Applied strength of materials pdf

Imagine cutting a thin-walled pipe lengthwise through the pressurized fluid and the pipe wall: the force exerted by the fluid must equal the force exerted by the pipe walls sum of the forces equals zero.

The stress in the walls of the pipe is equal to the fluid force divided by the cross-sectional area of the pipe wall.

This cross-section of one wall is the thickness of the pipe, t, times its length. Since there are two walls, the total L cross-sectional area of the wall is 2 t L. What is the hoop stress if the wall thickness is 0. Solution First, check if the pipe is thin-walled. The ratio of the pipe wall thickness to the internal radius is t 0. What if the pipe has a cap on the end? If the cap were loose, pressure would push the cap off the end. If the cap is firmly attached to the pipe, then a stress develops along the length of the pipe to resist pressure on the cap.

Imagine cutting the pipe and pressurized fluid transversely. The force p exerted by the fluid equals the force along the length of the pipe walls. If the pressure in a pipe exceeds the strength of the material, then the pipe will split along its length perpendicular to the hoop direction.

Does the shape of the cap affect the longitudinal stress in the pipe? No, because only the cross-sectional area of the pipe matters. Typically, pressure vessels have concave or convex p p domes, because flat caps tend to deflect under pressure, but the shape has no effect on longitudinal stress in the pipe walls. A welded steel outdoor propane tank typically consists of a tube with two convex hemispherical caps right. Hoop stress controls the design in the hoop stress tube portion.

Where the caps are welded to the tubes, longitudinal stress controls the design. If the steel is all the same thickness and the welds are perfect , then the tank will fail in the tube section because hoop stress is p p twice the longitudinal stress. A spherical tank far right only has longitudinal stress, so it can handle twice the pressure of a tubular tank.

Think of a spherical tank as two hemispheres welded together; the weld longitudinal stress prevents the two halves from separating. Solution First, calculate the allowable stress.

Next, rewrite the stress equation for longitudinal stress because this is a sphere, not a pipe to solve for wall thickness. The welds in real steel tanks contain defects, which reduce the strength of the welds. We can measure the strength of welded joints in the lab and compare them with the strength of the base metal. In the previous example, the allowable stress was Multiply this value by 0. We say that the stress is concentrated next to the hole. In general, the smaller the radius, the higher the stress.

For example, cracks have very high stress concentrations at their tips, exceeding the tensile strength of the material, even though the average stress in the part is well below the yield strength; this explains why cracks grow.

One way to prevent a crack from growing in a material is to drill a hole at its tip. Stress concentrations can occur anywhere Circular hole in a rectangular cross-section bar loaded in tension there is a change in geometry with a small radius, such as holes, fillets, and grooves. In the s, M. Frocht6 published a 2.

Peterson published 2. Peterson's K book on stress concentration factors7 is still 2. This graph is based on Frocht's original 2. Use a four-step process to solve stress concentration problems: 2. Step 3 Calculate the net cross-sectional area gross cross-sectional area minus the cross-sectional area of the hole. The easiest way to find this value is to multiply the net width by the thickness.

Pilkey, Peterson's Stress Concentration Factors, 3rd ed. The tensile load on the bar is N. Calculate the maximum stress in the bar due to the stress concentration. Report the result in MPa. The net cross-sectional area of the bar is the net width times the thickness. Step 2 Find the value of K from the graph on the following page.

Be sure to use the right curve! Step 3 Calculate the net cross-sectional area at the groove or fillet. The tensile load on the bar is 1 ton. Report the result in ksi. This equation is used for pipes. This equation is used for spherical pressure vessels. Use stress concentration graphs to find the value of K. They intersect at joints, which transmit load from one P member to the next. Large steel structures are typically bolted, riveted, or welded together.

The strength of the joint may determine the strength of the P structure. P Lap two steel plates, and pin them with a rivet or bolt. If you pull hard P enough, the plates will want to twist. You can prevent twisting by butting the two main plates, and joining them with lapped splice plates. P Bolted joints have several failure modes.

