Question

Picture Tools Help Acrobat Format Search AaBBCD AaBbCdx AaBbC AaBbcc AaB Aabber 1 Normal 1 No Spac... Heading 1 Heading 2 Title Subtitle Paragraph ng prooled. You may be able to get proofing tools for this language. Styles Download Never Show Again 3. [30 points] If the beam is subjected to a positive bending moment of M = 100 kN-m, determine the maximum and minimum bending stress. Also determine the shear stress at point. A which is 50 mm above from the bottom. The cross-section of the beam is I-shaped and shown in the figure 300 mm 30 mm 300 mm O Somm 30 mm 4. [30 points) The bar has a diameter of 40 mm. If it is subjected to a force of 800N as shown

Answer 1

Note : 1) cross section is not I section, as you mentioned in question but it's T section.

2) Shear force V is not given.

1 Given, 300 mm < 30 K N A 300 mm UT 워30k we M = lookNam know that that the maximum ben bending stress is induce at outer surfaces, 4. Maximum bending stress of 6 My T Ma 100 KN y = farest distance from neutral axis (centroid) I a moment of inertia Moment have to find neutral axis & moment of inertia of the section first have to find centroid of T bar. ave we

Exili Ateta! vi it ž= EY; Ai Atotal 300 here. ă=0 Az 2 30 Y, A + YEA ac = Y2 300 A+ Az X KA y ---> X 17 = 30 )x(30x300) + (300+ . ) *( 300x30) (300X30 J +(300730 O centroid y = 232.5 mm from bottom ã = ( symmetrical about yozis) got distance of neutral axis How mome sent of inertia. Ixx. Ixx = ? . Izc + I2c. we got

Now, Inc 7 I bh3 + Anda 12 [cy19-yil) (3 307300°)+(308300) (232-5 -150) Inc 12 12 -232-5 2 = 128756 250 mm 4 Izc = 1/2 bh²t A2d² 1x (300x30°) + ( 300x30)(300+15), 32-5) Izc 619 31250 mm Ich Ixx= Iict Ize 1915-232-53] 128756250 +61931250 Txx = 190687500 mm We know that minimum bending stress is induce at neutral aris.NA. and is on-m. because here y therefore minimum stress 5min = o Maximum bending stress бъ I forrest distance in becomes a bending stress 66 MY -- [ý is maximum this figure)

Ob (max) we know Q= SA.Y, ( 100x10°x103) (232-5) 190687500 121.9 N/mm² or Mpa to find shear stress at point A shear stress at A ZA ZA = QV I b V = Shear force (which is not given in question A, ; + A29,-) I - MO.I - 190687500 mm 4 at that section, & Shear force v is unknowns we can't solve further instead we can find 50 bending stress at point MYA GA [YA = 9-50 (100x10 = 1825 mm 190687500 GA 95-7 N/mm? or or Mpa b= Thickness as 232.5 we can I = 232-5-50 $1(182-5) =