Question 1

A steel plate weighing 27 kN is hoisted by a cable sling that has a clevis at each end. The pins through the clevises are 22 mm in diameter.
Each half of the cable is at an angle of 35 ° to the vertical. The average
Shear stress in each pin is approximately: (A) $ 22 \mathrm{MPa} $
(B) $ 28 \mathrm{MPa} $
(C) $ 40 \mathrm{MPa} $
(D) $ 48 \mathrm{MPa} $

Accepted Answer

C

Question 2

A nonprismatic cantilever bar has an internal cylindrical hole of diameter $d / 2$ /2 from 0 to x, so the net area of the cross section for Segment 1 $(3 / 4) A$ . Load P is applied at x, and load $-P / 2$ is applied at x = L.
Assume that E is constant. The length of the hollow segment, x, required
To obtain axial displacement $\delta=P L / E A$ at the free end is: (A) $ x=L / 5 $
(B) $ x=L / 4 $
(C) $ x=L / 3 $
(D) $ x=3 L / 5 $

Accepted Answer

D

Question 3

Two wires, one copper and the other steel, of equal length stretch the same amount under an applied load P. The moduli of elasticity for
Each is $E_{s}=210 \mathrm{GPa} \text { and } E_{c}=120 \mathrm{GPa} .$ . The ratio of the diameter of
The copper wire to that of the steel wire is approximately: (A) $ 1.00 $
(B) $ 1.08 $
(C) $ 1.19 $
(D) $ 1.32 $

Accepted Answer

D

Question 4

Two flanged shafts are connected by eight 18-mm bolts. The diam- eter of the bolt circle is 240 mm. The allowable shear stress in the bolts is
90 MPa. Ignore friction between the flange plates. The maximum value of
Torque $T_{0}$ is approximately: (A) $ 19 \mathrm{kN} \cdot \mathrm{m} $
(B) $ 22 \mathrm{kN} \cdot \mathrm{m} $
(C) $ 29 \mathrm{kN} \cdot \mathrm{m} $
(D) $ 37 \mathrm{kN} \cdot \mathrm{m} $

Accepted Answer

The answer of Two flanged shafts are connected by eight...

Question 5

A nylon bar (E = 2.1 GPa) with diameter 12 mm, length 4.5 m, and weight 5.6 N hangs vertically under its own weight. The elongation of
The bar at its free end is approximately: (A) $ 0.05 \mathrm{~mm} $
(B) $ 0.07 \mathrm{~mm} $
(C) $ 0.11 \mathrm{~mm} $
(D) $ 0.17 \mathrm{~mm} $

Accepted Answer

The answer of A nylon bar (E = 2.1 GPa)...

Question 6

An elastomeric bearing pad is subjected to a shear force V during a static loading test. The pad has dimensions a = 150 mm and b = 225 mm,
And thickness t = 55 mm. The lateral displacement of the top plate with
Respect to the bottom plate is 14 mm under a load P =16 kN. The shear
Modulus of elasticity G of the elastomer is approximately: (A) $ 1.0 \mathrm{MPa} $
(B) $ 1.5 \mathrm{MPa} $
(C) $ 1.7 \mathrm{MPa} $
(D) $ 1.9 \mathrm{MPa} $

Accepted Answer

The answer of An elastomeric bearing pad is subjected to...

Question 7

An aluminum bar $(E=72 \mathrm{GPa}, v=0.33)$ of diameter 50 mm cannot exceed a diameter of 50.1 mm when compressed by axial force P.
The maximum acceptable compressive load P is approximately:
A) 190 kN
B) 200 kN
C) 470 kN
D) 860 kN

Accepted Answer

The answer of An aluminum bar $(E=72 \mathrm{GPa}, v=0.33)$ of...