We can calculate the allowable P load of the joint for each failure mode; the lowest value is the limiting case, and determines the strength of the joint. Shear Failure The bolt could shear off, preserving the integrity of the plates; this is the least expensive type of failure because bolts are cheaper and easier to replace than plates. Bearing Failure The bolt could crush the plate where it bears against the plate. Gross Tensile Failure The plate could tear across its gross cross-sectional area thickness times overall width , away from the bolt holes.

Net Tensile Failure The plate could tear across its net cross-sectional area thickness times actual material width along a line of bolt holes. Now replace the glue with a bolt. The bolt has a much smaller cross-sectional area than the area of the glue, which is One shear plane PS compensated by the bolt's much higher shear strength. PS A bolt can carry the load through more than one shear plane. Appendix B3 lists allowable shear stresses for three bolt materials and two loading cases: either with the threads of the bolt in the shear plane, or with the smooth shank of the bolt in the shear plane.

A bolt is stronger if the threads are excluded from the shear plane because the threads act as stress concentration sites. The plates are 6 in. Bolt threads are excluded from the shear plane. Calculate PS 1. Report the result in kips. The problem is symmetrical, PS so either the four bolts in the left plate will fail in shear first, or the four bolts in the right plate will fail first.

Bearing Failure PP If one or more of the plates is crushed by the bolt, then the area that is crushed equals the bolt diameter, d, times the plate PP thickness, t. PP If the main plate is thicker than the sum of the thicknesses of two splice plates, then use the sum of the splice plate thicknesses for t.

The allowable bearing stress is listed in Appendix B4; it is 1. Example 2 Calculate the load that the joint in Example 1 can support in order to resist bearing failure. Solution From the Allowable Plate Stress table, A steel plates have an allowable bearing strength of 87 ksi. The two splice plates have a combined thickness of 1.

Gross Tensile Failure The gross cross-sectional area of the plate some distance away from the bolt holes is the width, b, times the thickness, t.

Check the gross cross-sectional areas of both the main plate and the splice plates, and use the smaller of the two. Solution From Appendix B4, A steel plates have an allowable gross tensile strength of Net Tensile Failure The net cross-sectional area of the plate at the bolt holes is the gross cross-sectional area, AG, minus the cross-sectional area of the holes, AH.

The plate will fail along the line of one set of holes. The allowable net tensile strength, listed in Appendix B4, is half of the ultimate tensile strength of the plate material. Example 4 Calculate the load that the joint in Example 1 can support in order to resist net tensile failure. Solution From Appendix B4, A steel plates have an allowable net tensile strength of 29 ksi. This flaw was one of several design and construction errors that led to the roof collapse. We can calculate the efficiency of the joint by dividing the minimum joint strength by the gross strength of the plate some distance away from the joint.

Example 5 Calculate the efficiency of the joint in the previous Examples. Report the result in percent. Solution The minimum joint strength is kips bolt shear failure. The gross tensile load of the plate is kips.

Joint kips efficiency is J. This is the better failure mode for repairability, because it is easier to replace a bolt than to replace an I-beam. Solution Method for Bolted Joints Use all of the previous steps. Solve for P based on bolt shear failure, bearing failure, gross tensile failure, and net tensile failure. The lowest value of P is the limiting case. Divide this result by PG to find the efficiency of the joint. The main plates are mm wide and 10 mm thick; the splice P plates are mm wide and 7 mm thick.

Calculate the load that the joint can support; report the result in kN. Calculate P 10 mm the joint efficiency; report the result in percent. Solution Solve for P based on shear, bearing, gross tensile, P and net tensile; the lowest value is the limiting case. Divide 7 mm this result by PG to find the efficiency of the joint. The splice and main plates have different thicknesses and widths, so calculate PG both ways.

Use this lower value for PG. Calculate PN both ways. If joined plates have to be watertight, such as the throat, h t plates in a tank or a ship, then welding is a better choice. Welding is Section A-A also a good choice for parts that will not be disassembled in the future. In welding, the weld metal and the base metal melt together. In electric arc welding, an electric arc is struck between A A an electrode and the base metal.

The arc melts the metals, and the electrode is gradually consumed. Welding problems in this chapter Pweld focus on continuous electric-arc fillet welds in lapped steel joints. The end return size l of a fillet weld is the length of a leg of the triangle. Overloaded fillet welds fail by shearing along the throat h.

E60 electrodes have an ultimate tensile strength of 60 ksi, while E70 electrodes have a tensile strength of 70 ksi. Appendix B5 shows joint strengths for common plate thicknesses. Use Appendix B5 for common weld and plate sizes; use the equation for a different electrode such as E90, or for a nonstandard plate size. The strengths in the table are given as load per unit length; multiply this value by the total length of the weld to find the shear load capacity of the welded joint.

For example, a 6 in. Weld strength is not the only consideration when you design a welded joint. Run the weld around the end of the top plate to minimize this effect. End returns are included in the total weld length, L. Lap Based on Thickness Two plates in a lap joint should lap at least five times the thickness of the thinner plate, and no less than 1 inch. Lap Based on Width This rule applies to welds that are parallel to the loading direction. Two plates in a lap joint should lap at least the width b of the narrower plate, unless the weld wraps fully around the end of the plate i.

Weld Size The leg of a fillet weld can be made as large as the thickness of the top plate, or smaller; AISC recommends ranges of size according to Appendix B6. Joint Efciency Calculate the efficiency of a welded joint the same way as for a bolted joint: divide the joint load by the gross tensile load in the plate with the smaller cross-section.

Also, Pweld calculate the efficiency of the joint. Each weld can support a load of 6 in. Pweld 4. Since there are two welds, the joint in. The gross plate strength is If the structure fails, the plate will crack while the joint remains intact.

Determine the joint strength, reporting the result in kips. Also, calculate the efficiency of the joint. Solution Appendix B5 does not include data for the E90 electrode; Pweld instead, use the weld strength equation. The length of each weld is 6 in. Pweld 7 in. The tensile strength of an E90 weld is 90 ksi. The joint can support a load of 7 in. From Appendix B4, A steel has an allowable gross tensile Pweld strength of Use the smaller plate cross-sectional area the Pweld top plate.

Pweld The joint could be strengthened in several ways: a larger weld bead, a stronger weld electrode, a continuous bead around the end of the top plate, or a weld bead underneath. Key Equations Check bolted joints for shear, bearing, gross tensile, and net tensile failure; the lowest result is the limiting case.

With a symmetrical spliced joint, erase half of the joint and solve. If standard plates are used with E60 or E70 electrodes, then use Appendix B5 to find the allowable tensile load. The efficiency of a bolted or welded joint equals the lowest allowable load divided by the allowable load of the weaker of the two plates some distance from the joint.

These properties are dimensions, area, centroid, centroidal x-x and y-y axes, moment of inertia, radius of gyration, and polar moment of inertia.

Dimensions and Area The depth of a beam is the distance from the top to the depth x x depth bottom the height of the beam. The width is the distance from the width front to the back, as you face the length long length of the beam the width thickness of the beam. The area is the cross-sectional area. For a rectangular beam, the area is the width times the depth. Centroid and Centroidal Axes y Carefully balance a cardboard rectangle on a sharp point so that the rectangle does not tip.

The location of centroid the point on the rectangle is called the centroid of the x x rectangle. The centroid of a two-dimensional shape is analogous to the center of gravity of a three- dimensional object.

Carefully balance the rectangle on a straightedge so y that two ends of the rectangle are parallel to the straightedge. The location of the balance line is a centroidal axis of the rectangle. Every shape has a centroid and centroidal axes. We will use these properties in beam problems. Set a beam on edge, define the x and y axes as horizontal and vertical, and we have x-x and y-y centroidal axes of the beam cross-sectional area.

In engineering mechanics, moment is the product of a quantity and the distance from that quantity to a given point or axis. This is the moment of a force. L We can also describe moments of areas. Consider a beam with a rectangular cross-section. The horizontal centroidal axis of this beam is the x-x axis in the drawing. Take a small area a within the cross-section at a distance y from the x-x y centroidal axis of the beam. The x and y in Ix and Iy y refer to the centroidal axis.

See Appendix C for moments of inertia of other simple shapes. What is the moment of inertia about the x-x centroidal axis? Report the answer in cm4. Estimates can get us close to the exact solution, but in this case, the math for the exact solution is less time-consuming one equation instead of Compound Beams Sharing a Centroidal Axis Laminated structural wood beams are manufactured by glueing planks together.

If we stand the planks vertically and glue them along their facing surfaces, we can increase the moment of inertia. Since bending strength is directly proportional I, if you double the moment of inertia you double the strength of the beam.

What is the moment of inertia about the x-x centroidal x x x x axis? What is the moment of inertia about the centroidal axis? Draw construction lines on a cross-section and analyze it as if it were glued together with rectangles sharing a common centroidal axis. Report the answer in in. Solution Subtract the moment of inertia of the outside dimensions from the moment of inertia of the hollow space. Sometimes we need to calculate the moment of inertia of a x x beam about a different, noncentroidal axis.

Note: the symbol d is also used for the x' x' diameter of a circle; these quantities are different, even though they share the same symbol. Example 6 5 in. A beam has a width of 5 inches and a depth of 6 inches. What is the moment of inertia about the base of the beam, marked as the x'-x' axis? Solution The moment of inertia about the centroidal axis is 6 in.

As long as the 8 in. For 1 3 in. Using the Transfer Formula, we can calculate the moment of inertia of each segment about the 12 in. Using a step process, we can calculate the moment of inertia of the compound beam. This approach is sometimes called a tabular method because you enter all of the required numbers in a table as you solve the problem.

Step 1 Divide the compound beam into simple shapes, and 1 label the segments. This compound beam can be divided into two segments; the method also works for complex shapes made up of many simple shapes. Enter the areas Seg- a and their sum into a table. Be sure to list the units, because in ment in. In 1 theory, you can select any axis, but in practice, the math is usually easier if you pick an axis along the top or bottom of the complex shape, or along the centroidal axis of one of the 2 segments.

Label these distances y1, y2, etc. Enter these values into the table. This is x1 1 x1 the distance from the centroidal axis of the segment to the centroidal axis of the complex shape. For other 2 compound beams, you will have to figure out the formulas for x x d2 d1, d2, d3, etc.

Enter the results into the table. Be sure to ment in. Step 9 Calculate I for each segment about its centroidal axis: Seg- a y ay d ad2 Io 3 ment in. Use the Transfer Formula to calculate I for the Step 10 compound shape. Let Segment 1 be the solid shape with no 2 cm hole , and Segment 2 be the hole. In all calculations, the area 6 cm of the hole and the moment of inertia of the hole are negative numbers.

Thus a1, a1y1, a1d12, and I1 are positive numbers; a2, a2y2, a2d12, and I2 are negative numbers. Segment 1 is a solid rectangle measuring 8 cm wide by 6 cm deep; Segment 2 is a hole measuring 6 cm wide by 2 cm deep. This is the distance from the centroidal axis of the segment to the centroidal axis of the complex shape.

Seg- a y ay d ad2 Io Note that I is negative for the hole. Instead of y terms in the table, we have x terms. Seg- a 2 ment in. Sum 18 Step 3 Pick a Reference Axis easiest along the left edge or along the right edge. Label Ref. In this problem, the y1-y1 and y3-y3 x1 1 2 centroidal axes are the same. Calculate this distance Ref.

Seg- a x ax d ad2 ment in. A B Consider a beam with two hollow sections. The moment x1 x1 of inertia could be calculated using the step Transfer Formula method with segments 1 large rectangle , 2 upper cavity , and 3 lower cavity , but it is easier to break it into rectangles sharing the same centroidal axis.

Instead of calculating these values, you can look them up in tables see Appendix D. Radius of Gyration Columns are tall, thin structures loaded in compression which fail at stresses below the expected yield strength of the material. Column analysis uses the radius of gyration to calculate failure loads.

Key Equations Moments of inertia of simple shapes about the x-x centroidal axis are in Appendix C. Power transmission parts are typically circular solid shafts or circular hollow shafts because these shapes are easy to manufacture and balance, and because the outermost material carries most of the stress. For a given maximum size, more material is available along the entire surface of a circle than at the four corners of a square.

Twisting means the material is deforming, so we have strain in the material. The strain varies linearly from the center to the surface of the shaft.

Therefore, the stress in the shaft also varies linearly from the center to the surface of the shaft. Consider a small area a at a distance r from the center of the circle. In many c Tc problems, we know the applied torque and dimensions; we need the stress. What is the shear stress? Report the answer in ksi. If the calculated stress is higher than the allowable stress, then we can either select a stronger material or use a larger diameter shaft.

The inside diameter is mm; the outside diameter is mm. What is the shear stress on the inside and outside surfaces? We use the same J for both the inside and the outside 32 shear stress calculations. Write the shear stress equation algebraically, substituting for J and for c. The math is the same; just the symbols have changed. The shaft is 10 cm in diameter and 1 m long. What is the angle of twist?

Report the answer in radians and in degrees. The shaft has an outside diameter of 1. In a tensile member, if you double the length the elongation doubles. Shear stress is independent on length, just as normal stress is independent of the length of a tensile member.

Many shafts in machines have different diameters in different r J locations, such as at the flange of an axle shaft, or adjacent to a bearing surface. A T sudden change in diameter creates a stress concentration. Other dimensional changes, such as keyways and transverse location of through-holes, also create stress concentrations in shafts. The 2. Use a five-step process for T calculating the maximum shear stress at d D the stress concentration site: 2.

Step 3 Using these two values, find the value of K from the graph. If your 1. For example, if 2. The weld bead is ground to a radius of 0. If the torque is ft. Use the 1. Loading the rod in tension parallel to its axis makes the rod a tensile member; loading it in compression parallel to its axis makes it P a compressive member. If you twist the steel rod with torque T, then we call it a torsional shaft. T If loading is perpendicular transverse to its axis so that the rod bends, then the rod is called a beam.

You can load a beam with point loads, uniformly distributed loads, or nonuniformly distributed loads. A swimmer standing P on the end of a diving board is an example of a point load: a force applied at a single point on the beam. The symbols for supports indicate the reactions that develop at the support. A pinned support may have vertical and horizontal reaction forces.

A roller support allows the beam to move freely horizontally; the symbol is a RAy RBy circle. A roller support has only a vertical reaction force. A beam supported by a pin at one end and a roller at the other end is called a simply-supported beam.

P A cantilever beam is embedded in a wall, so the beam has reaction forces as well MB as a reaction moment. The weight of a beam is a uniform distributed load.

The weight per unit length, w, RBy typically has units of lb. The U. Customary W-beam designation system has two numbers: the first is A B the nominal depth, and the second is the weight per unit length. In Canada, W-beams are specified in SI metric units. If the beam is 4 m long, then the total 1.

Report the answer in lb. The specific weight of steel is lb. Divide the total load on the beam by 2 to find the reaction forces.

Example 3 lb. Calculate the reaction forces RA and RB for a ft. Report the answer in kips. Solution Multiply the uniform distributed load by the length to find the total load on the 10 ft. The moment about a point is the force acting on an object times the perpendicular distance from the MA A B force to the pivot point. A RA You can pick a pivot point at either end of a simply-supported beam.

Most P students find it easier to select the left end of the beam, point A. Since moment has a magnitude and a direction clockwise or counterclockwise , we need to establish a A B convention for positive and negative moments.

The load acts at a x distance x from point A. Think of point A as a hinge point…the load causes the beam to RA RB rotate clockwise about point A, so the moment is negative. The reaction force RB acts at a L distance L from point A, and causes the beam to rotate counterclockwise about point A, so the moment is positive. Use the sum of the forces in the vertical direction to calculate the other reaction force.

A B Solution Redraw the diagram, marking the distances to all loads and reactions from point 7m 3m A. A B You can check the answer by solving the sum of the moments about point B. RA 12kN Most of the applied load is supported by the left end of the 40kN beam.

Intuitively, this makes sense because the load is closer A B to the left end of the beam. Flip the beam upside down and it looks like two children on a see-saw: the pivot point has to 12kN 28kN be closer to the heavier child in order to balance the see-saw. Calculate the reaction forces RA and RB for a beam with two point loads.

A B Solution Redraw the diagram, marking the distances to all loads and reactions from point A. This step may seem to be a waste of time, but as the loading conditions become more 3" 3" 4" complicated, it becomes more important to redimension the drawing, in order to keep RA RB track of the distances used in the Sum of the Moments calculation.

Use the equivalent load diagram for calculating the reaction forces. Report the result in N. Load diagram Equivalent load diagram Use the same approach for a nonuniformly distributed load. Again, the location of the equivalent load is at the w W centroid of the distributed load. If the beam has point loads and distributed loads, draw an equivalent load diagram with the applied point loads and the equivalent point loads, then solve like Example 5. Reactions for Overhanging and Cantilever Beams A simply-supported beam is supported by a pinned connection at one end and a roller support at the other; all applied loads lie between these two points.

An overhanging beam extends beyond one or both supports. The solution method is the same as for simply-supported beams: use the sum of the moments about one of the support points to find the reaction at the other support point. Notice the 50 kN load produces a positive counterclockwise moment about point A, 50 kN 30 kN while the 30 kN load produces a negative clockwise moment about point A.

Solve for the reactions to a distributed load on an overhanging beam the same way as for a distributed load on a simply- supported beam: draw an equivalent load diagram, then use the sum of the moments and the sum of the forces to find the reactions. We do not need the dimensions to the ends of the overhangs, because there are no loads outside of the 3 ft.

The distributed load runs for 9 ft. RB Shear Diagrams When we calculate reaction forces and torques on tension members, torsion members, and beams, we are calculating external forces and torques. Unless the material has no strength at all, the material resists these external loads by developing internal loads. Apply a torque of 25 ft.

Apply a tensile force of N to each end of a 2 cm diameter rod, and a resisting tensile force of N exists within the rod all along its length.

Beams in bending also develop internal forces to resist external forces. Since the external forces on beams are transverse perpendicular to the axis of the beam , the internal resisting forces are also transverse forces. Imagine a simply-supported beam with a point load at the mid-span. Cut the beam to the left of the point load, and P V draw a free-body diagram of the beam segment.

In a free- A B A body diagram, forces must balance. Therefore, a downward force at the cut edge balances the support reaction RA. We call this shear force V. To counteract this tendency to spin, a moment M RA P V develops within the beam to prevent this rotation. The moment equals the shear force times its distance from point A.

A M Cut the beam to the right of the point load, and draw the free-body diagram. Since P is larger than RA, force V points upwards. RA Example 9 30 kN Calculate the shear forces in this beam to the left and to the right of the 30 kN point load.

Draw vertical construction lines below the load diagram 30 kN wherever the applied loads and reactions occur. Draw a horizontal A B construction line, indicating zero shear load. Next, draw the value of V along the length of the beam, as follows: 4m 4m 15 kN 15 kN Step 1Starting at the left side of the shear diagram, go up 15 kN, because RA is 15 kN upwards. V Step 5 kN Finish the shear diagram by shading the areas between your line and the 15 kN horizontal zero shear line.

Mark all significant points anywhere the Finished V shear shear line changes direction. A point load at the midspan of a simply-supported beam produces identical reaction forces and a symmetric shear diagram with two rectangles. The fully updated Sixth Edition. Built around an educational philosophy that stresses active learning, consistent reinforcement of key concepts, and a strong visual component, Applied Strength of Materials, Sixth Edition continues to offer the readers the most thorough and understandable approach to mechanics of materials.

Mott Summary Strength Of Materials Pdf Download This book discusses key topics in strength of materials,emphasizing applications, problem solving, and design of structural members, mechanical devices, and systems.

Untener Book Summary: 6Elasticity is a form of materials response that refers to immediate and time-independent deformation upon loading, and complete and instant recovery of the original geometry upon removal of the load.

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