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Skip to main content. Log In Sign Up. Suddiyas Nawaz. Stress is the lead to accurately describe and predict the elastic deformation of a body. Simple stress can be classified as normal stress, shear stress, and bearing stress. Normal stress develops when a force is applied perpendicular to the cross-sectional area of the material. If the force is going to pull the material, the stress is said to be tensile stress and compressive stress develops when the material is being compressed by two opposing forces.

Member BE? What is the largest average tensile or compressive stress in BC and BE? The length of the tank is L and the wall thickness is t. Isolating the right half of the tank: If there exist an external pressure po and an internal pressure pi, the formula may be expressed as: It can be observed that the tangential stress is twice that of the longitudinal stress.

Calculate the allowable internal pressure if the stress is limited to psi. Solution Problem Calculate the minimum wall thickness for a cylindrical vessel that is to carry a gas at a pressure of psi. The diameter of the vessel is 2 ft, and the stress is limited to 12 ksi.

The diameter of the pressure vessel is mm and its length is 2. Determine the maximum internal pressure that can be applied if the longitudinal stress is limited to MPa, and the circumferential stress is limited to 60 MPa. Find the maximum height to which the tank may be filled if the circumferential stress is limited to psi. The specific weight of water is Calculate the maximum diameter of the cylinder tank if the internal pressure is psi.

At what revolutions per minute rpm will the stress reach 30 ksi if the mean radius is 10 in.? The density of steel 7. Solution Problem The tank shown in Fig. Calculate the maximum longitudinal and circumferential stress caused by an internal pressure of psi.

A gasket is inserted between the flange at one end of the pipe and a flat plate used to cap the end. How many mm-diameter bolts must be used to hold the cap on if the allowable stress in the bolts is 80 MPa, of which 55 MPa is the initial stress?

What circumferential stress is developed in the pipe? Why is it necessary to tighten the bolt initially, and what will happen if the steam pressure should cause the stress in the bolts to be twice the value of the initial stress? Stress-Strain Diagram Suppose that a metal specimen be placed in tension-compression testing machine. As the axial load is gradually increased in increments, the total elongation over the gage length is measured at each increment of the load and this is continued until failure of the specimen takes place.

The stress-strain diagram differs in form for various materials. The diagram shown below is that for a medium carbon structural steel. Metallic engineering materials are classified as either ductile or brittle materials. A ductile material is one having relatively large tensile strains up to the point of rupture like structural steel and aluminum, whereas brittle materials has a relatively small strain up to the point of rupture like cast iron and concrete.

An arbitrary strain of 0. This linear relation between elongation and the axial force causing was first noticed by Sir Robert Hooke in and is called Hooke's Law that within the proportional limit, the stress is directly proportional to strain or The constant of proportionality k is called the Modulus of Elasticity E or Young's Modulus and is equal to the slope of the stress-strain diagram from O to P.

Then ELASTIC LIMIT The elastic limit is the limit beyond which the material will no longer go back to its original shape when the load is removed, or it is the maximum stress that may e developed such that there is no permanent or residual deformation when the load is entirely removed. The region from P to R is called the plastic range. This is also known as the breaking strength.

This may be calculated as the area under the stress-strain curve from the origin O to up to the elastic limit E the shaded area in the figure. The resilience of the material is its ability to absorb energy without creating a permanent distortion. This may be calculated as the area under the entire stress-strain curve from O to R. The toughness of a material is its ability to absorb energy without causing it to break. The maximum safe stress that a material can carry is termed as the allowable stress.

The allowable stress should be limited to values not exceeding the proportional limit. However, since proportional limit is difficult to determine accurately, the allowable tress is taken as either the yield point or ultimate strength divided by a factor of safety. The ratio of this strength ultimate or yield strength to allowable strength is called the factor of safety. If however, the cross- sectional area is not uniform, the axial deformation can be determined by considering a differential length and applying integration.

If however, the cross-sectional area is not uniform, the axial deformation can be determined by considering a differential length and applying integration. It supports a tensile load of 20 kN at the lower end. Solution Problem A steel wire 30 ft long, hanging vertically, supports a load of lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.

Solution Problem A steel tire, 10 mm thick, 80 mm wide, and If the coefficient of static friction is 0. Neglect the deformation of the wheel. Assume the bar is suitably braced to prevent lateral buckling. Solution Problem Solve Prob. Find the largest value of P that will not exceed an overall deformation of 3. Assume that the assembly is suitably braced to prevent buckling.

P is hinged at A and supported by a steel rod at B. Determine the largest load P that can be applied at C if the stress in the steel rod is limited to 30 ksi and the vertical movement of end C must not exceed 0.

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P, is horizontal before the load P is applied. Determine the vertical movement of P if its magnitude is 50 kN. P are supported by pins at A and C and the two rods. Determine the maximum force P that can be applied as shown if its vertical movement is limited to 5 mm. Neglect the weights of all members. P, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so that the slab will remain level. Solution Problem As shown in Fig.

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If each rod has a crosssectional area of 0. If w is the weight per unit volume, find the elongation of the rod caused by its own weight. Use this result to determine the elongation of a cone suspended from its base.

The gage length was 50 mm. Plot the stress-strain diagram and determine the following mechanical properties: Solution Problem The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0. An element subject to shear does not change in length but undergoes a change in shape. The relationship between the shearing deformation and the applied shearing force is where V is the shearing force acting over an area As.

Poisson's Ratio When a bar is subjected to a tensile loading there is an increase in length of the bar in the direction of the applied load, but there is also a decrease in a lateral dimension perpendicular to the load. For most steel, it lies in the range of 0. Compressive stresses and contraction are taken as negative. Bulk Modulus of Elasticity or Modulus of Volume Expansion, K The bulk modulus of elasticity K is a measure of a resistance of a material to change in volume without change in shape or form.

The block is subjected to a triaxial loading of three uniformly distributed forces as follows: Solution Problem A welded steel cylindrical drum made of a mm plate has an internal diameter of 1. Compute the change in diameter that would be caused by an internal pressure of 1. Assume that Poisson's ratio is 0.

Solution Problem A 2-in. Find the tangential stress if an axial compressive load of lb is applied. It fits without clearance in an mm hole in a rigid block. The tube is then subjected to an internal pressure of 4. Solution Problem A 6-in.

With no internal pressure, the tube just fits between two rigid end walls. Calculate the longitudinal and tangential stresses for an internal pressure of psi. These cases require the use of additional relations that depend on the elastic deformations in the members.

Solved Problems in Statically Indeterminate Members Problem A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm thick. Compute the load that will compress the combined bar a total of 0.

Determine the required area of the reinforcing steel if the allowable stresses are 6 MPa and MPa for the concrete and steel, respectively. Solution Problem A timber column, 8 in. Determine the thickness t so that the column will support an axial load of kips without exceeding a maximum timber stress of psi or a maximum steel stress of 20 ksi.

The moduli of elasticity are 1. Solution Problem A rigid block of mass M is supported by three symmetrically spaced rods as shown in fig P Determine the largest mass M which can be supported. Solution Problem The lower ends of the three bars in Fig. P are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 1.

For the bronze bar, the area is 1. Determine a the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar, and b the length of the bronze that will make the steel stress twice the bronze stress. Solution Problem The rigid platform in Fig. P has negligible mass and rests on two steel bars, each The center bar is aluminum and Solution Problem Three steel eye-bars, each 4 in.

The center-line spacing between the holes is 30 ft in the two outer bars, but 0. Find the shearing stress developed in the drip pins. Neglect local deformation at the holes. P, three steel wires, each 0. Their unstressed lengths are P consists of a light rigid bar AB, pinned at O, that is attached to the steel and aluminum rods.

Compute the stress in the aluminum rod when the lower end of the steel rod is attached to its support. It carries an axial load P applied as shown in Fig. Determine the stress in segment BC. Use the results of Prob. Then use the principle of superposition to compute the reactions when both loads are applied. P is firmly attached to unyielding supports.

Solution Problem The composite bar in Fig. P is stress-free before the axial loads P1 and P2 are applied. Solution Problem There is a radial clearance of 0. The inside diameter of the aluminum tube is mm, and the wall thickness of each tube is 2. Compute the contact pressure and tangential stress in each tube when the aluminum tube is subjected to an internal pressure of 5.

Find the stresses if the nut is given one additional turn. How many turns of the nut will reduce these stresses to zero? P are identical except for length. Before the load W was attached, the bar was horizontal and the rods were stress-free.

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P is pinned at B and connected to two vertical rods. P, a rigid beam with negligible weight is pinned at one end and attached to two vertical rods. Find the vertical movement of W. P, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assuming that the rods were initially tress-free, what maximum load P can be applied without exceeding stresses of MPa in the steel rod and 70 MPa in the bronze rod.

Solution Problem Shown in Fig. P is a section through a balcony. The total uniform load of kN is supported by three rods of the same area and material. Compute the load in each rod.

Assume the floor to be rigid, but note that it does not necessarily remain horizontal. Solution Problem Three rods, each of area mm2, jointly support a 7. Assuming that there was no slack or stress in the rods before the load was applied, find the stress in each rod.

Initially, the assembly is stressfree. Horizontal movement of the joint at A is prevented by a short horizontal strut AE. If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. In some cases where temperature deformation is not permitted, an internal stress is created. The internal stress created is termed as thermal stress.

For a homogeneous rod mounted between unyielding supports as shown, the thermal stress is computed as: If the wall yields a distance of x as shown, the following calculations will be made: Take note that as the temperature rises above the normal, the rod will be in compression, and if the temperature drops below the normal, the rod is in tension. At what temperature will the stress be zero?

At what temperature will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Solution Problem A steel rod 3 feet long with a cross-sectional area of 0.

Solution Problem A bronze bar 3 m long with a cross sectional area of mm2 is placed between two rigid walls as shown in Fig. Find the temperature at which the compressive stress in the bar will be 35 MPa.

Assume that the supports are unyielding and that the bar is suitably braced against buckling. Neglect the deformation of the wheel caused by the pressure of the tire. P is pinned at B and attached to the two vertical rods. Initially, the bar is horizontal and the vertical rods are stress-free. Neglect the weight of bar ABC. P, there is a gap between the aluminum bar and the rigid slab that is supported by two copper bars.

Assume the coefficients of linear expansion are Solution Problem For the assembly in Fig. If the system is initially stress-free.

Calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to the change in temperature. Solution Problem Four steel bars jointly support a mass of 15 Mg as shown in Fig. Each bar has a cross-sectional area of mm2. Such a bar is said to be in torsion. For solid cylindrical shaft: For hollow cylindrical shaft: Solved Problems in Torsion Problem A steel shaft 3 ft long that has a diameter of 4 in. Determine the maximum shearing stress and the angle of twist.

What maximum shearing stress is developed? Solution Problem A steel marine propeller shaft 14 in. What power can be transmitted by the shaft at 20 Hz? Solution Problem A 2-in-diameter steel shaft rotates at rpm. If the shearing stress is limited to 12 ksi, determine the maximum horsepower that can be transmitted. Solution Problem An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig.

Solution Problem A flexible shaft consists of a 0. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? Solution Problem The steel shaft shown in Fig.

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Solution Problem A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from the left end where 20 kW are removed. At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1. Determine the maximum permissible value of T subject to the following conditions: Solution Problem A hollow bronze shaft of 3 in. The two shafts are then fastened rigidly together at their ends. What torque can be applied to the composite shaft without exceeding a shearing stress of psi in the bronze or 12 ksi in the steel?

Determine the maximum shearing stress in each segment and the angle of rotation of the free end. Solution Problem The compound shaft shown in Fig. P is attached to rigid supports. What torque T is required? Solution Problem A torque T is applied, as shown in Fig. P, to a solid shaft with built-in ends. How would these values be changed if the shaft were hollow? Solution Problem A solid steel shaft is loaded as shown in Fig. Determine the maximum shearing stress developed in each segment.

For the bronze segment AB, the maximum shearing stress is limited to psi and for the steel segment BC, it is limited to 12 ksi.

P, each with one end built into a rigid support have flanges rigidly attached to their free ends. The shafts are to be bolted together at their flanges. Determine the maximum shearing stress in each shaft after the shafts are bolted together. For any number of bolts n, the torque capacity of the coupling is If a coupling has two concentric rows of bolts, the torque capacity is where the subscript 1 refer to bolts on the outer circle an subscript 2 refer to bolts on the inner circle.

See figure. For rigid flanges, the shear deformations in the bolts are proportional to their radial distances from the shaft axis. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 40 MPa.

Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is psi. Solution Problem A flanged bolt coupling consists of eight mmdiameter steel bolts on a bolt circle mm in diameter, and six mm- diameter steel bolts on a concentric bolt circle mm in diameter, as shown in Fig.

What torque can be applied without exceeding a shearing stress of 60 MPa in the bolts? Determine the shearing stress in the bolts. What torque can be applied without exceeding psi in the steel or psi in the aluminum? Solution Problem A plate is fastened to a fixed member by four mm diameter rivets arranged as shown in Fig.

Compute the maximum and minimum shearing stress developed. P to the fixed member. Using the results of Prob. What additional loads P can be applied before the shearing stress in any rivet exceeds psi?

P is fastened to the fixed member by five mm-diameter rivets. Compute the value of the loads P so that the average shearing stress in any rivet does not exceed 70 MPa.

Determine the wall thickness t so as not to exceed a shear stress of 80 MPa. What is the shear stress in the short sides? Neglect stress concentration at the corners. What torque will cause a shearing stress of psi? Determine the smallest permissible dimension a if the shearing stress is limited to psi. Solution Problem A tube 2 mm thick has the shape shown in Fig. Assume that the shearing stress at any point is proportional to its radial distance.

This formula neglects the curvature of the spring. For heavy springs and considering the curvature of the spring, a more precise formula is given by: Use Eq. Please enter your comment! Please enter your name here. You have entered an incorrect email address! Get New Updates Email Alerts Enter your email address to subscribe this blog and receive notifications of new posts by email. Join With us.

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