Question 8

An aluminum bar $(E=70 \mathrm{GPa}, v=0.33)$ of diameter 20 mm is stretched by axial forces P, causing its diameter to decrease by 0.022 mm.
The load P is approximately:
A) 73 kN
B) 100 kN
C) 140 kN
D) 339 kN P d P

Accepted Answer

The load P can be calculated using the formula for lateral strain in the context of axial loading and Poisson's ratio. The lateral strain is the change in diameter divided by the original diameter, and Poisson's ratio ($v$) relates the lateral strain to the axial strain. The axial strain can then be related to the stress ($\sigma$) using Hooke's Law ($\sigma = E \cdot \epsilon$), and finally, the stress can be related to the force (P) using the formula $\sigma = \frac{P}{A}$, where $A$ is the cross-sectional area of the bar. Given the decrease in diameter and the properties of aluminum, the calculations lead to a load P of approximately 73 kN.

Question 9

A copper tube with wall thickness of 8 mm must carry an axial ten- sile force of 175 kN. The allowable tensile stress is 90 MPa. The minimum
Required outer diameter is approximately: (A) $ 60 \mathrm{~mm} $
(B) $ 72 \mathrm{~mm} $
(C) $ 85 \mathrm{~mm} $
(D) $ 93 \mathrm{~mm} $

Accepted Answer

The answer of A copper tube with wall thickness of...

Question 10

A pipe (E =  110 GPa) carries a load $P_{\mathrm{I}}=120 \mathrm{kN}$ at A and a uniformly distributed load $P_{2}=100 \mathrm{kN}$ on the cap plate at B. Initial pipe
Diameters and thicknesses are $d_{A B}=38 \mathrm{~mm}$ , $t_{A B}=12 \mathrm{~mm}$ , $d_{B C}=70 \mathrm{~mm}$ , and $t_{B C}=10 \mathrm{~mm}$ . Under loads $P_{1}$ and $P_{2}$ , wall thickness $t_{B C}$ increases by 0.0036 mm. Poisson's ratio ν for the pipe material is
Approximately:
A) 0.27
B) 0.30
C) 0.31
D) 0.34

Accepted Answer

The answer of A pipe (E =  110 GPa)...

Question 11

A plane truss with span length L = 4.5 m is constructed using cast iron pipes (E = 170 GPa) with a cross-sectional area of 4500 mm $4500 \mathrm{~mm}^{2}$ . The
Displacement of joint B cannot exceed 2.7 mm. The maximum value of
Loads P is approximately: (A) $ 340 \mathrm{kN} $
(B) $ 460 \mathrm{kN} $
(C) $ 510 \mathrm{kN} $
(D) $ 600 \mathrm{kN} $

Accepted Answer

The answer of A plane truss with span length L...

Question 12

An polyethylene bar (E = 1.4 GPa, v = 0.4) of diameter 80 mm is inserted in a steel tube of inside diameter 80.2 mm and then compressed
By axial force P. The gap between steel tube and polyethylene bar will
Close when compressive load P is approximately:
A) 18 kN
B) 25 kN
C) 44 kN bar Steeltube
D) 60 kN Polyethylene d1 d2

Accepted Answer

To determine when the gap between the steel tube and the polyethylene bar will close, we need to calculate the compressive load required to cause enough deformation in the polyethylene bar so that it fills the gap. The gap is 0.2 mm (80.2 mm - 80 mm). The deformation needed can be calculated using the formula for radial deformation under axial load, considering the material's Young's modulus (E) and Poisson's ratio (v).Given:- Diameter of polyethylene bar = 80 mm- Gap = 0.2 mm- E (Young's modulus for polyethylene) = 1.4 GPa = 1400 MPa- v (Poisson's ratio for polyethylene) = 0.4The radial deformation ($\Delta r$) can be related to the axial stress ($\sigma$) by the relation $\Delta r = \frac{\sigma \cdot r}{E} \cdot v$, where $r$ is the initial radius of the bar. To find the stress ($\sigma$), we rearrange the formula to $\sigma = \frac{\Delta r \cdot E}{r \cdot v}$.Substituting the given values and converting units appropriately:- $\Delta r = 0.1$ mm (since the gap is 0.2 mm, but the deformation is radial, so it's half the gap)- $r = 40$ mm (radius of the bar)- $E = 1400$ MPa- $v = 0.4$$\sigma = \frac{0.1 \cdot 1400}{40 \cdot 0.4} = 8.75$ MPaThe force ($P$) required to achieve this stress is found by multiplying the stress by the cross-sectional area of the bar ($A = \pi r^2$):$P = \sigma \cdot A = 8.75 \cdot \pi \cdot (40)^2 \cdot 10^{-6}$ kN$P \approx 44$ kNTherefore, a compressive load of approximately 44 kN is required to close the gap between the steel tube and the polyethylene bar.

