Pomodoro
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Difficulty
A steel bar carries a tensile force of $24\ \text{kN}$ and has a cross-sectional area of $600\ \text{mm}^2$.
What is the average normal stress in the bar?
Solution
Use the axial stress formula:
Substitute the values:
So the normal stress is
A bar with original length $750\ \text{mm}$ elongates by $1.5\ \text{mm}$.
What is the engineering strain?
Solution
Use the strain definition:
Substitute the values:
So the engineering strain is
A linearly elastic material has Young's modulus $E = 210\ \text{GPa}$. If it experiences a normal stress of $105\ \text{MPa}$, what is the axial strain?
Solution
Use Hooke's law:
Solve for strain:
Substitute:
So the axial strain is
A steel rod has length $2.5\ \text{m}$ and coefficient of thermal expansion
If the temperature increases by $20^\circ\text{C}$ and the rod is free to expand, how much does its length change?
Solution
Use free thermal expansion:
Substitute:
Convert to millimeters:
A solid circular shaft with diameter $60\ \text{mm}$ carries a torque of $600\ \text{N}\cdot\text{m}$.
What is the maximum shear stress in the shaft?
Solution
Use the torsion formula at the outer surface:
For a solid circular shaft,
Substitute $T = 600\,000\ \text{N}\cdot\text{mm}$ and $d = 60\ \text{mm}$:
A solid circular shaft has length $1.5\ \text{m}$, diameter $50\ \text{mm}$, shear modulus $G = 80\ \text{GPa}$, and applied torque $900\ \text{N}\cdot\text{m}$.
What is the angle of twist?
Solution
Use the torsion deformation formula:
For a solid circular shaft,
Substitute in consistent units:
Then
In degrees, this is about
A beam section is rectangular with width $100\ \text{mm}$ and height $150\ \text{mm}$. The internal bending moment at a section is $8\ \text{kN}\cdot\text{m}$.
What is the maximum bending stress?
Solution
Use the flexure formula:
For a rectangle,
Here,
and
Convert the moment:
Now compute:
A rectangular beam has width $50\ \text{mm}$ and height $150\ \text{mm}$. The internal shear force is $15\ \text{kN}$.
What is the maximum shear stress at the neutral axis?
Solution
For a rectangular section, the maximum shear stress is
where
Substitute:
So the maximum shear stress is
A beam's shear-force diagram is being drawn from left to right. At one point, the beam crosses a $12\ \text{kN}$ downward concentrated load.
What happens to the shear-force diagram at that point?
Solution
A downward concentrated load causes a downward jump in the shear diagram by the same magnitude.
So the shear-force diagram drops by
at that point.
A pinned-pinned column has
What is the critical buckling load?
Solution
Use Euler buckling with $K = 1$ for pinned-pinned ends:
Substitute in consistent units:
So
Difficulty
A steel bar has two segments in series. Segment 1 has length $400\ \text{mm}$ and area $200\ \text{mm}^2$. Segment 2 has length $600\ \text{mm}$ and area $300\ \text{mm}^2$. The bar carries a tensile force of $12\ \text{kN}$ throughout, and $E = 200\ \text{GPa}$.
What is the total elongation?
Solution
Use the deformation formula on each segment:
For segment 1:
For segment 2:
So the total elongation is
A stepped steel bar is in tension with an axial force of $18\ \text{kN}$. Its segments are:
Take $E = 200\ \text{GPa}$.
Find the stress in each segment and the total elongation.
Solution
The axial force is the same in every segment, so the stress in each segment is
Segment 1:
Segment 2:
Segment 3:
Now compute elongation segment by segment:
A steel bar with length $2\ \text{m}$ is fixed between rigid walls. Its coefficient of thermal expansion is
and $E = 200\ \text{GPa}$. If the temperature rises by $35^\circ\text{C}$, what thermal stress develops?
Solution
For a fully restrained bar, the total strain is zero, so the thermal stress is
Substitute:
Because the bar is prevented from expanding, the stress is compressive.
