Solved Examples Based On Normal And Shearing Stresses
A bar of uniform cross-section 1.3 cm2 is added upon by an axial compressive load of 15 kN. Find the normal and shearing stresses on a plane inclined at 30o to the direction of loading, neglecting the possibility of buckling of the bar.
Solution: Analytical Method
Area of plane AB = 1.3 cm2
Applied load ‘P’ \ 15 kN
Area of plane AC = A/cos (90 – 30) = 1.3 / cos 60o
Stress on plane AC are
σ’ = P/1.3 cos 60o = 15/1.3 cos 60o
σ = σ’ sin 30o
t = σ’ cos 30o
= (15/1.3 cos 60o).sin 30o = 2.88 kN/cm2 (compressive)
t = (15/1.3 cos 60o) cos 30o = 5 kN/cm2
Graphical Method (Mohr’s circle):
Step 1: Take origin ‘O’ and draw a horizontal line OX.
Step 2: Cut-off OA equal to σx, by taking some suitable scale.
Step 3: Bisect OA at C.
Step 4: With center C and radius CA draw a circle.
Step 5: At C draw a line CP at an angle 2Θ with OX, meeting the circle at P, where q is angle made by oblique plane with minor principal stress, here zero.
Step 6: Through P draw perpendicular to OX, intersect OX at Q. Measure PQ and OQ as t and s respectively.
σx = P/A = –15/1.3 = –11.5 kN/cm2
From figure σ = OQ = –1.45 × 2 = 2.90 kN/cm2
t = PQ = 2.5 × 2 = 5 kN/cm2
Direction of measuring angle is reversed as the stress is compressive in nature.
Email Based, Online Homework Assignment Help in Normal And Shearing Stresses
Transtutors is the best place to get answers to all your doubts regarding solved examples based on normal and shearing stresses. Transtutors has a vast panel of experienced in normal and shearing stresses mechanical engineering tutorswho can explain the different concepts to you effectively. You can submit your school, college or university level homework or assignment to us and we will make sure that you get the answers related to normal and shearing stresses.