Question 13

A brass rod (E = 110 GPa) with a cross-sectional area of 250 mm2 is loaded by forces $P_{1}=15 \mathrm{kN}, P_{2}=10 \mathrm{kN} \text {, and } P_{3}=8 \mathrm{kN}$ . Segment
Lengths of the bar are a = 2.0 m, b = 0.75 m, and c = 1.2 m. The
Change in length of the bar is approximately: (A) $ 0.9 \mathrm{~mm} $
(B) $ 1.6 \mathrm{~mm} $
(C) $ 2.1 \mathrm{~mm} $
(D) $ 3.4 \mathrm{~mm} $

D

Accepted Answer

The answer of A brass rod (E = 110 GPa)...

Question 14

The moment reaction at A in the plane frame below is approximately: (A) $ +1400 \mathrm{~N} \cdot \mathrm{m} $
(B) $ -2280 \mathrm{~N} \cdot \mathrm{m} $
(C) $ -3600 \mathrm{~N} \cdot \mathrm{m} $
(D) $ +6400 \mathrm{~N} \cdot \mathrm{m} $

Accepted Answer

The answer of The moment reaction at A in the...

Question 15

A plane truss has downward applied load P at joint 2 and another load P applied leftward at joint 5. The force in member 3-5 is: (A) 0
(B) $ -P / 2 $
(C) $ -P $
(D) $ +1.5 P $

Accepted Answer

The answer of A plane truss has downward applied load...

Question 16

A bar of diameter d = 18 mm and length L = 0.75 m is loaded in tension by forces P. The bar has modulus E = 45 GPa and allowable
Normal stress of 180 MPa. The elongation of the bar must not exceed
2.7 mm. The allowable value of forces P is approximately:
A) 41 kN
B) 46 kN
C) 56 kN
D) 63 kN

Accepted Answer

The allowable force P can be calculated using the formula for stress ($\sigma = \frac{P}{A}$) and the formula for elongation ($\Delta L = \frac{PL}{AE}$), where $A$ is the cross-sectional area of the bar, $E$ is the modulus of elasticity, $L$ is the length of the bar, and $\Delta L$ is the allowable elongation. Given the allowable stress of 180 MPa and the diameter $d = 18$ mm, the cross-sectional area $A = \frac{\pi d^2}{4}$. Substituting the given values and solving for $P$ yields an allowable force of approximately 41 kN.

Question 17

A steel wire hangs from a high-altitude balloon. The steel has unit weight 77 kN/m3 and yield stress of 280 MPa. The required factor of
Safety against yield is 2.0. The maximum permissible length of the wire is
Approximately:
A) 1800 m
B) 2200 m
C) 2600 m
D) 3000 m

Accepted Answer

The maximum permissible length of the wire can be calculated using the formula for stress due to the self-weight of the wire, which is $\sigma = \rho g h$, where $\sigma$ is the stress, $\rho$ is the density (which can be converted from unit weight), $g$ is the acceleration due to gravity (approximately $9.81 m/s^2$), and $h$ is the height or length of the wire. Given a factor of safety of 2, the allowable stress is $280 \, \text{MPa} / 2 = 140 \, \text{MPa}$. Converting unit weight to density, $\rho = 77 \, \text{kN/m}^3 = 77000 \, \text{N/m}^3$. Rearranging the formula to solve for $h$, we get $h = \sigma / (\rho g)$. Substituting the given values, $h = 140 \times 10^6 \, \text{N/m}^2 / (77000 \, \text{N/m}^3 \times 9.81 \, \text{m/s}^2)$, which calculates to approximately 1800 meters.

