2.  Solids are often placed under stress - for example, a weight is hung from a steel cable, or a steel column supports the weight of a building or a floor.  Structural engineers, who make sure that a structure such as a plane, a bridge, or a sky scraper, is safe and functional, need to study the properties of solids. Topic 2.2 Extended I – Statics : Solids and elastic moduli  We first define stress, which is a measure of the magnitude of a load that is placed on a material. stress = FAFA Stress  There are three types of stress: tension is a lengthening stress, compression is a shortening stress and shear is a cutting or bending stress. tension compression shear FYI: All three types of stress are measured in newtons / meter 2 but all have different effects on solids.

3.  Structural engineers can measure the amount of strain, which is the amount of deformation a solid is exhibiting under the action of a stress. strain = Topic 2.2 Extended I – Statics : Solids and elastic moduli change in length original length = LL0LL0 Strain  Engineers study stress-strain curves, which are characteristic of each material.  To create a stress-strain curve, engineers subject a material to ever-greater stress forces, and measure the strain (deformation): L0L0 LL A F FYI: Typical building materials return to their original dimensions when the stress is removed, as long as that stress does not exceed a particular limit, characteristic of the material.

4. Topic 2.2 Extended I – Statics : Solids and elastic moduli plastic region yield region elastic region elastic limit failure region 70000 60000 50000 40000 30000 20000 10000 0 0 50 100 150 200 250 300 350 400 strain (  10 -8 ) s t r e s s ( N / m 2 ) permanent set yield point breaking point FYI: If a material is stressed beyond its YIELD POINT, it will become permanently deformed. As the stress is removed, it will follow the PERMANENT SET line, as illustrated: FYI: Structural engineers NEVER design a structural component to exceed its ELASTIC LIMIT. FYI: Since the region of the graph up to the elastic limit is linear, we can characterize a material by the slope: stress / strain. We call this ratio the ELASTIC MODULUS. elastic modulus = stress strain FYI: For this particular material the elastic modulus is elastic modulus = stress strain 40000 n/m 2 50  10 -8 = = 8  10 10 n/m 2 FYI: Thus tables showing properties of materials can simply show a single number for that material, rather than a graph.

5. Y OUNG'S M ODULUS - C HANGE IN L ENGTH Topic 2.2 Extended I – Statics : Solids and elastic moduli  Both tension and compression act to change the length of a material - tension stretches, and compression shortens.  The elastic modulus for tension and compression (not shear) is called Young's modulus (Y). Thus Y = stress strain Y = F/A  L/L 0 Y = FL 0 A  L Young's Modulus (for tension and compression)  We can solve the above equation for  L to get a useful relationship: L =L = FL 0 AY Deformation Under Compression or Tension

6. Y OUNG'S M ODULUS - C HANGE IN L ENGTH Topic 2.2 Extended I – Statics : Solids and elastic moduli A 15.0-meter long steel cable with a diameter of 1 cm has a 500 metric ton mass hanging from it. How much does it stretch under this tension load? Assume Young's modulus for steel is Y = 40  10 10 n/m 2. A =  r 2 L L0L0 LL A A A =  d2d2 2.01 2 2 A = 7.854  10 -5 m 2 F = mg F = (500)(1000)(10) F = 5  10 6 n L =L = FL 0 AY = (5  10 6 )(15) (7.854  10 -5 )(40  10 10 )  L = 2.4 m FYI: Since the cable is under TENSION it STRETCHES so that its length under the load is 15.0 + 2. 4 = 17. 4 m. FYI: The typical stretches for cables (and compressions for columns) is extremely small.

7. Y OUNG'S M ODULUS - C HANGE IN L ENGTH Topic 2.2 Extended I – Statics : Solids and elastic moduli Material Density  (kg/m 3) Young's Modulus Y (10 9 n/m 2) Ultimate Strength S u (10 6 n/m 2) Steel a 7860 400 Aluminum2710 110 70 Glass2190 50 b 65 Concrete c 2320 40 b 30 Wood d 525 50 b 13 Bone1900 170 b 9b9b a Structural steel (ASTM-A36) b In compression c High strength d Douglas fir FYI: The ULTIMATE STRENGTH is just the stress at the ELASTIC LIMIT. FYI: For the previous example, the stress is given by F / A = 5  10 6 / 7.854×10 -5 = 6.37  10 10 n/m 2. This is clearly beyond the elastic limit of 400×10 6. This structure would be deemed UNSAFE.

8. S HEAR M ODULUS - C HANGE IN S HAPE Topic 2.2 Extended I – Statics : Solids and elastic moduli  Suppose you apply a sideways force to a stack of tiles:  Note the original and the final shapes are quite different.  h A  h A  We define the shear modulus S like this: x x S = stress strain S = F/A x/h FYI: Since x / h is tan , and since x will be very small for a solid object, we can use the approximation that tan    (in radians). Thus x / h  . = F/A  Shear Modulus (for shear) FYI: Strain, remember, is a unitless quantitiy which gives a value for the amount of deformation. Since the deformation is perpendicular to the height, it makes sense to use an angle.  A more useful working definition looks like this:  = = F AS Deformation Under Shear

9. S HEAR M ODULUS - C HANGE IN S HAPE Topic 2.2 Extended I – Statics : Solids and elastic moduli A 20-meter span of steel supports a 600-metric ton hanging mass from its center. The span is 25-cm high and 10-cm wide. If the shear modulus of this particular steel is 10  10 10 n/m 2, what is  and what is x - the amount the span sags due to the hanging mass? h = 10 m  Each half span supports half the weight, or F = 3  10 6 n. W = (600)(1000)(10) W = 6  10 6 n  The area parallel to the force is A = (.25)(.10) = 0.025 m 2.  = = F AS = 3  10 6 (0.025)(10  10 10 )   = 0.0012 rad FYI: Since x / h  , and since h is 10 m, we can easily find out how much the span sags: x = h . x = h  x = (10)(.0012) x = 0.012 m

10. B ULK M ODULUS - C HANGE IN V OLUME Topic 2.2 Extended I – Statics : Solids and elastic moduli  Suppose you apply a force to an object equally in all directions, as shown.  The object will lose volume and become denser:  The volume strain is given by -  V/V 0.  We define pressure p as the force per unit area, or p = FAFA Definition of Pressure  The bulk modulus B is then defined as B = stress strain B = F/A -  V/V 0 Question: Why is the volume strain NEGATIVE? = p-V/V0p-V/V0 Bulk Modulus (Volume Change)  V = - V 0 pBpB FYI:  p is called the VOLUME STRESS and is the difference in pressure between original volume and final volume. Deformation Under Pressure

11. B ULK M ODULUS - C HANGE IN V OLUME Topic 2.2 Extended I – Statics : Solids and elastic moduli  Solids, liquids and gases all have bulk moduli. ONLY SOLIDS HAVE YOUNG'S AND SHEAR MODULI. Why? MaterialY (10 10 N/m 2 )S (10 10 N/m 2 )B (10 10 N/m 2 ) Moduli for Various Materials Aluminum7.02.57.0 Brass9.03.57.5 Copper113.812 Iron156.012 Steel208.215 Alcohol, Ethyl0.10 Glycerin0.45 Mercury2.6 Water0.22

12. B ULK M ODULUS - C HANGE IN V OLUME Topic 2.2 Extended I – Statics : Solids and elastic moduli What is the change in pressure needed to compress 15 cubic meters of water by 2%?   V/V 0 = -0.02 B = p-V/V0p-V/V0  p = -B  V/V 0  p = -(.22  10 10 )(-0.02)  p = 4.4  10 7 n/m 2. FYI: As we will find out, this is an EXTREMELY high pressure. Water is not very much compressed, even at the bottom of the deepest ocean trenches.