Regular low-alloy steel has an allowable shearing stress of approximately 20.0 ksi (Fv = 20.0 ksi). The beam shown has an unusual loading pattern that will cause very high shearing forces. If this beam is a W30 X 173, will it be adequate in shear? ### Transcription and Explanation of Structural Diagram#### Left Diagram: W30 X 173 I-Beam Cross-sectionThis is a technical drawing of an I-Beam with the following dimensions:- **Depth (Height):** 30.44 inches- **Flange Width:** 14.985 inches- **Flange Thickness:** 1.065 inches- **Web Thickness:** 0.655 inches- **Designation:** W30 X 173The I-Beam is commonly used in construction due to its strength and load-bearing capability. The "W30 X 173" designation indicates the nominal depth (30 inches) and weight per foot (173 pounds).#### Right Diagram: Load Distribution and Shear DiagramThis diagram shows the load distribution on a beam and the corresponding shear forces:- **Top Section:** The beam is subjected to point loads of 250 kips each. The loads are spaced as follows: - 2 feet from the ends of the beam. - 6 feet between each pair of loads. - **Load Distribution Details:** - Three point loads are applied on the beam. - **Shear Force Diagram:** - The vertical lines indicate positions where loads and reactions occur. - **Reactions at Supports:** 375 kips at each end. - **Shear Force Values:** - Initial shear force: 375 kips (upward). - Drops by 250 kips for each load, altering the shear along the beam. - Shear drops to -125 kips and further decreases with each subsequent load to -375 kips.This educational content is designed to inform about the importance of understanding beam specifications and the effects of load distribution on structural components.

Answer In order to check the adequacy of given beam for allowable shearing stress fv = 20 ksi we need to calculate horizontal shear stress for given load. We know that Horizontal stress formula is given by fv = VQI*b V = External shear due to loading V = 375 K Q = Statistical moment of that portion of the section lying outside (either above or below) the line on which 'fv' is desired, taken about the neutral axis. We know that maximum shear stress occcur at neutral axis(N.A) hence find Q about N.A Q = A1*y1+A2*y2 A1 = 14.985*1.065 A1 = 15.96 in2 A2 =30.44-2*1.0652*0.655 A2 = 9.27 in2 y1 = centroid of A1 area from N.A y1 = 30.442 - 1.0652 y1 = 14.6875 in y2 = centroid of A2 area from N.A y2 = (30.44-2*1.065)4 y2 = 7.1 in Now Q = 15.96*14.6875+9.27*7.1 Q = 15.96*14.6875+9.27*7.1 Q = 300.23 in3I = Moment of inertia about Neutral axis Since the Neutral axis is passing through the half the section due to symmetry Neutral axis from top = 30.442=15.22 in I = 2 * 14.985*1.065312+14.985*1.065*15.22-1.06522 + 0.655*(30.44-2*1.065)312 I = 8126.92 in4 b = Width of cross section where the unit stress is desired. b = 0.655 in (width at neutral axis location) Substitute the values to calculate 'fv' fv = VQI*b fv = 375*300.238126.92*0.655 fv = 21.16 ksi > 20 ksi Not Adequate " W 30 x 173 is not Adequate "