FAR = 5.56 kN tension 80 kN FAE = 75.56 kN tension %3D FBC = 4.45 kN tension FBE = 3.34 kN compression FcD = 88.87 kN compression 1m 1m 1m FCE = 5.57 kN tension Fer = 50 kN tension FDr = 71.11 kN tension 50 kN FFE = 71.11 kN tension 0.75 m

At which point on the design P-M interaction diagram is the design flexural strength largest? O Pure compression Pure tension Balanced failure point (CCL limit) O TCL limit Beam condition (Pn = 0) %3D //img00.zhaopin.cm

Answer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 in

4. Find the maximum allowable value of P for the column. The cross- sectional areas and working stresses () are shown in Figure 3. 2P FIGURE 3 Steel: A=500 mm² Qw=120 MPa Timber: A=2000 mm² Ωw =12 MPa Timber: A=8000 mm² Ωw =16 MPa

The structural frame below is loaded as shown. Determine the required area for members AB, ED,CD, AF, and FG considering the material’sproperty below : E= 200GPa, Allowable Stress for Compression is 120MPa and 315MPa for tension. Note : all units in meters.

6. 20ft P= 20K Mmax CA A max =

Please solve this question queckly

1.)Calculate the height of the neutral axis measured from the extreme compression fiber. 2.) Calculate the strength reduction factor for the given beam. 3.) Calculate the design moment for the given beam. Given  b=9.25mm d=152.5mm d(prime)=32.5mm L=1.1m WsubD=6.25kn/m WsubL=8.25kn/m 4 25mm tension steel bars 4 12mm compression steelbars FsubC(prime)=28Mpa FsubY=420Mpa

Find the allowable bending stress for the following shapes. Assume adequate lateral support for the compression flange. a.) W30 x132 of A36 Steel