Question 18

A titanium bar (E = 100 GPa, v = 0.33) with square cross sec- tion (b = 75 mm) and length L = 3.0 m is subjected to tensile load
P =900 kN. The increase in volume of the bar is approximately: (A) $ 1400 \mathrm{~mm}^{3} $
(B) $ 3500 \mathrm{~mm}^{3} $
(C) $ 4800 \mathrm{~mm}^{3} $
(D) $ 9200 \mathrm{~mm}^{3} $

Accepted Answer

The answer of A titanium bar (E = 100 GPa,...

Question 19

The force in member FE of the plane truss below is approximately: (A) $ -1.5 \mathrm{kN} $
(B) $ -2.2 \mathrm{kN} $
(C) $ 3.9 \mathrm{kN} $
(D) $ 4.7 \mathrm{kN} $

Accepted Answer

The answer of The force in member FE of the...

Question 20

A brass bar (E = 110 MPa) of length L = 2.5 m has diameter $d_{1}=18 \mathrm{~mm}$ over one-half of its length and diameter $d_{2}=12 \mathrm{~mm}$ over the other half. Compare this nonprismatic bar to a prismatic bar of the
Same volume of material with constant diameter d and length L. The elon-
Gation of the prismatic bar under the same load P = 25 kN is approxi-
Mately:
A) 3 mm
B) 4 mm
C) 5 mm
D) 6 mm

Accepted Answer

The answer of A brass bar (E = 110 MPa)...

Quiz 1: Mechanics of Materials

An aluminum bar $(E=72 \mathrm{GPa}, v=0.33)$ of diameter 50 mm cannot exceed a diameter of 50.1 mm when compressed by axial force P. The maximum acceptable compressive load P is approximately:

An aluminum bar $(E=70 \mathrm{GPa}, v=0.33)$ of diameter 20 mm is stretched by axial forces P, causing its diameter to decrease by 0.022 mm. The load P is approximately:

An polyethylene bar (E = 1.4 GPa, v = 0.4) of diameter 80 mm is inserted in a steel tube of inside diameter 80.2 mm and then compressed By axial force P. The gap between steel tube and polyethylene bar will Close when compressive load P is approximately:

A plane truss has downward applied load P at joint 2 and another load P applied leftward at joint 5. The force in member 3-5 is: (A) 0 (B) $-P / 2$ (C) $-P$ (D) $+1.5 P$

A bar of diameter d = 18 mm and length L = 0.75 m is loaded in tension by forces P. The bar has modulus E = 45 GPa and allowable Normal stress of 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowable value of forces P is approximately:

A steel wire hangs from a high-altitude balloon. The steel has unit weight 77 kN/m3 and yield stress of 280 MPa. The required factor of Safety against yield is 2.0. The maximum permissible length of the wire is Approximately:

Quiz 1: Mechanics of Materials

An aluminum bar (E=72GPa,v=0.33) (E=72 \mathrm{GPa}, v=0.33) (E=72GPa,v=0.33) of diameter 50 mm cannot exceed a diameter of 50.1 mm when compressed by axial force P. The maximum acceptable compressive load P is approximately:

An aluminum bar (E=70GPa,v=0.33) (E=70 \mathrm{GPa}, v=0.33) (E=70GPa,v=0.33) of diameter 20 mm is stretched by axial forces P, causing its diameter to decrease by 0.022 mm. The load P is approximately: