GTU BE03000231 Machine Drawing and Elements of Machine Design Summer 2026 Solved Paper & Solutions

A support holds a beam in place and gives it reaction forces. There are three common types.

1. Roller support: it allows the beam to move along the surface and to rotate. It gives only one reaction, at right angles to the surface.
2. Hinged (pin) support: it allows rotation but no movement. It gives two reactions, horizontal and vertical.
3. Fixed support: it allows neither movement nor rotation. It gives a horizontal reaction, a vertical reaction and a fixing moment.

Factor of safety (FOS) is the ratio of the stress at which a part fails to the stress it is allowed to carry in service.

For a ductile material (mild steel) the failure stress is the yield stress. For a brittle material (cast iron) it is the ultimate stress. FOS is always more than 1, so the part keeps a reserve of strength.

Factors affecting the FOS:

1. Material: how uniform it is, and how far its strength is known from tests.
2. Type of load: a steady load needs a lower FOS than a shock or a fatigue load.
3. Accuracy of the load and the stress calculation: if they are doubtful, FOS is raised.
4. Manufacturing: the quality of the work, surface finish and the size of errors.
5. Service conditions: high temperature, corrosion and wear.
6. Consequence of failure: danger to human life needs a higher FOS.
7. Cost and weight: a very high FOS makes the part heavy and costly.

1. Tensile stress: the internal resistance per unit area when a load tries to stretch the member, $\sigma_t = \dfrac{P}{A}$. The load acts away from the cross-section.
2. Compressive stress: the internal resistance per unit area when a load tries to shorten the member, $\sigma_c = \dfrac{P}{A}$. The load acts towards the cross-section.
3. Bearing pressure: the contact pressure between two parts that press on each other, such as a pin in a hole. It is the load divided by the projected area of contact, $p = \dfrac{P}{d \times l}$, where $d$ is the diameter and $l$ the length of contact.
4. Lateral strain: when a bar is stretched, its width becomes smaller. The change in width divided by the original width is the lateral strain. It is opposite in sign to the longitudinal strain.
5. Hooke's law: up to the elastic limit, the stress is directly proportional to the strain, $\sigma \propto \varepsilon$. So $\sigma = E\,\varepsilon$.
6. Poisson's ratio ($\mu$): the ratio of lateral strain to longitudinal strain within the elastic limit, $\mu = \dfrac{\text{lateral strain}}{\text{longitudinal strain}}$. It is about 0.25 to 0.33 for most metals.
7. Modulus of elasticity ($E$): the ratio of stress to strain within the elastic limit, $E = \dfrac{\sigma}{\varepsilon}$. It measures stiffness: a larger $E$ means the material stretches less for the same stress. For steel it is about 200 GPa.

A designer thinks about the following points when designing a machine component.

1. Function: the part must do the job it is made for, such as transmit power.
2. Strength and rigidity: it must not break or bend too much under the load.
3. Material: the right strength, wear resistance and cost, and easy to machine.
4. Manufacture and cost: a shape that is cheap to make, with standard sizes.
5. Safety: a suitable factor of safety, with guards where needed.
6. Maintenance: easy to assemble, lubricate and replace.
7. Weight and size: as small as the strength allows.

Elasticity is the property by which a body returns to its original size after the load is removed. Plasticity is the property by which it keeps a permanent deformation after the load is removed.

Take the left bottom corner as origin. Divide the plane into a rectangle ($60 \times 45$), a triangle (base 15, height 45) and subtract the hole (diameter 20). The total height is $30 + 15 = 45$ mm. The hole centre is 30 mm from the right edge, so it is at $x = 75 - 30 = 45$ mm, $y = 15$ mm.

The triangle has its vertices at $(0,0)$, $(15,0)$ and $(15,45)$, so its centroid is at $x = \dfrac{0+15+15}{3} = 10$ and $y = \dfrac{45}{3} = 15$.

Use $I = I_G + A h^2$ for each part, where $h$ is the distance of its centroid from $\bar{y}$.

Aesthetic considerations are about the look of the part: shape, proportion, colour and surface finish. A pleasing, tidy product sells better and shows good quality. Smooth lines, matching colours and a good finish also help to protect against rust.

Ergonomic considerations are about the user. The machine should fit the human body and its limits:

1. controls, handles and levers are within easy reach and need little force;
2. dials and signs are easy to see and read;
3. noise, vibration and heat are low;
4. the operator is safe and does not get tired.

The parallel axis theorem states: the moment of inertia of a plane area about any axis is equal to its moment of inertia about a parallel axis through the centroid, plus the product of the area and the square of the distance between the two axes.

Proof. Let $G$ be the centroid and $AB$ an axis parallel to the centroidal axis, at a distance $h$. Take a small strip of area $dA$ at a distance $y$ from the centroidal axis. Its distance from $AB$ is $(y + h)$.

The first term is $I_G$. The second term is zero because $\int y\,dA$ is the first moment about the centroid. The third term is $A h^2$. Hence $I_{AB} = I_G + A h^2$.

Use: it gives the moment of inertia of a built-up section, such as an I section, about its neutral axis.

Given: overall depth 300 mm, flange 200 mm wide and 20 mm thick, web 20 mm thick, span $L = 7$ m, UDL $w = 5$ kN/m and a point load $W = 20$ kN at mid-span. The weight of the girder is not given, so it is ignored.

The section is a large rectangle $200 \times 300$ minus two side rectangles. Each is $90$ mm wide and $260$ mm high (the web is $20$ mm, so the gap on each side is $(200 - 20)/2 = 90$ mm, and the clear height between flanges is $300 - 40 = 260$ mm).

The load is at mid-span, so the maximum moment is at the centre.

The extreme fibre is at $y = 150$ mm from the neutral axis.

A cotter joint connects two coaxial rods that carry an axial pull or push. It is a temporary joint that can be taken apart. It has these parts.

1. Socket: the enlarged end of one rod, with a hole for the spigot and a slot for the cotter.
2. Spigot: the end of the second rod. It enters the socket and has a slot and a collar that stops it going in too far.
3. Cotter: a flat steel wedge with a small taper. It is driven through the slots and locks the two rods.

Simple bending is the bending of a beam under a pure moment $M$, with the beam in the elastic range.

Assumptions:

1. the beam is straight and the material is uniform (homogeneous) and isotropic;
2. Hooke's law holds, and $E$ is the same in tension and compression;
3. plane sections stay plane after bending;
4. the radius of curvature is large compared with the depth, and each layer bends freely;
5. the loads act in the plane of symmetry.

Derivation. A layer at a distance $y$ from the neutral axis has the original length $L$. Let $R$ be the radius of curvature of the neutral layer. After bending, the length of that layer is $(R + y)\,\theta$, while the neutral layer has $R\,\theta = L$. So

The moment of the force on a strip $dA$ about the neutral axis is $\sigma\,dA\,y$. Adding up all strips:

Fatigue loading has a steady mean stress $\sigma_m$ and a varying (amplitude) stress $\sigma_v$. The Soderberg line is a safe design line on the $\sigma_m$ and $\sigma_v$ diagram. It joins the endurance limit $\sigma_e$ on the $\sigma_v$ axis to the yield strength $\sigma_y$ on the $\sigma_m$ axis.

Derivation. Take the point $P(\sigma_m, \sigma_v)$ on the safe line. The point $P$ lies on the line through $(0, \sigma_e)$ and $(\sigma_y, 0)$, scaled down by the factor of safety $N$. By similar triangles on the line,

When the allowable values $\dfrac{\sigma_y}{N}$ and $\dfrac{\sigma_e}{N}$ are used, the equation is

If there is a stress concentration, it is applied to the variable part only, through the fatigue stress concentration factor $K_f$:

Application to different loadings:

1. Steady load ($\sigma_v = 0$): $\sigma_m = \dfrac{\sigma_y}{N}$, a static design on the yield strength.
2. Completely reversed load ($\sigma_m = 0$): $\sigma_v = \dfrac{\sigma_e}{N}$, a design on the endurance limit.
3. Repeated load ($\sigma_{min} = 0$): $\sigma_m = \sigma_v = \dfrac{\sigma_{max}}{2}$.
4. Fluctuating load: use the full equation with $\sigma_m = \dfrac{\sigma_{max} + \sigma_{min}}{2}$ and $\sigma_v = \dfrac{\sigma_{max} - \sigma_{min}}{2}$.

For shear (torsion), use $\tau_m$, $\tau_v$, $\tau_y$ and $\tau_e$ in the same equation.

Limits: the two extreme sizes between which the actual size of a part must lie. The larger is the upper limit and the smaller is the lower limit.

Tolerance: the difference between the upper and lower limits, the permitted variation in size. For a shaft with limits 25.00 and 24.96 mm the tolerance is 0.04 mm.

Fit: the looseness or tightness between a shaft and its hole.

1. Clearance fit: the shaft is always smaller than the hole, so it turns freely.
2. Interference fit: the shaft is always larger, so it is pressed in.
3. Transition fit: the shaft may be a little smaller or larger.

Take a shaft that carries only a twisting moment $T$. The maximum shear stress at its outside surface is given by the torsion equation $\dfrac{T}{J} = \dfrac{\tau}{r}$.

Solid shaft of diameter $d$: $J = \dfrac{\pi}{32}d^4$ and $r = \dfrac{d}{2}$.

Hollow shaft with outer diameter $d_o$, inner diameter $d_i$ and $k = \dfrac{d_i}{d_o}$:

When the shaft carries power $P$ (in watts) at $N$ rpm, the torque is found from

Here $\tau$ is the allowable shear stress of the shaft material. If the twist must also be limited, check it with $\theta = \dfrac{TL}{GJ}$.

Given: thickness $t = 15$ mm, width $w = 50$ mm, length $100$ mm, axial load $P = 20$ kN, holes of diameter $8$, $16$ and $24$ mm. The figure shows each hole on its own cross-section. A brittle material does not yield, so the peak stress at a hole must be found with the stress concentration factor $K_t$.