Engineering Questions with Answers - Multiple Choice Questions
Elasticity – Boundary Condition Equations
1 - Question
The boundary condition equation for X, where X is the component of the surface force in x-direction per unit area is ___________
a) X = σy m+τyz n+τxy l
b) X = σz n+τzx l+τzy m
c) X = σx l+τyx m+τzx n
d) X = σy l+τyx m+τzx n
View Answer
Answer: c
Explanation: Let the area of ABC=ds.
area OBC=ds cos(N,x)=ds.l
area OAB=ds cos(N,y)=ds.m
area OAC=ds cos(N,z)=ds.n
Resolving all the forces in x-direction and equating them to zero, we get,
X = σx l+τyx m+τzx n.
2 - Question
The boundary condition equation for Y, where Y is the component of the surface force in y-direction per unit area is ___________
a) Y = σy m+τyz n+τxy l
b) Y = σz n+τzx l+τzy m
c) Y = σx l+τyx m+τzx n
d) Y = σy l+τyx m+τzx n
View Answer
Answer: a
Explanation: Let the area of ABC=ds.
area OBC=ds cos(N,x)=ds.l
area OAB=ds cos(N,y)=ds.m
area OAC=ds cos(N,z)=ds.n
Resolving all the forces in y-direction and equating them to zero, we get,
Y = σy m+τyz n+τxy l.
3 - Question
The boundary condition equation for Z, where Z is the component of the surface force in z-direction per unit area is ___________
a) Z = σy m+τyz n+τxy l
b) Z = σz n+τzx l+τzy m
c) Z = σx l+τyx m+τzx n
d) Z = σy l+τyx m+τzx n
View Answer
Answer: b
Explanation: Let the area of ABC=ds.
area OBC=ds cos(N,x)=ds.l
area OAB=ds cos(N,y)=ds.m
area OAC=ds cos(N,z)=ds.n
Resolving all the forces in z-direction and equating them to zero, we get,
Z = σz n+τzx l+τzy m.
4 - Question
The matrix form of the boundary condition equations is _____________
a) ⎡⎣⎢⎢X¯¯¯¯Y¯¯¯¯Z¯¯¯¯⎤⎦⎥⎥=⎡⎣⎢σxxτyxτzxτxyσyyτzyτxzτyzσzz⎤⎦⎥⎡⎣⎢lmn⎤⎦⎥
b) ⎡⎣⎢⎢X¯¯¯¯Y¯¯¯¯Z¯¯¯¯⎤⎦⎥⎥=⎡⎣⎢σzzτyxτzxτxyσyyτzyτxzτyzσxx⎤⎦⎥⎡⎣⎢lmn⎤⎦⎥
c) ⎡⎣⎢⎢X¯¯¯¯Y¯¯¯¯Z¯¯¯¯⎤⎦⎥⎥=⎡⎣⎢σxxτyxτzxτzzσyyτzyτxzτyzσzz⎤⎦⎥⎡⎣⎢lmn⎤⎦⎥
d) ⎡⎣⎢⎢X¯¯¯¯Y¯¯¯¯Z¯¯¯¯⎤⎦⎥⎥=⎡⎣⎢σxxτyxτzxτyyσyyτyyτxzτyzσzz⎤⎦⎥⎡⎣⎢lmn⎤⎦⎥
View Answer
Answer: a
Explanation: The boundary condition equations are given by,
X = σx l+τyx m+τzx n
Y = σy m+τyz n+τxy l
Z = σz n+τzx l+τzy m,
∴ The matrix form is given by,
⎡⎣⎢⎢X¯¯¯¯Y¯¯¯¯Z¯¯¯¯⎤⎦⎥⎥=⎡⎣⎢σxxτyxτzxτxyσyyτzyτxzτyzσzz⎤⎦⎥⎡⎣⎢lmn⎤⎦⎥
5 - Question
The partial differential of normal stress in y-direction in terms of effective stress is given by __________
a) ∂σz‘∂z
b) ∂σz‘∂z−γw∂h∂z
c) ∂σz‘∂z+γw∂h∂z
d) ∂σz‘∂z∗γw∂h∂z
View Answer
Answer: c
Explanation: Since, the normal stress in y-direction in terms of effective stress is given by,
σz= σz’+γw(h-he)
differentiating partially with respect to z,
∂σz∂z=∂σz‘∂z+γw∂(h−he)∂z
But since ∂he∂z=0,
∴ ∂σz∂z=∂σz‘∂z+γw∂h∂z.
6 - Question
The equilibrium equation in X-direction in terms of effected stress for a saturated soil body is given by __________
a) ∂σx‘∂x+∂τyx∂y=0
b) ∂τxy∂x+∂σy‘∂y+∂τzy∂z+γw∂h∂x=0
c) ∂τxz∂x+∂τyz∂y+∂σz‘∂z+γw∂h∂x=0
d) ∂σx‘∂x+∂τyx∂y+∂τzx∂z+γw∂h∂x=0
View Answer
Answer: d
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on x-direction will include seepage force. Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
∂σx∂x+∂τyx∂y+∂τzx∂z=0.
Since σx= σx’+γw(h-he) and ∂σx∂x=∂σx‘∂x+γw∂h∂x
∴ ∂σx‘∂x+∂τyx∂y+∂τzx∂z+γw∂h∂x=0.
7 - Question
The equilibrium equation in Y-direction in terms of effected stress for a saturated soil body is given by __________
a) ∂σx‘∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy‘∂y+∂τzy∂z+γw∂h∂y=0
c) ∂τxz∂x+∂τyz∂y+∂σz‘∂z+Z=0
d) ∂σx‘∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: b
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on y-direction will include seepage force. Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on y-direction is,
∂τxy∂x+∂σy∂y+∂τzy∂z=0.
Since σy= σy’+γw(h-he) and ∂σy∂y=∂σy‘∂y+γw∂h∂y
∴ ∂τxy∂x+∂σy‘∂y+∂τzy∂z+γw∂h∂y=0
8 - Question
The equilibrium equation in Z-direction in terms of effected stress for a saturated soil body is given by __________
a) ∂σx‘∂x+∂τyx∂y+∂τzx∂z++γw∂h∂z=0
b) ∂τxy∂x+∂σy‘∂y+∂τzy∂z+γw∂h∂z=0
c) ∂τxz∂x+∂τyz∂y+∂σz‘∂z+γ′+γw∂h∂z=0
d) ∂σx‘∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: c
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on z-direction will include seepage force. Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on z-direction is,
∂τxz∂x+∂τyz∂y+∂σz∂z+γ=0.
Since σz= σz’+γw(h-he) and ∂σz∂z=∂σz‘∂z+γw∂h∂z
∴ ∂τxz∂x+∂τyz∂y+∂σz‘∂z+γ′+γw∂h∂z=0