Bending and Shear Stress in Beam and Torsion
Exam Duration: 45 Mins Total Questions : 25
Moment of inertia of a semi-circle about its XX-axis is given by
- (a)
0.22 r3
- (b)
0.11 r4
- (c)
0.2 r4
- (d)
0.14 r4
The strength of the beam mainly depends on
- (a)
bending moment
- (b)
CG of the section
- (c)
section modulus
- (d)
All of these
If the bending moment is consistent there will be no stresses.
- (a)
compressive
- (b)
tensile
- (c)
shearing
- (d)
All of these
In the case of an I-section beam, maximum shear stress is at
- (a)
top of flange
- (b)
junction of flange and web
- (c)
middle of web
- (d)
None of the above
A shaft is subjected to torsion as shown in figure.
Which of the following figures represent the shear stress on the element LMNOPQRS?
- (a)
- (b)
- (c)
- (d)
The strength of a hollow shaft for the same length, material and weight is __________ a solid shaft.
- (a)
more than
- (b)
less than
- (c)
equal to
- (d)
None of these
In case of a circular section, the maximum shear stress is _______ per cent more than the mean shear stress
- (a)
20
- (b)
10
- (c)
66
- (d)
33
By pure bending of a beam, we meant that
- (a)
shear force in the beam is uniform throughout
- (b)
bending moment in the beam is uniform throughout
- (c)
bending moment in beam is zero
- (d)
shear force in beam is zero
Neutral axis of a beam is the axis at which
- (a)
the shear force is zero
- (b)
the moment of inertia is zero
- (c)
the bending stress is maximum
- (d)
the bending stress is zero
When a beam is loaded transversally the maximum compressive stress shall develop at
- (a)
the bottom fibres
- (b)
the neutral axis
- (c)
the top fibres
- (d)
None of these
Two shafts having same length and material are joined in series. If the ratio of their diameters is 2, then the ratio of their angles of twist shall be
- (a)
16
- (b)
8
- (c)
2
- (d)
4
For two shafts connected in series, find which statements is true
- (a)
torque in each shaft is the same
- (b)
shear stress in each shaft is the same
- (c)
torsional stiffness of each shaft is same
- (d)
angle of twist of each shaft is same
A solid shaft of diameter d is replaced by a hollow shaft of the same material and length and having outside diameter \(\frac{2}{\sqrt{3}}d\) and inside \(\frac{1}{\sqrt{3}}d\) . The weight of the hollow shaft will be-the weight of solid shaft
- (a)
\(\frac{2}{3}\)
- (b)
\(\frac{5}{3}\)
- (c)
twice
- (d)
same as
Solid | Hollow |
G | G |
l | l |
d | \(d_0=\frac{2}{\sqrt{3}}d, d_i=\frac{d}{\sqrt{3}}\) |
Weight of hollow shaft = A x l x ρ
=\(\frac{\pi}{4}(d_0^2-d_i^2)\times l\times p\)
=\(\frac{\pi}{4}(\frac{4}{3}d^2-\frac{d^2}{3})\rho l\)
=\(\frac{\pi}{4}d^2\rho l\)
Weight of solid shaft
=\(\frac{\pi}{4}d^2\rho l\)
If shafts are connected in parallel, then which option will be true?
- (a)
torque in each shaft is the same
- (b)
shear stress in each shaft is the same
- (c)
torsional stiffness of each shaft is same
- (d)
angle of twist of each shaft is same
T=T1+T2
θ1=θ2
A beam is in triangular section with a hole of 20 mm diameter
Moment of inertia of this cross-section about its neutral axis is
- (a)
2.87x106 mm4
- (b)
3.74x106 mm4
- (c)
4.01x106 mm4
- (d)
1.29x106 mm4
Centre of gravity of triangle from base = \(\frac{h}{3}=\frac{120}{3}=40\)mm while centre of gravity of hole lies at this point. Hence, position of neutral axis of section is at 40 mm above the base.
