A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C, plate A-C is supported by a19mm-diameter pin (double shear). Find the axial stress in AE and the shear stress in pin C.

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

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