Question

The beam shown (Figure 1) is supported by a pin at A and a cable at B. Two loads P = 13 kN are applied straight down from the centerline of the bottom face. Determine the state of stress at the point shown (Figure 2) in a section 2 m from the wall. The dimensions are w = 5.2 cm , h = 10.5 cm , L = 0.8 m , a = 1.5 cm , and b = 4 cm .
Review Learning Goal: To calculate the normal and shear stresses at a point in a beam subjected to a combined loading The state of stress at a point is a description of the normal and shear stresses at that point. When a structural element is subjected to multiple kinds of loads, the effects of each can be combined using the principle of superposition, as long as the deflections remain small, the response is elastic, and the material is homogenous. The beam shown (Figure 1) is supported by a pin at A and a cable at B. Two loads P = 13 kN are applied straight down from the centerline of the bottom face. Determine the state of stress at the point shown (Figure 2) in a section 2 m from the wall. The dimensions are w = 5.2 cm , h = 10.5 cm , L = 0.8 m, a = 1.5 cm , and b = 4 cm Part A - Normal stress due to the axial load What is the normal stress at the point due to the axial load in the beam? Let a compressive stress be negative Express your answer with appropriate units to three significant figures. View Available Hint(s) Oaxial = 4.12 MPa Submit Previous Answers Correct Part B - Normal stress due to bending Figure < 1 of 2 The section 2 m from the wall also has a bending moment, which causes a normal stress at points that are not on the neutral axis. What is the normal stress at the point due to the bending moment? Let a compressive stress be negative. Express your answer with appropriate units to three significant figures. View Available Hint(s) Obending = 41.5 MPa 300 Submit Previous Answers Correct B The total normal stress at the point is the sum of the stress due to bending and the stress due to the axial load. Notice that the bending stress is significantly larger, which is typical for most points in the cross section. Points near the neutral axis are the exception . P P L Activate Windows Go to Settings to activate Windows. L Part C-Shear stress L

Answer 1

solution >T HA L - A VA A) Here, compressive force = HAS T60530°: Now, taking moment about. A and equating to zeso; T sinoo x 3L = Px2L+ PxL. 7 x3 = 3P T= 2 p. = 2x13= 20KN. Caxial - - TCOS 30° = -26X1000 x Cos 30° wxh (52 x D x(10-SXID) Paxiay = -4,124. mpu Bencing moment at given point =m. M = (Tsin3°) * 3L-2) = (26 X1000) XL x(0.4x1000) me 52x105 N-mm. win(512X10) x (10.5X10 39 Now, moment of Inertias I: W63 12 I = 5016375 mmy.

Now Elr beneling g but y= b = 4cm = 40mm. 52x105 bendling 5016375 40. bending = 41.464 mpa © Here, Shear force at given section =V HE V T Sin 300 ل 13 KN 15. 14--HHH I(92-b) .لبا Now, shear stress at given point = TE VA IB where, A=(1-blaws (1935-4)*5.2 come A- 6.5 cm² 6.5x100 mm² 9 = 5+( ར 4 + 10:5 105 4 ģ= 4.625 cm = 46.25 mm. B= w=5.2 cm a 52 mm ::ta (13x1000) x 6.5x100 x 46:25 DJ 1.4.98 mPa. SO 16375x52 > = 1:498 mea