Two bars connect rigid plates in parallel. Each bar is $800\ \text{mm}$ long and has cross-sectional area $400\ \text{mm}^2$. One bar is steel with $E = 200\ \text{GPa}$, and the other is aluminum with $E = 70\ \text{GPa}$. The plates are pulled by a total tensile load of $27\ \text{kN}$.
How much force does each bar carry?
Solution
Because the bars are in parallel between rigid plates, they have the same elongation. So the force in each bar is proportional to its stiffness:
Since $A$ and $L$ are the same, the force ratio is
The total load is $27\ \text{kN}$, so there are $27$ parts total.
Steel bar:
Aluminum bar:
The common elongation is
A point in a member is under the plane stress state
Find the principal stresses and the maximum in-plane shear stress.
Solution
Use the plane stress transformation results:
Compute the average stress:
and the radius:
So the principal stresses are
The maximum in-plane shear stress is
Using the same plane stress state
find the normal stress and shear stress on a plane rotated $45^\circ$ counterclockwise.
Solution
Use the stress transformation equations:
With $\theta = 45^\circ$, we have
So
and
The normal stress on the rotated plane is $78\ \text{MPa}$, and the shear stress is $-24\ \text{MPa}$.
A rectangular beam has width $50\ \text{mm}$ and height $150\ \text{mm}$. The internal shear force is $15\ \text{kN}$.
Find the maximum shear stress at the neutral axis using the beam shear formula.
Solution
For a rectangular section, the maximum shear stress occurs at the neutral axis and can be found from
Here,
At the neutral axis, the area above the cut is
and its centroid is $37.5\ \text{mm}$ from the neutral axis, so
Using $t = b = 50\ \text{mm}$, the result simplifies to
with
So
Difficulty
A cantilever beam has length $2\ \text{m}$ and an end load of $4\ \text{kN}$. The beam has $E = 200\ \text{GPa}$ and $I = 8.0 \times 10^6\ \text{mm}^4$.
Find the tip deflection using the moment-curvature relation.
Solution
Take $x$ from the fixed end. The bending moment is
and the beam equation is
So
Integrate once:
At the fixed end, the slope is zero, so $C_1 = 0$.
Integrate again:
At the fixed end, the deflection is zero, so $C_2 = 0$.
At $x = L$,
Substitute $P = 4000\ \text{N}$, $L = 2000\ \text{mm}$, $E = 200\,000\ \text{N/mm}^2$, and $I = 8.0 \times 10^6\ \text{mm}^4$:
So the tip deflection is about $6.7\ \text{mm}$ downward.
A motor delivers $12\ \text{kW}$ at $900\ \text{rpm}$ to a solid circular shaft. If the allowable shear stress is $40\ \text{MPa}$, what minimum shaft diameter is required?
Solution
First convert power and speed to torque:
The angular speed is
So the torque is
Now use the torsion stress formula for a solid shaft:
Solve for $d$:
Substitute $T = 127\,300\ \text{N}\cdot\text{mm}$ and $\tau_{allow} = 40\ \text{N/mm}^2$:
So the minimum practical diameter is
A member carries a compressive axial load of $30\ \text{kN}$ and an internal bending moment of $6\ \text{kN}\cdot\text{m}$. Its cross-section is a rectangle with width $80\ \text{mm}$ and height $160\ \text{mm}$.
Find the normal stress at the top fiber and at the bottom fiber.
Solution
First compute the axial stress:
This is compressive.
Next compute the bending stress:
For a rectangle,
and
Convert the moment:
So
If the moment puts the top fiber in compression, then:
So the top fiber is more highly compressed, and the bottom fiber is in tension.
A steel column has length $3\ \text{m}$, pinned-pinned ends, Young's modulus $E = 200\ \text{GPa}$, and a circular cross-section with diameter $40\ \text{mm}$. The yield strength is $250\ \text{MPa}$.
Compute the Euler critical load and the yield load. Which one is smaller?
