Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
77 Cards in this Set
- Front
- Back
|
Tensile stress formula |
S = F /A |
|
|
For solid circular cross -section formula for tensile and compressive |
A = (π/4) D² |
|
|
For hollow circular cross-section formula for tensile and compressive |
A= π/4 (Do² -Di²) |
|
|
For rectangular cross-section formula for tensile and compressive |
A = base x height |
|
|
Shearing stress formulas |
S = F/A For single rivet : A =( π/4 )D² For double rivited joint :A =2 ( π/4 )D² |
|
|
Shearing due to punching of hole formulas |
S = F/A A = πDt ( for punching a hole ) A = 4St ( for square hole )
Where = t is plate thickness S = length of side of square |
|
|
Formula for force , pressure needed to punch a hole |
F = d x t x 80 tons
Where: d = hole diameter , in t = thickness , in |
|
|
Formula for working strength stress to screwed up tightly to packed joint |
W = St (0.55d² -0.25 d )
Where: St= working stress in psi d = bolt diameter , in² |
|
|
Formulas for bearing stress |
S = F/A Where : A = DL |
|
|
Formulas for factor of safety |
A. Based on yield strength Fₛ = Sᵧ / Sₐₗₗ
B. Based on ultimate strength Fₛ = Sᵤ / Sₐₗₗ
Where : Sᵧ = yield strength Sₐₗₗ = allowable stress Sᵤ = ultimate stress |
|
|
Formulas for pure torsional shear stress(Ss) |
Sₛ = Tc / J
Where = T = torque J = polar moment of inertia c = r ( for circular cross.section) C = distance from neutral axis to the farthest fiber |
|
|
Formulas for bending stress |
Sf = Mc / I
Where : M = moment I = moment of inertia about the neutral axis I = bh³ / 12 ( for rectangular cross section ) Z = section modulus = I/C = M / Sf |
|
|
Formulas for thermal stress |
Thermal stress = Eδ/ L Thermal stress = α E ( ∆t )
Where : δ = elongation due to applied load L = Original length α = coefficient of thermal stress E = modulus of elasticity |
|
|
Formulas for thermal elongation |
δ = αL (∆t) δ = αL (∆t)δ = FL / AE |
|
|
Formulas for relation between shearing and tensile stress based on theory of failure |
Stmax = Sty Ssmax = Sty/ 2 |
|
|
Formula for variable stress |
1/FS = Sm/Sy + Sa/Sn Where: Fₛ = factor of safety Sy = yield point Sn = endurance limit Sm = mean stress Sm = (Smax + Smin)/2 Sa = variable component of stress Sa = (Smax - Smin ) / 2 |
|
|
Formulas for Poisson's Ratio |
u = E/2G ( -1) Where : G = shear modulus of elasticity E = modulus of elasticity E = strain = F / AE u = Lateral strain / longitudinal strain = Eᵧ/Eₓ Where: Eₓ = ( L₂-L₁) /L₁ Eᵧ = (W₂ - W₁)/ W₁ Eᵧ= (t₂-t₁) /t₁ |
|
|
(prob.1) A steel rod 30 mm & 800 mm long has an allowance elongation not to exceed 1.5 mm, find the allowable load in kN. given: for steel, E = 30x10⁶ psi = 206,785,714 kpa |
274 kN |
|
|
(prob.2) Compute the induced/compressive stress, in kpa, of a steel of a steel solid shafting of 50 mm diameter and 800 mm in length that is subjected to an increase of temperature by 80 °C. given: for steel: E = 30x10⁶ psi = 206,785,714 kPa G = 12x10⁶ psi = 82,714,286 Kpa k = thermal elongation = 11.7x10⁻⁶ m/m°C |
193,551.4 kPa |
|
|
(prob.3) Compute for the load in kN on a 3 cm diameter, 100 cm long steel rod its maximum elongation exceed 0.12cm |
178 kN |
|
|
(prob.4) Compute for the polar section modulus of a SAE 1060 shafting having a diameter of 3 inches. Use a factor of safety 2 and design stress at 800 psi. |
5.30 in³ |
|
|
(prob.5) Compute the allowable load in kN on a 20 mm x 120 cm long steel rod with a maximum elongation must not exceed 1 mm. Given : for steel, E = 30x10⁶ psi = 206,785,714 kpa |
55 kN |
|
|
(prob.6) Determine the load in kW on a 25 mm diameter x 1200 mm long steel maximum elongation exceeds 1mm. Given : for steel, E = 30x10⁶ psi = 206,785,714 kpa |
85 kN |
|
|
(prob.7) The total weight of steel plates 3/4" x 3' x 2' is: Given density of steel = 0.284 lb/in³ |
1840 lbs |
|
|
(prob.8) What pressure is required for punching a hole 2 inches in diameter thru a 1/4 in steel plate? |
40 tons |
|
|
(prob.9) A link has a load factor of 0.8 , the surface is 0.92 and the endurance strength is 28,000 psi. Compute the alternating stress of the link if its is subjected to a reversing load. Assume a factor of safety of 3. |
9,333.33 psi |
|
|
(prob.10) Determine the minimum mean diameter of a taper pin for use to fix a lever to a shaft, if it is transmit a maximum torque of 700 in-lbs. The shaft diameter is 2 inches and the material allowable stresses is 15,000 psi. Use factor of safety of 2. |
7.0 mm |
|
|
(porb.11) How many 5/16 inch holes can be punch in one motion in a steel plate made of SAE 1010 steel. 7/16 inch thick using a force of 55 tons. The ultimate strength for shear is 50 ksi and use 2 factor of safety. |
No. of holes = 5.12 = says = 5 |
|
|
(prob.12) The shaft whose torque varies from 2000 to 6000 in-lbs has 1 1/2 inches in diameter and 60,000 psi yield strength. Compute for the shaft mean average stress. |
6036 psi |
|
|
(prob.13) A journal bearing with diameter of 76.2 mm is subjected to a load of 4900 N while rotating at 200 rpm, If it is coefficient of friction is 0.02 and L/D = 2.5, find its projected area in mm². |
14,516 mm² |
|
|
(prob.14) A steel tie rod on bridge must be made withstand a pull of 5000 lbs. Find the diameter(inch) of the rod assuming a factor of safety of 5 and ultimate stress of 64,000 psi. |
0.71 in |
|
|
(prob.15) If the weight of 6" diameter by 48" long SAE 1030 shafting is 174.5 kg. then what will be the weight of chromium SAE 51416 of the same size? Note: the major component of different steels is iron, therefore their densities do not differ much. |
384.4 lbs |
|
|
(prob.16) if the ultimate shear strength of a steel plates is 42,000 psi, what force is necessary to punch a 0.75 inch diameter hole in a 0.625 inch thick plate? |
61,850 lbs |
|
|
(prob.17) What is the modulus elasticity if the stress is 4,000 psi and unit strain of 0.00105? |
41.905 x 10⁶ psi |
|
|
(prob.18) A 2.5 inch diameter by 2 inch long journal bearing is to carry a 5500-lb load at 3600 rpm using SAE 40 lube oil at 200° F through a single hole at 25 psi. Compute the bearing pressure. |
1100 psi |
|
|
(prob.19) It is a problem of expansion and shrinkage of steel material so that the slightly bigger shafting of 2" diameter can be inserted/fitted to the slightly smaller hole of a steel bushing of 1.999" diameter with the following process/material/data to apply: Coefficient of expansion of carbon steel = 0.0000068"/°F temperature raised by gas heating = 24.5°F. Cooling media to use dry ice with boiling point of -109.3°F (-78.5°C). Shrinkage rate below boiling point is 0.00073 in/in. Determine the final clearance between the expanded steel brushing hole against the shrinkage of the steel shaft. |
0.000793" |
|
|
(prob.20) What modulus of elasticity in tension is required to obtain a unit of deformation of 0.00105 m/m from a load producing a unit tensile stress of 44,000 psi? |
41.905 x 10⁶ psi |
|
|
(prob.21) What force "P" is required to punch a 1/2 inch, hole on a 3/8 inch, thick plate if the ultimate shear strength of the plate is 42,000 psi? |
24,740 lbs |
|
|
(prob.22) What pressure is required to punch a hole 2" diameter through a 1/4" steel plate? |
40 tons |
|
|
(prob.23) Compute the working strength of 1" bolt which is screwed up tightly in packed joint when the allowable working stress is 13,000 psi. |
3900 lbs |
|
|
(prob.24) What is the working strength of a 2" bolt which is screwed up tightly in a packed jointly when the allowable working stress 12,000 psi? |
20,400 lbs |
|
|
(prob.25) Determine the estimated weight of an A-36 steel plate size 1/2" x 4' x 8 ' Given: density of steel = 0.284 lb/in³ |
297 kgs |
|
|
(prob.26) Determine the estimated weight of an A-36 steel plates size 3/16 x 6' x 20'. Given: density of steel = 0.284 lb/in³ |
920 lbs |
|
|
(prob.27) A 19 mm stud bolts is used to fastened on a 250 mm diameter cylinder head of diesel engine. If there are 10 stud bolts, determine the pressure inside the cylinder if bolt stress is limited to 50 Mpa. |
288.8 kPa |
|
|
(prob.28) A column supports a compressive load of 250 kN. Determine the outside diameter(in mm) of column if inside diameter is 185 mm and compressive stress of 50 Mpa. |
201.47 mm |
|
|
(prob.29) A steel hallow tube is use to carry a tensile load of 500 kN at a stress of 140 Mpa. If outside diameter is 10 times the tube thickness, find the thickness (in mm) of the tube? |
11.24 mm |
|
|
(prob.30) A 20 mm diameter rivet is used to fastened two 25 mm thick plate. If the shearing stress is 80 Mpa. What tensile force applied each plate to shear the bolt? |
25.13 kN |
|
|
(prob.31) what force is necessary to punch a 30 mm hole in 12.5 mm thick plate if ultimate shear stress is 410 Mpa? |
483 kN |
|
|
(prob.32) Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter. If the plates is subjected to 50 kN tension, find the bearing stress in bolts. |
33,333.33 kPa |
|
|
(prob.33) A 2.5 inches shaft is subjected to 3 kN-m torque. Find the stress developed. |
59.68 Mpa |
|
|
(prob.34) A shaft when subjected to pure torsion developed a stress of 50 Mpa. If polar moment of inertia is 6. 1359 x 10⁻⁷ m⁴, determine the maximum torque(in kN-m) the shaft could handle. |
1.23 kN-m |
|
|
(prob.35) What is the modulus of elasticity if stress is 300 Mpa and strain of 0.00138 ? |
217.39 x 10³ Mpa |
|
|
(prob.36) In a 2.0 m cantilevered L- beam, a 2 Mton weight is applied at free end. If the allowable stress in beam is 110 Mpa, determine the section modulus. |
21.77 in³ |
|
|
(prob.37) A 5 kN force acting at the end of a 3 m cantilever beam. If section modulus of the beam is 10 in³, what is the stress induced? |
91,535 kPa |
|
|
(prob.38) A 6 mm steel wire is 5m long and stretches 8 mm under a given load. If modulus of elasticity is 200 Gpa, find the load applied. |
9kN |
|
|
(prob.39) A steel wire 10 m long, hanging vertically supports a tensile load of 2 kN. Neglecting the weight of wire, determine the required diameter if the stress is not to exceed 140 Mpa and the total elongation is not to exceed 5 mm Assume E = 200Gpa. |
5 mm |
|
|
(prob.40) An iron rod 4 m long and 0.5 cm² in cross section stretches 1 mm when a mass of 225 kg is hang on it. Compute the modulus of elasticity of the iron. ( in Gpa) |
176.58 Gpa |
|
|
(prob.41) A 20 m rod is stretches to a strain of 0.001. Determine the deflection of the rod. |
20 mm |
|
|
(prob.42) A rail having a coefficient of linear expansion of 11.6 x 10⁻⁶ m/m- °C increases its length when heated from 70°F to 113°F. Determine the strain. |
2.77 x 10⁻⁴ |
|
|
(prob.43) What temperature will the rails touch if steel railroad is 10m long are laid with clearance of 3 mm at initial temperature of 15°C? Use k=11.7x10⁻⁶ m/m- °C. |
40.64°C |
|
|
(prob.44) For a given material, the modulus of elasticity is 15,000,000 psi in tension and 6,000,000 psi in shear. Determine the poisson's ratio for this material. |
0.25 |
|
|
(prob.45) A 1000 lb force acts at the end of a 10ft. cantilever beam. The section modulus for the steel beam is 8.0 in³. How much stress is induced? |
15,000 psi |
|
|
(prob. 46) For a given material, the modulus of elasticity is 100 GN/m² in tension and 40 GN/m² in shear. Find the Poisson's ratio. |
0.25 |
|
|
(prob.47) A 25 mm shaft is keyed to a 300 mm diameter pulley and transmits 3 kw of power. The keyed assembly rotates at 1725 rpm. What is the tangential force at the key? |
1.33 kN |
|
|
(prob.48) What force is necessary to punch a 1 in hole a 1/8 in steel plate if the ultimate shearing stress is 60,000 psi and the ultimate compressive stress is 80,000 psi? |
23,562 lbs |
|
|
(prob.49) A U-bolt supports a load of 6000 lb. The cross-section of the bolt has a diameter of 1/2 inch. How much stress is induced in the sides of the bolt? |
15,278.84 psi |
|
|
(prob.50) A 1 inch diameter shaft has a 2 inch diameter collar resting on a support. The axial load on the shaft is 10,000 lb and the thickness of the collar is 1/2 inch. How much shearing stress is induced? |
6,366.20 psi |
|
|
(prob.51) A hallow rivet has an outside diameter of 5 mm and and inside diameter of 3 mm. If the allowable shearing stress is 400 x 10⁶ N/m², what maximum shearing force can the rivet sustain if subjected to double shear? |
10.05 kN |
|
|
(prob.52) A vertical load of 400 Newtons acts at the end of a horizontal rectangular cantilever beam 2m long and 25 mm wide. If the allowable bending stress is 130 MN/m², find the depth of the beam in mm. |
38.43 mm |
|
|
(prob.53) A simply supported timber beam is 50 mm by 200 mm in cross section and 4m long. If the fiber stress is not to exceed 8.3 Mpa and the beam weight is neglected, find the maximum mid-span concentrated load that the beam can support if the 200 mm dimension is vertically oriented. |
2,766.67 N |
|
|
(Prob.54) A lap joint consists of steel plate 250 mm by 18 mm in thickness connected by 4-20 mm diameter rivets. Compute the bearing capacity of the rivet connection if the allowable bearing stress is 210 Mpa. |
302,400N |
|
|
(prob.55) A steel hanger of 2m length is to carry an axial load of 18 kN. If the tensile stress is restricted to 104 Mpa and the elongation caused by this load is restricted to 1 mm. What is the minimum cross-sectional area the member can have? E = 200,000 Mpa |
180 mm² |
|
|
(prob.56) A short hollow steel cylinder with a wall thickness of 38 mm is to carry a compression load, applied uniformly on the end, of 7,800 kN. If the allowable working stress in steel in compression is 138 Mpa, then the minimum outside diameter of the cylinder required to safety support the load is: |
511.55 mm |
|
|
(prob.57) A 25mm diameter steel bar is loaded in double shear until failure, the ultimate load is found to be 446 kN. If the allowable stress is to be based on a factor of safety 3 what must be the diameter of a pin designed for an allowable load of 26 kN in a single shear be? |
14.79 mm |
|
|
(prob.58) A steel bar, initially free of stress, is held between rigid supports. Determine the stress in the bar if temperature drops 130°F. K= 6.5 x 10⁻⁶ in/°F, E = 30 x10⁶ psi |
25,350 psi |
|
|
(Prob.59) A steel rod with a cross-sectional area of 160 mm² is to be attached between two fixed points 1.25 m apart. If the rod is too short by 0.25 mm, find the stress applied to put the rod back to fitness. E= 200,000 Mpa |
40.008 Mpa |
|
|
Formula for Modulus of Elasticity |
E = S / ε Where : ε = deformation or strain |
