(c) direct tensile stress in one plane accompanied by a shear stress
(d) all of the above
(e) none of the above.
90. Fig. 8.20 shows the Mohr's circle of stress for two unequal unlike principal stresses. The two
unequal unlike principal stresses are given by
(a) AC and AD
(b)AC and AB
(c) CE and CD
(d) CO and OB.
91. In Fig. 8.20, radius of Mohr's circle is
(a) sum of two principle stresses
(c) difference of two principle stresses
(c) half of difference of principle stresses
(d) half of sum of two stresses.
92. In Fig. 8.20, the normal stress is given by
(a) ED (b) AE (c) AD (d) CE.
93. In Fig. 8.20, the tangential stress is given by
a) ED (b)AE (c)AD (d) CE.
94.In Fig. 8.20, the resultant stress is given by
(a) ED (b) AD (c)AE (d) CE.
95. In Fig. 8.20, obliquity is represented by
(a) angle ECD (b) angle EOD (b) angle EAD (d) angle AED.
96. In Fig. 8.20, the angle of oblique plane on which normal and tangential stresses are determined,
is given by
(a) angle ECD (b) angle AD (c) angle EOD (d) angle AED.
97.In Fig. 8.20, the maximum shear stress is equal to
(a) radius of Mohrs circle
(b) diameter to Mohr's circle
(c) circumference of Mohr's circle
(d) half of the radius of Mohr's circle.
98. Fig. 8.21 shows a body subjected lo two unequal like direct stresses (p1 and p2) in two mutually
perpendicular planes along with a simple shear stress (q). The maximum normal stress will be
-
a)
+
+
b)
+
+
c)
+
d)
+
99.In question 98, the minimum normal
stress will be
a)
+
+
b)
+
+
c)
+
d)
+
100. In question 98, the maximum shear stress will be
a)
+ b)
+ c) ()
+
d) (+) +
101. A body is subjected to a direct tensile stress of 300 N/cm2 in one plane accompanied by a
simple shear stress of 200 N/cm2. The maximum normal stress will be
(a) 250 N/cm2
(b) 400 N/cm2
(c) -100 N/cm2
(d) 300 N/cm2
102. In question 101, the minimum normal stress will be
(a) 250 N/cm2
(b) 400 N/cm2
(c) -100 N/cm2
(d) 300 N/cm2
103. In question 10], the maximum shear stress will be
(a) 250 N/cm2
(b) 400 N/cm2
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(c) -100 N/cm2
(d) 300 N/cm2
104. A body is subjected to a tensile stress of 1200 N/cm, on one plane and a tensile stress of 600
N/cm2 on the other right-angled plane together with shear stresses of 400 N/cm2 on the same
planes. The maximum normal stress will be
(a) 900 N/cm2 (b) 1400 N/cm2 c) 400 N/cm2 d) 500 N/cm2
105. In question 104, the minimum normal stress would be
(a) 900 N/cm2 (b) 1400 N/cm2 c) 400 N/cm2 d) 500 N/cm2
106. In question 104, the greatest shear stress would be
(a) 900 N/cm2 (b) 1400 N/cm2 c) 400 N/cm2 d) 500 N/cm2
107. A strained element is subjected to principal stresses p1 and p2 in two mutually perpendicular
directions. The strain (e1) in the direction of principal stress (p1) is given by
a) =
+
b) =
c) =
+
d) =
108. In question 107, the strain in the direction of principal stress (e2 ) is given by
a) =
+
b) =
c) =
+
d) =
109. In question 107, the strain energy stored per unit volume is equal to
a)
+
b)
+
c) both a and b d) none of the above
110. A strained element is subjected to two principal stresses of 800 N/cm2, and -200N/cm2. If the
Poisson ratio = 0.25 and E= 2 x 10& N/cm2, the strain in the direction of principal stress (800 N/cm2)
will be
(a)0.000425 (b) 0.0002 (c) .0004 (d) 0.0001.
111. In question 110, the strain in the direction of principal stress (- 200 N/cm2) will be
(a)0.000425 (b) 0.0002 (c) .0004 (d) 0.0001.
112. In question 110, the strain energy stored per unit volume would be
(a) 0.125 N/cm2 (b) 0.250 N/cm2 (c) 038 N/cm2 (d) 1.0 N/cm2
113. If p1, p2 and p3 be the principal stresses at a point in a strained material, then strain energy
stored per unit volume will be
a)
+ +
b)
+ +
+
c)
+ +
d) None of the above