Homework Help is Here – Start Your Trial Now!
Engineering
Civil Engineering
With the dimensions already given, determine the allowable value of the tensile load P, considering the following parameters (The figure shown below is a hanger): 1. The allowable normal stress at the two vertical bolts (bolts 1 and 2) is 160 MPa. 2. The allowable tensile stress in the hanger (see label) is 110 MPa. 3. The allowable tensile stess in the hanger considering the cross section through the horizontal bolt is 75 MPa. 4. The allowable shear stess at the horizontal bolt is 45 MPa. 5. The allowable bearing stress between the washer and the clip angle at the vertical bolts is 65 MPa. 6. The allowable bearing stress at the "shank" of the horizontal bolt is 180 MPa. 7. The allowable shear stress through the clip angle at the vertical bolts is 35 MPa

With the dimensions already given, determine the allowable value of the tensile load P, considering the following parameters (The figure shown below is a hanger): 1. The allowable normal stress at the two vertical bolts (bolts 1 and 2) is 160 MPa. 2. The allowable tensile stress in the hanger (see label) is 110 MPa. 3. The allowable tensile stess in the hanger considering the cross section through the horizontal bolt is 75 MPa. 4. The allowable shear stess at the horizontal bolt is 45 MPa. 5. The allowable bearing stress between the washer and the clip angle at the vertical bolts is 65 MPa. 6. The allowable bearing stress at the "shank" of the horizontal bolt is 180 MPa. 7. The allowable shear stress through the clip angle at the vertical bolts is 35 MPa

Residential Construction Academy: House Wiring (MindTap Course List)
Residential Construction Academy: House Wiring (MindTap Course List)
5th Edition
ISBN:9781337402415
Author:Gregory W Fletcher
Publisher:Gregory W Fletcher
Chapter3: Tools Used In Residential Wiring
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

With the dimensions already given, determine the allowable value of the tensile load P, considering the following parameters (The figure shown below is a hanger):

1. The allowable normal stress at the two vertical bolts (bolts 1 and 2) is 160 MPa.

2. The allowable tensile stress in the hanger (see label) is 110 MPa.

3. The allowable tensile stess in the hanger considering the cross section through the horizontal bolt is 75 MPa.

4. The allowable shear stess at the horizontal bolt is 45 MPa.

5. The allowable bearing stress between the washer and the clip angle at the vertical bolts is 65 MPa.

6. The allowable bearing stress at the "shank" of the horizontal bolt is 180 MPa.

7. The allowable shear stress through the clip angle at the vertical bolts is 35 MPa.

Note: Draw the Free Body Diagram, include the proper units and round-off the answers to 3 decimal places.

