Engineering Mechanics and Ground Control

Truss member force analysis

A plane truss is analysed by assuming that every member is a two-force member joined by frictionless pins, so each member carries only axial force, with tension taken as positive and compression as negative. The usual starting point is a joint where only two unknown members meet, or a joint with no external load acting on it. At each joint, \sum F_x = 0 and \sum F_y = 0 are applied in turn. Zero-force member rules shorten the work: at an unloaded joint with only two non-collinear members, both forces are zero, and at an unloaded joint where one member is perpendicular to two collinear members, that perpendicular member carries no force. A vertical member at a joint whose other members all lie horizontally, with no vertical load, must have zero vertical force, so it is a zero-force member. Vertical equilibrium at such a joint alone settles the result without computing support reactions.

When the permissible load on a truss is limited by the capacity of several members, the force in each member is first found per unit applied load. The permissible load is then the smallest limit divided by the corresponding force per unit load, because the member that reaches its limit first governs the design. For a truss where a vertical load acts at a joint, the support reactions are obtained first from overall equilibrium of the truss. Inclined forces are then resolved into horizontal and vertical components using the member geometry, and the joints are solved one by one, usually starting from a joint with only two unknown members.

Constant deceleration kinematics

When a body moves in a straight line with constant deceleration a, the velocity changes uniformly with time. The standard relations are v = u - at, s = ut - \tfrac{1}{2}at^2 and v^2 = u^2 - 2as, where u is the initial speed, v the final speed and s the distance travelled. Setting v = 0 in the last relation gives the stopping distance s = u^2/(2a), which is a convenient way to find the deceleration from a measured stopping distance. For uniform change, the average velocity is (u+v)/2, so the distance covered equals this average multiplied by the time.

When a wheel locks and a vehicle skids on level ground, the only horizontal retarding force is kinetic friction, \mu_k N, and N = mg. Newton's second law then gives ma = \mu_k mg, so \mu_k = a/g, with a obtained from the kinematic relation.

The same relations apply to rotation with angular speed \omega, angular acceleration \alpha and angle turned \theta, giving \theta = \omega_1 t + \tfrac{1}{2}\alpha t^2. Since \omega = 2\pi N when N is in revolutions per second, the number of revolutions equals \theta/2\pi. For uniform change, this count equals the average of the initial and final revolutions per second multiplied by the time.

Axial deformation of stepped bars

For an axially loaded bar made of segments that differ in length, cross-sectional area or internal force, the total change in length is the algebraic sum of the segment deformations, \Delta L = \sum \dfrac{N_i L_i}{A_i E_i}. The internal force N_i in each segment is found by cutting a free body through that segment and applying equilibrium, with tension positive and compression negative. Moving along the bar, each external load is added algebraically to give the force in the next segment. A common error is to use an applied load in place of the internal force, or to forget that a compressed segment shortens and therefore contributes a negative term.

When two members are linked by a rigid bar and must deform by the same amount, compatibility requires \delta_1 = \delta_2. Since \delta_i = F_i L/(A E_i), equal lengths and areas make the force in each member proportional to its modulus of elasticity, so the stiffer member carries the larger share. Moment equilibrium of the rigid bar then locates the applied force relative to the members.

Longwall shield support capacity

A hydraulic shield in a longwall face is characterised by its yield capacity, which is the load at which the leg yields. In design, the shield is set at a lower load called the setting load, which is a fixed percentage of the yield capacity. The setting resistance per unit area is the setting load divided by the roof area supported by one shield, p = P_{set}/A, where A is the shield width multiplied by the effective span. Because the span depends on the distance from the canopy tip to the coal face and on the web depth, the setting pressure varies along the face. In the standard calculation used here, the maximum span is the canopy length plus the tip-to-face distance plus the web depth, whereas the minimum span is the canopy length plus the tip-to-face distance. A shorter span therefore gives a higher setting pressure for the same setting load.

Beam bending moment under triangular load

A beam carrying a distributed load that varies linearly, such as a triangular load with zero intensity at one end, has a total load equal to the area of the load diagram. This total acts at the centroid of the triangle, which lies two-thirds of the span from the zero-intensity end. Support reactions follow from moments about one support and force equilibrium.

The maximum bending moment occurs where the shear force is zero, that is, where the shear diagram crosses the axis. Because the load intensity varies with distance, the shear force is a quadratic function of position. The location of zero shear is found by equating the shear expression to zero, and the moment at that section is obtained by summing the moments of all forces on one side of it.

Darcy's law hydraulic conductivity

Darcy's law states that the volumetric flow rate through a porous medium is proportional to the hydraulic gradient and to the cross-sectional area normal to the flow, Q = kiA. Here k is the hydraulic conductivity, and the gradient is i = \Delta h/L, where \Delta h is the head loss over the flow length L. Hydraulic conductivity has units of velocity, such as mm/min or m/s, and it depends on the grain size, porosity and the viscosity of the fluid.

To find k, the relation is rearranged as k = Q/(iA) after the gradient is computed from the head loss over the length of the column. Units must be consistent, since mixing a head difference with a column length, or mixing area in mm² with flow in mm³/min, gives wrong results. Darcy's law applies to laminar flow in the medium. The Darcy velocity Q/A is an apparent velocity, and the actual seepage velocity is larger by the porosity factor.

Friction on inclined plane

On an inclined plane, the weight of a body resolves into a component W\sin\alpha along the slope and W\cos\alpha normal to it. Kinetic friction opposes relative sliding and has magnitude \mu_k N. Static friction can take any value up to the limit \mu_s N, and the limiting value is reached just before slipping. When a body is pushed up the slope at constant velocity, the net force along the slope is zero, so the applied force balances the gravity component together with the friction force.

When a pushing member transmits force by friction at a contact, the friction at the limiting state equals the coefficient times the normal force at that contact, and it acts to oppose the sliding. The normal force at each contact is found from equilibrium perpendicular to the surface. Taking friction at its limit then gives the maximum normal load the contact can carry, and this is the quantity to be determined in such problems.

Constant velocity lifting power

A body moving at constant velocity has zero acceleration, so by Newton's second law the net force is zero. The lifting force therefore equals the total weight that must be raised. For a vehicle carrying a payload, the total mass is the payload plus the self-weight, and the thrust must balance mg for this total mass.

Power is the rate of doing work, and for a force acting along the direction of constant motion it is P = Fv. The speed is v = s/t, so the power for a vertical lift is P = mgv. The result is in watts, and 1 kW equals 1000 W. Students often forget to include the self-weight of the vehicle or to convert the answer to kilowatts.

Superelevation and maximum safe speed on curve

Superelevation is the transverse rise of the outer edge of a curved road above its inner edge. It is expressed as the rate of superelevation, the rise divided by the width of the carriageway, which equals \tan\theta for the cross-slope angle \theta. On a banked curve, the component of weight along the slope supplies part of the centripetal force. When lateral friction is neglected, the speed at which the banking alone balances the curve is v = \sqrt{gR\tan\theta}, where R is the radius of the curve.

The overturning limit is reached when the resultant of weight and centrifugal force passes through the outer edge of the road. Under the same assumptions, this gives the maximum safe speed on the curve, and it depends on the radius and the rate of superelevation. Units must be kept consistent, with g in m/s² and R in metres.