The solid 1.05-in.-diameter rod is subjected to a uniform axial distributed loading along its length of w = 800 lb/ft. Two concentrated loads also act on the rod: P= 2300 lb and Q = 1400 lb. Assume a = 12 in. and b = 24 in. Determine the normal stress in the rod at the following locations: (a) x = 8 in. (b) x = 24 in. → → -→ P B a Answer: Ox=8in. = psi Ox=24in. = psi

Q: The rod has a diameter of 40 mm. Determine the state of stress at point A and draw the resulting 3-D…

A:

Q: The solid 1.15-in.-diameter rod is subjected to a uniform axial distributed loading along its length…

A:

Q: The following shaft consists of a 3.50 m long steel shaft and a 2.00 m long aluminum tube. The steel…

A:

Q: For the stress block shown, what is OP2 if OP1 = 140 MPa (T)? A.60 MPa (T) 40 MPa B.60 MPa (C) C.20…

A: Calculations:σx=+120mpaσy=-40Mpaσ1=140mpagivenσ1=σx+σy2+σx-σy2+ζxy2140=120-402+120+4022+ζxy2140=40+8…

Q: 2. The assembly in Figure consists of a light rigid bar AB, pinned at 0, that is attached to the…

A:

Q: Two forces P1 and P2, of magnitude P1 = 15 kN and P2 = 18 kN, are applied as shown to the end A of…

A: Given:P1=15 kN     =15000 NP2=18 kN     =18000 NRadius c=20 mmThe axial force and bending moment…

Q: a b 45 MPa 60° a

A: Given: Shear stress  To find: Normal stress and principal stress on inclined plane

Q: A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is…

A:

Q: 1) The copper pipe below has an outer diameter of 50 mm, inner diameter of 30 mm, and G = 45 GPa. It…

A:

Q: 2-29 The nonuniform loading causes a normal strain in the shaft that can be expressed as e, = kr²,…

A:

Q: Consider the given state of stress. Take A= 16 MPa and B= 48 MPa. B 60 MPa

A: A=16 MPa B=48 MPa τ=60 MPa Formula used to find the principal planes tan2θ=2τA-B Here, θ is the…

Pressure and Stress