Fundamental Equations of Mechanics of Materials Axial Load Shear Normal Stress Average direct shear stress V A Tavg A Displacement P(x)dx Transverse shear stress 8 = A (x)E VQ It PL AE Shear flow VQ q = Tt = δ- α ΔΤL = a Torsion Stress in Thin-Walled Pressure Vessel Shear stress in circular shaft Тр Cylinder pr pr 2t where Sphere pr = solid cross section 01 = 02 = 2t J = - (c,“ – c,") Stress Transformation Equations tubular cross section 0x + 0y O, – 0, + y cos 20 + Power sin 20 2 P = Tw = 2™fT Ox – 0y Angle of twist sin 20 + Try cos 20 2 T(x)dx Principal Stress J(x)G Txy tan 26, TL $ = E JG (ox – 0,)/2 Ox + 0y 1,2 0,- 0, y Average shear stress in a thin-walled tube + Tây 2 T Tavg 21A m Maximum in-plane shear stress (0, – 0,)/2 Shear Flow tan 20, T Txy q = Tavg! 2A m Tmax + Txy 2 Bending Ox + 0y O avg Normal stress Мy Absolute maximum shear stress I σ. max Unsymmetric bending Tabs max for ởmax, O min same sign 2 Mạy max Ở min tan a Tabs max for omax, Omin Opposite signs tan 0 2 Geometric Properties of Area Elements Material Property Relations Poisson's ratio -A = bh Elat 1, = bh !, = hb long C %3D Generalized Hooke's Law :[0, - vo, + o.)] Rectangular area %3D Ex E [0, – v(o, + 0,)] -A = bh E Ez [0. - v(0, + 0,)] E 1 Tyz, Yzx 1 %3D Yxy Txy Yyz Triangular area where E G = 2(1 + v) A = a (9 + D)w : Relations Between w, V, M h (2a + b\ a + b dM = V dx dV b w(x), dx Trapezoidal area Elastic Curve 1 M -A = EI d*v EI =w(x) !, = }ar* %3D d'v El = V(x) Semicircular area dv El dx? M(x) -A = Tr² Buckling Critical axial load 1, = fart !, = fart T²EI Per (KL)² Critical stress VIJA O cr Circular area (KL/r)²* Secant formula ес P Ở max 1 + sec A V EA žab A = Energy Methods Conservation of energy U. = U; zero slope Strain energy N²L Semiparabolic area U; = 2AE constant axial load "M²dx U, = bending moment %3D 2EI A = U; = transverse shear %3D 2GA zero slope - pLT°dx torsional moment %3D 2GJ Exparabolic area
Transcribed Image Text:Fundamental Equations of Mechanics of Materials Axial Load Shear Normal Stress Average direct shear stress V A Tavg A Displacement P(x)dx Transverse shear stress 8 = A (x)E VQ It PL AE Shear flow VQ q = Tt = δ- α ΔΤL = a Torsion Stress in Thin-Walled Pressure Vessel Shear stress in circular shaft Тр Cylinder pr pr 2t where Sphere pr = solid cross section 01 = 02 = 2t J = - (c,“ – c,") Stress Transformation Equations tubular cross section 0x + 0y O, – 0, + y cos 20 + Power sin 20 2 P = Tw = 2™fT Ox – 0y Angle of twist sin 20 + Try cos 20 2 T(x)dx Principal Stress J(x)G Txy tan 26, TL $ = E JG (ox – 0,)/2 Ox + 0y 1,2 0,- 0, y Average shear stress in a thin-walled tube + Tây 2 T Tavg 21A m Maximum in-plane shear stress (0, – 0,)/2 Shear Flow tan 20, T Txy q = Tavg! 2A m Tmax + Txy 2 Bending Ox + 0y O avg Normal stress Мy Absolute maximum shear stress I σ. max Unsymmetric bending Tabs max for ởmax, O min same sign 2 Mạy max Ở min tan a Tabs max for omax, Omin Opposite signs tan 0 2 Geometric Properties of Area Elements Material Property Relations Poisson's ratio -A = bh Elat 1, = bh !, = hb long C %3D Generalized Hooke's Law :[0, - vo, + o.)] Rectangular area %3D Ex E [0, – v(o, + 0,)] -A = bh E Ez [0. - v(0, + 0,)] E 1 Tyz, Yzx 1 %3D Yxy Txy Yyz Triangular area where E G = 2(1 + v) A = a (9 + D)w : Relations Between w, V, M h (2a + b\ a + b dM = V dx dV b w(x), dx Trapezoidal area Elastic Curve 1 M -A = EI d*v EI =w(x) !, = }ar* %3D d'v El = V(x) Semicircular area dv El dx? M(x) -A = Tr² Buckling Critical axial load 1, = fart !, = fart T²EI Per (KL)² Critical stress VIJA O cr Circular area (KL/r)²* Secant formula ес P Ở max 1 + sec A V EA žab A = Energy Methods Conservation of energy U. = U; zero slope Strain energy N²L Semiparabolic area U; = 2AE constant axial load "M²dx U, = bending moment %3D 2EI A = U; = transverse shear %3D 2GA zero slope - pLT°dx torsional moment %3D 2GJ Exparabolic area
Bolt I Bo 2 Thread Nut Runout Shank Radjus Head Bo3 75 mm Washer -4-95 mm -Thread length- – Nominal length- Cip angle -Grip length- Hanger I-13 mm
Transcribed Image Text:Bolt I Bo 2 Thread Nut Runout Shank Radjus Head Bo3 75 mm Washer -4-95 mm -Thread length- – Nominal length- Cip angle -Grip length- Hanger I-13 mm
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
System of units
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Residential Construction Academy: House Wiring (M…
Residential Construction Academy: House Wiring (M…
Civil Engineering
ISBN:
9781337402415
Author:
Gregory W Fletcher
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS