Engineering Mechanics and Solid Mechanics

Plane stress transformation

Plane stress describes a state in which the stresses normal to one plane and the shear stresses on that plane are zero, so a point is fully defined by \sigma_x, \sigma_y and \tau_{xy}. When the axes are rotated by an angle \theta, measured counter-clockwise from x to X, the normal and shear stresses on the rotated faces are \sigma_{XX}=\sigma_x\cos^2\theta+\sigma_y\sin^2\theta+2\tau_{xy}\sin\theta\cos\theta and \sigma_{XY}=(\sigma_y-\sigma_x)\sin\theta\cos\theta+\tau_{xy}(\cos^2\theta-\sin^2\theta). The angle must be taken to the normal of the face in question. If a problem gives the angle to the face itself, add 90^\circ, which interchanges \cos^2\theta and \sin^2\theta in the first relation. Tension is positive, and if one normal stress on an inclined face is known, the unknown \tau_{xy} can be found by solving this linear equation and then checked against the shear relation. For example, with \sigma_x=60 MPa, \sigma_y=20 MPa and \tau_{xy}=0, the normal stress on a face whose normal lies at 30^\circ to x is 60(0.75)+20(0.25)=50 MPa. The invariance \sigma_{XX}+\sigma_{YY}=\sigma_x+\sigma_y gives a quick check on the computed values.

In plane stress problems, the displacement of a deforming element is often interpolated from its corner values. For a square element whose edges remain straight, a bilinear interpolation is consistent with the straight-edge condition, and each bilinear shape function equals one quarter at the centre. The displacement at the centre is therefore the arithmetic mean of the four corner displacements.

The principal stresses follow from \sigma_{1,2}=\dfrac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}, and the maximum principal stress is the one of greater absolute value. For \sigma_x=10, \sigma_y=-4 and \tau_{xy}=6 MPa, the centre is at 3 MPa and the radius is \sqrt{7^2+6^2}\approx9.2 MPa, so the principal stresses are about 12.2 and -6.2 MPa. Stress values read from the faces of a loaded element should first be checked against equilibrium before this formula is used.

When stresses are inferred from forces on the faces of a small triangular element, equilibrium should be applied rather than assumed values. The traction on an inclined face is t_i=\sigma_{ij}n_j, so for an outward unit normal (n_x,n_y) the components are t_x=\sigma_x n_x+\tau_{xy}n_y and t_y=\tau_{xy}n_x+\sigma_y n_y. If the faces normal to x and y carry no shear, the force on each axis face balances the inclined-face force, giving \sigma_x a=t_x c and \sigma_y b=t_y c, where a, b and c are the face lengths. These relations give \sigma_x and \sigma_y directly from the given traction.

A Mohr's circle is fixed by its centre \left(\dfrac{\sigma_x+\sigma_y}{2},0\right) and its radius \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2}, so matching stress states to circles is done by reading these features. Uniaxial tension s gives a circle from 0 to s on the normal-stress axis, with centre s/2 and radius s/2. Uniaxial compression gives the mirror-image circle from -s to 0. Pure shear \tau gives a circle centred at the origin with radius \tau. Equal biaxial stress s gives a zero-radius circle, which is a point at s.

Stress states can be superposed by adding their components, and subtracting one state from another is the same operation with a negative sign. A uniaxial tension s along a unit normal \mathbf{n} corresponds to the tensor s\,\mathbf{n}\otimes\mathbf{n}, whose components are s n_x^2, s n_y^2 and s n_x n_y. When the mean stress of the combined state is zero, its Mohr circle is centred at the origin, and the radius bounds the principal stresses. The largest compressive normal stress on any plane perpendicular to the plane of the stress state is then the magnitude of the more negative principal stress.

Beam deflection and curvature

The curvature of a plane curve y(x) is \kappa=\dfrac{y''}{[1+(y')^2]^{3/2}}, which reduces to \kappa\approx y'' for small slopes. At a point where the slope y' is zero, the curvature equals the second derivative directly. For y=2x^2-4x at x=1, the slope is zero and \kappa=y''=4 in units of reciprocal length.

A couple applied to a beam produces a pure bending moment, and the reactions of a statically determinate beam depend only on the net applied force and net applied moment. If two equal and opposite couples act on a hinged-and-roller beam, the net moment about the supports is zero, so both reactions vanish. With no transverse load, the shear force is zero at every section. The bending moment, however, is zero only where no couple has acted and takes the constant value of the couple in the region between them. A constant moment gives a constant curvature M/(EI), so the beam bends into an arc and deflects even though its shear force is zero.

Superposition is valid for linear elastic behaviour. A load on one half of a span can be split into a uniform load of half the intensity over the whole span plus an antisymmetric load of the same intensity, positive on one half and negative on the other. For a simply supported beam under a uniform load w, the midspan deflection is \dfrac{5wL^4}{384EI}, so the uniform part gives half of this value at midspan. An antisymmetric load produces a deflection that is antisymmetric about midspan, so it contributes nothing at midspan.

A rigid segment transmits its loads to a flexible part as an equivalent force and moment at the junction, and the rigid part adds no deflection of its own. For a cantilever of length L and flexural rigidity EI, a tip force P gives deflection \dfrac{PL^3}{3EI} and a tip moment M gives deflection \dfrac{ML^2}{2EI}, and the two add when they act in the same sense. The resultant of a distributed load on a rigid segment acts at its centroid, and its moment about the junction must be included in the calculation.

When a beam has segments of different flexural rigidity, the bending moment and EI must be applied segment by segment. The second derivative of the deflection is \dfrac{d^2v}{dx^2}=\dfrac{M(x)}{EI(x)} in the segment considered, with the sign convention of the bending moment kept consistent. Reactions produced by a couple on a pinned and roller system are found from moment equilibrium, and the bending moment varies linearly wherever no transverse load acts.

Torsional shear stress in circular shafts

For a solid circular shaft, the shear stress at radius r is \tau=Tr/J, and its maximum at the surface is \tau_{max}=\dfrac{16T}{\pi d^3}. For a shaft with several torque segments, the internal torque in each segment is found by sectioning and summing the applied moments on one side. The maximum shear is the largest value of 16T/(\pi d^3) over all segments, because both the torque and the diameter can differ between segments. For example, a shaft with T=1 kN·m and d=0.1 m has \tau_{max}=\dfrac{16(1000)}{\pi(0.1)^3}\approx5.1 MPa. The computed value is then compared with the yield shear strength to check safety.

For a circular shaft fixed at one end and loaded by an end torque T, the angle of twist at the free end is \theta=\dfrac{TL}{GJ}, and the strain energy stored is U=\dfrac{T^2L}{2GJ}=\dfrac{T\theta}{2}. Both expressions follow from the linear torque–twist relation. For T=2 kN·m, L=4 m and GJ=20000 kN·m²/rad, the twist is \theta=0.0004 rad and the strain energy is U=0.0004 kN·m. Statements about rotation and about energy must be checked against these two formulas separately.

A shaft fixed at both ends is statically indeterminate, because the two support torques cannot be found from statics alone. Their sum equals the applied torque, T_P+T_R=T, and the compatibility condition is that the twist of the two segments meeting at the loaded section is the same, so \dfrac{T_PL_1}{GJ_1}=\dfrac{T_RL_2}{GJ_2}. The support torque ratio is therefore \dfrac{T_P}{T_R}=\dfrac{L_2J_1}{L_1J_2}, so the support at the end of the stiffer segment takes the larger share of the torque. Confusing the length ratio with the polar moment ratio is a common error.

Equilibrium of rigid bodies

A rigid body in coplanar equilibrium gives three independent equations, \sum F_x=0, \sum F_y=0 and \sum M=0. Inclined forces must be resolved into components before the sums are taken, and the moment of a force about a point equals its perpendicular distance from that point multiplied by its magnitude. Choosing a moment centre at a point where an unknown reaction acts eliminates that reaction from the moment equation. For a 6 m simply supported beam carrying a 12 kN load at 2 m from the left support, moments about the right support give R_A=\dfrac{12\times4}{6}=8 kN.

A rigid body supported on springs that translates without rotating has the same displacement at every spring, so each spring force is K_i\delta and the spring forces are proportional to their stiffnesses. Moment equilibrium about one support fixes one spring force, which determines the common displacement, and vertical equilibrium then gives the remaining spring force and its stiffness. The no-rotation condition is an essential part of this compatibility argument.

A rigid bar supported by axial links is a constrained body whose displacements vary linearly along its length. Each link force is found from equilibrium, and each link elongation is \delta=\dfrac{FL}{AE}. Once one link's movement is known, the rotation of the bar is fixed, and the displacement of any other point follows by linear extrapolation, which is the basis of the similar-triangle geometry used in such problems.

Principal stresses and maximum shear

When the stress tensor is diagonal, its diagonal entries are the principal stresses. They are ordered algebraically as \sigma_1\ge\sigma_2\ge\sigma_3. The maximum shear stress at a point is \tau_{max}=\dfrac{\sigma_1-\sigma_3}{2}, which uses the largest and smallest principal stresses and not any pair of them. For diagonal entries of -30, 20 and 5 MPa, \tau_{max}=\dfrac{20-(-30)}{2}=25 MPa.

A thin spherical shell of radius R and wall thickness t under internal pressure p is in membrane action. Both meridional and hoop stresses equal \sigma=\dfrac{pR}{2t}, and both are tensile. Because the two in-plane principal stresses are equal, the in-plane shear stress is zero, and the stress through the thickness is negligible when t\ll R. For p=0.5 MPa, R=0.5 m and t=10 mm, the membrane stress is \sigma=12.5 MPa.

Pure shear is a state whose normal stresses on perpendicular faces are equal in magnitude and opposite in sign, so the mean stress is zero and the Mohr circle is centred at the origin. An element with \sigma_x=40 and \sigma_y=-40 and no shear is equivalent to pure shear of magnitude 40, with principal stresses \pm40 on planes at 45^\circ. To identify pure shear, therefore, check that \sigma_x+\sigma_y=0 and that no net tension or compression is present.

Friction on wedges and blocks

Friction acts tangentially at a contact and opposes impending motion. Its limiting value is F=\mu N, where N is the normal force, and a frictionless surface transmits only normal force. For a wedge pushed under a block, the impending motion of the block is governed by the vertical component of the reaction on the inclined surface, and the friction force on that surface acts along the slope at its limiting value. Equilibrium of each body is written separately, and the normal forces are eliminated to give the required force ratio.

When a frictionless vertical wall supports a block, the wall gives only a horizontal reaction, so the block can move only in the vertical direction. Such a constraint simplifies the force balance because one component of the contact force is known to be zero.

When two blocks are joined by a rigid rod and rest against a wall and a floor, friction can act at both contacts. The procedure is to write equilibrium of the whole system to relate the normal reactions, then equilibrium of one block to find the rod force and its inclination, and finally to impose the limiting friction condition at both contacts. At the point of impending sliding, the condition often reduces to a quadratic equation in \mu, and only the positive root is physically admissible.