\(\frac{bh^3}{36}-\frac{\pi d^4}{64}\)
=\(\frac{60\times120^3}{36}-\frac{\pi\times20^4}{64}\)
=2.87x106mm4
A beam is in triangular section with a hole of 20 mm diameter
If maximum stress in beam section not to exceed 100 N/mm2, then maximum bending moment on beam is
- (a)
3.6 kN-m
- (b)
10 kN-m
- (c)
4 kN-m
- (d)
5.6 kN-m
We know that at neutral axis bending stress is zero
\(\therefore\frac{\sigma_1}{80}=\frac{\sigma_2}{40}\Rightarrow\sigma_1=2\sigma_2\)
At top stress is maximum.
By bending formula,
\(\frac{\sigma}{Y}=\frac{M}{l}\)
M=\(\frac{100}{80}\times2.87\times10^6\)
=3.6x106 N-mm
=3.6 kN-m
Position of centroid of an area as shown in figure if r1=180 and r2=135 is
- (a)
\(\bar{X}=103,\bar{Y}=0\)
- (b)
\(\bar{X}=0,\bar{Y}=0\)
- (c)
\(\bar{X}=0,\bar{Y}=101\)
- (d)
\(\bar{X}=103,\bar{Y}=103\)
\(\bar{X}=0\)
\(\bar{Y}=\frac{A_1y_1-A_2y_2}{A_1-A_2}\)
\(y_1=\frac{4r_1}{3\pi}, y_2=\frac{4r_2}{3\pi}\)
\(\bar{Y}=\frac{\frac{\pi r_1^2}{2}\frac{4r_1}{3\pi}-\frac{\pi r_2^2}{2}\frac{4r_2}{3\pi}}{\frac{\pi}{2}(r_1^2-r_2^2)}\)
=\(\frac{\frac{2}{3\pi}(r_1^3-r_2^3)}{\frac{1}{2}(r_1^2-r_2^2)}\)=101mm
A timber beam of 3 m span carries a uniformly distributed load of 5 kN/m and a point load 1 kN at the centre of the span. If the permissible bending stress be 100 N/mm2, determine the section taking depth as twice the breadth
- (a)
90x45 mm2
- (b)
55x40 mm2
- (c)
46x92 mm2
- (d)
30x70 mm2
Given d=2b
σmax=100 N/mm2
Maximum bending moment due to point load = \(\frac{wl}{4}\)
=\(\frac{1\times3}{4}\)=0.75 kN-m
Maximum bending moment due to UDL
=\(\frac{wl^2}{8}\)
=\(\frac{5\times(3)^2}{8}\)=5.62 kN-m
Total maximum bending moment=0.75 + 5.62 = 6.375 kN-m
\(I=\frac{bd^3}{12}=\frac{b(2b)^3}{12}=\frac{8b^4}{12}\)
By bending equation,
\(\frac{\sigma}{Y}=\frac{M}{I}\)
\(\Rightarrow \sigma=\frac{M}{I}(\frac{d}{2})\)
100=\(\frac{5.62\times1000\times1000}{\frac{8}{12}(b)^4}\times b\)
b=46 mm
d=92 mm
A cantilever 2.5 m long carries a UDL of 20 kN/m run. The breadth of the section remains constant and is equal to 100 mm. Determine the depth of the section at the middle of the length of the cantilever if stress remains the same throughout and equal to 120 MN/m2.
- (a)
70 mm
- (b)
90 mm
- (c)
85 mm
- (d)
88 mm
Given, b = 100 mm
To find d at middle, we have to determine bending moment at middle
MC=20x1.25x\(\frac{1.25}{2}\)
=15.625 kN-m
Now, by bending equation
\(\frac{\sigma}{y}=\frac{M}{I}, I=\frac{bd^3}{12}, Y=\frac{d}{2}, \sigma=120\)N/mm2
M=15.624 kN-m
Solving it, we get d=88.3 mm
In the given figure, angIe of twist in each part is (G = 80 GPa = 80 x 103 MPa)
- (a)
0.003 rad
- (b)
0.005 rad
- (c)
0.006 rad
- (d)
0.001 rad
Its a case of indeterminate.