Solution
First compute the second moment of area for a solid circle:
With $d = 40\ \text{mm}$,
Euler buckling load:
Substitute:
Now compute the yield load:
Since $27.6\ \text{kN} < 314\ \text{kN}$, Euler buckling governs.
A ductile part has the plane stress state
If the yield strength is $75\ \text{MPa}$, find the von Mises stress and decide whether yielding is predicted.
Solution
Use the plane-stress von Mises formula:
Substitute:
Because $74.2\ \text{MPa} < 75\ \text{MPa}$, yielding is not predicted. The part is just barely safe.
Difficulty
A solid circular shaft has diameter $50\ \text{mm}$ and a shoulder with stress concentration factor $K_t = 1.6$. If the allowable maximum shear stress is $50\ \text{MPa}$, what is the largest torque the shaft can carry?
Solution
The local maximum shear stress is
So the nominal stress must satisfy
For a solid circular shaft,
Solve for $T$:
Substitute $d = 50\ \text{mm}$:
So the largest torque is
Two bars connect rigid plates in parallel. Each bar is $800\ \text{mm}$ long and has cross-sectional area $400\ \text{mm}^2$. One bar is steel with $E = 200\ \text{GPa}$ and $\alpha = 12 \times 10^{-6}/^\circ\text{C}$. The other is aluminum with $E = 70\ \text{GPa}$ and $\alpha = 23 \times 10^{-6}/^\circ\text{C}$.
The temperature rises by $10^\circ\text{C}$, and the plates are also pulled by a total tensile load of $18\ \text{kN}$.
Find the force in each bar.
Solution
Because the plates are rigid, both bars have the same total elongation. Let $\delta$ be that common elongation.
For each bar,
Compute the stiffnesses:
Compute the free thermal expansions:
Force equilibrium gives
and
So
Now compute the forces:
So the steel bar carries about $16.2\ \text{kN}$ and the aluminum bar carries about $1.8\ \text{kN}$.
A solid circular shaft with diameter $60\ \text{mm}$ carries a bending moment of $300\ \text{N}\cdot\text{m}$ and a torque of $500\ \text{N}\cdot\text{m}$.
Find the principal stresses at the outer surface and the von Mises stress. If the yield strength is $40\ \text{MPa}$, is the shaft safe?
Solution
At the outer surface, the bending stress is
and the torsional shear stress is
Substitute $d = 60\ \text{mm}$, $M = 300\,000\ \text{N}\cdot\text{mm}$, and $T = 500\,000\ \text{N}\cdot\text{mm}$:
Now treat this as a plane stress state with $\sigma_x = 14.1\ \text{MPa}$, $\sigma_y = 0$, and $\tau_{xy} = 11.8\ \text{MPa}$.
The principal stresses are
So
The von Mises stress is
Since $24.8\ \text{MPa} < 40\ \text{MPa}$, the shaft is safe by the von Mises criterion.
A member has alternating stress and mean stress
The endurance limit is $S_e = 240\ \text{MPa}$ and the ultimate tensile strength is $S_{ut} = 600\ \text{MPa}$.
What factor of safety does the Goodman relation predict?
Solution
Use the Goodman design relation:
Substitute the values:
So
which gives
A solid steel column has diameter $30\ \text{mm}$, length $3\ \text{m}$, and modulus $E = 200\ \text{GPa}$. It carries a compressive load of $8\ \text{kN}$ applied with eccentricity $15\ \text{mm}$.
Find the maximum compressive stress and the Euler buckling load. If the allowable compressive stress is $70\ \text{MPa}$, which check is more restrictive?
Solution
The eccentric load creates both axial stress and bending stress.
Area:
Axial stress:
Bending moment from eccentricity:
For a solid circular section,
and
Bending stress:
So the maximum compressive stress is
Now compute Euler buckling:
with $L = 3000\ \text{mm}$:
Since the applied load is $8\ \text{kN}$, the column is below the Euler load but only by a small margin.
The compressive stress check is also below the $70\ \text{MPa}$ allowable. The more restrictive check is buckling, because $8\ \text{kN}$ is close to $8.7\ \text{kN}$.
