Question

Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum. The width d = 9.5 mm .

Answer 1

Consider the free-body diagram of the beam as follows: 3 kN/m 2.5 kN/m Ax 2 m 2 m 0.6 m By Ay Consider the moment equilibrium at point A ΣΜ.-0 2-3/24-2-2) 2.5x22 0 1 -x2 x3x| 2.6+-x2 B,x4.6x2x 2 By x 4.6 14.8 By 3.22 kN Consider the forces acting along the vertical direction ΣΕ-0 1 AyBy 2.5x2+-x3x2 2 4, +3.22 8 A, = 4.78 kN Consider the forces acting along the horizontal direction ΣΕ-0 A 0 Calculate the shear force at various distance as follows: Shear force at point A = A Shear force at point A = 4.78 kN Shear force at 2m distance from point A = A, -2.5x2 Shear force at 2m distance from point A 4.78-5 Shear force at 2m distance from point A -0.22 kN

Shear force at 2.6m distance from point A =-0.22 kN 1 Shear force at point B A, -2.5 x2-x3x2 2 1 Shear force at point B 4.78-2.5x2-x3x2 2 Shear force at point B -3.22 kN Draw the shear force diagram as follows: 3 kN/m 2.5 kN/m Ax 2 m 2 m 0.6 m By Ay 4.78 kN 0.22 kN -3.22 kN Obtain the maximum shear force from the diagram as follows: Vmax =4.78 kN Consider the cross section of the beam as follows 00 mn m 18 mm 9.5 mm 150 mm y 18 mm 100 mm

Calculate the moment of inertia as follows: 1 bh I 12 3 1 100 x186 12 1 ')(100-9.5)x150) 3 I: 12 4 I 28170675 mm Calculate the value y as follows: Ay1Ay3 A3ys y= A1 A2+A = 93 mm (from bottom) Hence, neutral axis is at 93mm from bottom Consider the cross-sectional area above the neutral axis as follows: 100 mm 18 mm yn2 75 mm 9,5 mm Neutral axis 93 mm 18 mm 100 mm

Calculate the value of Q as follows: Q AnAn2 n2 75 18 Q -(9.5x 75)x+(100x18) x 75+ 2 2 Q-177918.75 mm3 Calculate the maximum shear stress as follows: Vmax xQ max Ixd 4.78x10 x177918.75 28170675x9.5 max =3.178 N/mm Tmax 3.178 MPa