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

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

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

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⎤⎦⎥

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) σzz
b) σzzγwhz
c) σzz+γwhz
d) σzzγwhz

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,
σzz=σzz+γw(hhe)z
But since hez=0,
∴ σzz=σzz+γwhz.

6 - Question

The equilibrium equation in X-direction in terms of effected stress for a saturated soil body is given by __________
a) σxx+τyxy=0
b) τxyx+σyy+τzyz+γwhx=0
c) τxzx+τyzy+σzz+γwhx=0
d) σxx+τyxy+τzxz+γwhx=0

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,
σxx+τyxy+τzxz=0.
Since σx= σx’+γw(h-he) and σxx=σxx+γwhx
∴ σxx+τyxy+τzxz+γwhx=0.

7 - Question

The equilibrium equation in Y-direction in terms of effected stress for a saturated soil body is given by __________
a) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz+γwhy=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

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,
τxyx+σyy+τzyz=0.
Since σy= σy’+γw(h-he) and σyy=σyy+γwhy
∴ τxyx+σyy+τzyz+γwhy=0

8 - Question

The equilibrium equation in Z-direction in terms of effected stress for a saturated soil body is given by __________
a) σxx+τyxy+τzxz++γwhz=0
b) τxyx+σyy+τzyz+γwhz=0
c) τxzx+τyzy+σzz+γ+γwhz=0
d) σxx+τyxy+τzxz=0