Shaft first of all draw free body diagram at junction
TA+TB=T=120 ...(i)
Angle of twist in each section will be same
θA=θB
\(\Rightarrow\frac{T_AL_A}{GJ_A}=\frac{T_BL_B}{GJ_B}\)
\(\Rightarrow\frac{T_A\times400}{(200)^4}=\frac{T_B\times200}{(120)^4}\)
TA=3.85TB ...(ii)
Solving Eqs. (i) and (ii), we get
TA = 95.3 kN-m
TB = 24.7 kN-m
=3.03x10-3 rad
If maximum permissible shear stress is 80 MPa, then value of P is
- (a)
4.7 kN
- (b)
5.2 kN
- (c)
3.9 kN
- (d)
7.1 kN
P will produce torque at point
T = P x (200 + 200)
= 400P
TA + TB = Te = 400 P ...(i)
\({ \theta }_{ AC }={ \theta }_{ CB }=\frac { { T }_{ A }{ L }_{ A } }{ { GJ }_{ A } } =\frac { { T }_{ B }{ L }_{ B } }{ { GJ }_{ B } } \)
As JA = JB
So, TALA = TBLB
Solving Eqs. (i) and (ii), we get
TB = 211.64 P ....(ii)
Solving Eqs. (i) and (ii), we get
TB = 211.64 P
As
\({ \tau }_{ max }=\)= 80 MPa = 80 N/mm2
By torsion equation,
\(\frac { I }{ J } =\frac { \tau }{ R } \)
We get, P = 4.7 kN
A hollow shaft has 50 mm outside diameter and 30 mm internal diameter. An applied torque of 1.5 N-m is found to produce and angular twist of 0.4°, mesured on a length of 0.2 m the shaft. The value of the modulus of rigidity
- (a)
70 kN/m2
- (b)
90 kN/m2
- (c)
60 kN/m2
- (d)
80 kN/m2
\(J=\frac{\pi}{32}(d_0^4-d_i^4)\)
=\(\frac{\pi}{32}(0.05^4-0.03^4)\)
\(\theta=0.4^0=0.4^0\times\frac{\pi}{180^0}\)
=0.00698 rad
l=0.2m
T=1.5N-m
By torsion equation
\(\frac{T}{J}=\frac{\tau}{r}=\frac{G\theta}{l}\)
We get,
G=80.74 kN/m2
The maximum power which could be transmitted by the shaft at 2000 rpm, if maximum allowable shearing stress is 70 MN/m2.
- (a)
300 kW
- (b)
310 kW
- (c)
313kW
- (d)
318 kW
Maximum power
P=\(\frac{2\pi NT_{max}}{60\times1000}\)
Now, we have to determine Tmax
\(\frac{T}{J}=\frac{\tau}{r}\)
If \(\tau=\tau_{max}\)
Then, T=Tmax
ஃ Tmax=\(\frac{J}{r}\tau_{max}\)
=1495.4 N-m
P=\(\frac{2\pi\times2000\times1495.4}{60\times1000}\)
=313.2 kW
A hollow shaft is subjected to a bending moment 30 kN-m and a torque of 40 kN-m. If tmax = 80 MN/m2 and d0 = 2di, then
- (a)
d0 = 75mm
- (b)
d0 = 100 mm
- (c)
d0 = 180 mm
- (d)
d0 = 150 mm
M = 30 kN-m, T=40 kN-m
ஃ Equivalent torque
Te=\(\sqrt{M^2+T^2}\)=50 kN-m
\(\frac{T_e}{J}=\frac{\tau}{R}\)
\(\Rightarrow\frac{50}{\frac{\pi}{32}(d_0^2-d_i^2)}=\frac{80\times1000}{(\frac{d_0}{2})}\)
Putting d0=2di
⇒ d0=150 mm
On the basis of given figure, select the right option.
- (a)
TAB = 3300, TBC = 2200
- (b)
TAB = 6380, TBC = 3080
- (c)
TAB = 880, TBC = 2200
- (d)
TAB = 0, TBC = 1100
Draw free body diagram
Net torque at any point should be according to question