- Problem 6 Hinged Beams Design For The Set Steel Beam Fig 10 Tab 10 Constructed Schedules Of Cross Section Forces An 1 (93.74 KiB) Viewed 23 times
HINGED BEAMS DESIGN Design a steel beam, loaded with the forces: P=50kN; q=-20kN/m; and bending moment M=15kN·m as shown on Fig. 10.1. The beam is manufactured of steel 50.. Plot a shear diagram and bending moment's diagram. The combined cross-section should be considered as a single whole. Distance between support 1=4m.Coefficient "=0.5. Safety factor " =1.5₁ Cross-section: 'I' Double-T No 18 nl 9 1² 0,51 Fig. 10.1 Agenda 1. Work out the equilibrium equation and to determine reactions at supports. 2. Write down expressions Q(x) and (x) on each part of the frame and plot the shearing forces diagram Q and bending moments diagram M. Verity bending moments diagram with the help of differential dependences.
3. Determine the drag torque cross-section concerning the main central axes and it is rational to arrange it. 4. Check up durability in dangerous cross-section on the elastic flexure formulas; to specify a degree under load or an overload of a frame (%). 5. To give recommendations concerning rational number of a structure cross-section. Solution 1. We shall work out the equilibrium equation of a frame and we shall determine reactions at support. ΣΜ.-0. ΣΜ.=4-2-1 2+P.2-R, 4-M =0 R₁ = q·2+P.2-M 20-2+50-2-15 4 = 31.25 4 KN; 2-P.2-M = 0 ΣΜ, =0 ΣΜ, =R-4-9-2-4-12)- - P.₁² S = q·2·3+ P.2+M 20-6+50-2+15 4 = 58.75 4 KN; Y=R₁ q.2 P+R₁=0 Y=58.75-20-2-50+31.25= 0 2. Plot the shearing forces diagram Q and bending moments diagram M. Part 1: 0≤x≤ 2 Q(x) = R₁-qx Q(0) = R₁ = 58.75 KN; Q(2) R₁-q-2-58.75-20-218.75 A KN; x² M(x) = R₁- -9 M (0)=0. M (2) 58.75-20-2-77.5 kN-m. Part 2: 0≤x≤ 2 Q(x)=R₁-q-2-P=58.75-20-2-P=-31.25 KN;
R, 58,75 KH 58,75 X - q=20 kH/M 2 M (1) + P=50 KH M=15 KH-M 2 4 M 18,75 77,5 62,5 (+) (0) R₁-31,25 kH Q KH 31,25 MKH M Fig. 10.2 3. Determine the moments of resistance cross-section concerning the main central axes. Geometrical characteristics of a double-T No 18 [1]: -height: h=180mm; the moment of inertia width: b=95mm; Jx₁ = Jy =1290-10*mm* the area: F = 2340mm²; J₁ J₁₂ = 82,6-10* mm*. thickness of a strut: d = 5,1mm;
To find the drag torque, we execute such actions. Find position of the main central axes cross-section. 3.1. Find position of the main central axes of inertia cross-section (Fig. 10.3). YA h=180 MM d=5,1MM b=95MM h=180 MM C F て sec AMMWWWWWW** 1,-46,275MM a=46,275MM b=95MM PHC. 10.3 Given cross-section has an axis of symmetry Yo, and is the main central axis The second main axis will pass through the center of weight cross-section perpendicular to it. To find position of the center of weight cross-section on an axis Yc, we shall choose auxiliary system of coordinates yo. The area-moment ratio cross-section concerning an auxiliary axis z is estimated under the formula S. = yc-Fsec where Fec - The area of all cross-section, y. - distance from its center of weight to an axis z. From here S₂ Y₁ == y=136,275MM
On the other hand the static moment complex cross-section concerning any axis is equaled to the sum of the static moments its compound concerning this axis. For ours cross-section we shall have S. = S + S² = y®F₁ + y2F₂- Where y..y. - distances of the centers of weight accordingly the first and second doublets from an auxiliary axis z: h 180 2.0) = = 90mm; y(2) = h +0,5d = 180+0,5-5,1=182,55mm; F₁ = F₂ = 2340mm²- Their areas cross-section. S. = 90-2340+182,55-2340-637767mm³; sec=F₁+F₂=2340+2340=4680mm² Then S. 637767 = Ye = 136,275mm 4680 sec Designate the center of weight section a point C and it is spent through it to the second the main axis of inertia Z.. 3.2. Find the moments of inertia section concerning the main central axes. Concerning an axis Ye Jy = JQ+JQ²). Here and 2 the moments of inertia accordingly the first and second doublets concerning the main central axis section Yc. Taking into account, that the central axis section is simultaneously and the central axis for each of doublets, we shall receive Concerning an axis Z = Jy₂ = J + J(2²) = J₁₁ +Jy2=826000+ 12900000-13726-104 mm4 31
J₂ = JQ) + JQ); - Here and the moments of inertia accordingly the first and second doublets concerning the main central axis section Z. Those them to calculate, it is necessary to take advantage of formulas for definition of the moments of inertia concerning parallel axes: J2) = J₂ + a²F₁ Here JZ, - the moment of inertia of the first doublet concerning its central axis Z₁ (fig. 10.3); q- distance between an axis and the central axis of all section. z h a₁ = y₂ =136,275-90= 46,275mm; 2 Then J=12900000+ 46,275²-2340=179108-10*mm*. Similarly for the second doublete J(²2) = J-2+ a²F₂; a₂ = (h+ 0,5d) - y = (180+ 0,5-5,1)—136,275= 46,275mm J(2) = 826000+ 46.275²-2340= 583,68-10 mm*. The total moment of inertia section concerning axis Z: Jzc=1791,08-104 +583,68-104 =2374,76-10*mm*. 3.3. Determine the moments of resistance for assigned section: Jr. Jy 13726-104 WIC -= 152,51-10³ mm³; = == Zmax 0,5h 90 Jzc Jz. 2374,76-104 -=174,26-10³ mm³ Ymax Yo 136,275 W zo -- = ==
3.4. That it is rational to arrange section concerning working loading, we compare sizes of the moments of resistance concerning the main axes of inertia. The field line in section should be perpendicular to an axis, concerning which moment of resistance the greater (see a strength condition). As in our case preliminary accepted orientation section is correct as the field line coincides with axis Yc, and perpendicular to axis Z.. 4. Check up durability in dangerous section applying the elastic flexure formulas. M. =<[0]. max W. The greatest bending moment, it agrees diagram (fig. 10.2). Mmax=77.5kN-m. Admissible stress for steel 50: [0]= 380 -253,3 MPa. nr 1,5 The greatest pressure in dangerous cross-section: o max 77,5-106 174,26-10³ =444,74MPa ▷ [o]. As we see, the strength condition is not carried out. Determine a degree of an overload: - [o] 100= 444.74-253.3 253.3 [o] 5. According to calculation the beam is overloaded. That it could sustain the set loading, it is necessary to pick up doublets with big number of a structure according to an assortment. max