(a) Explanation Given information: The thickness of the cross section is h = 10 mm . The width of the cross section is b = 80 mm . The centric load applied to the welded plates is 100 kN. The shearing stress parallel to the weld is τ w = 30 MPa . Calculation: Show the welded steel plates as in Figure 1. Find the area of the weld ( A w ) using the relation. A w = h × b cos β Substitute 10 mm for h , 80 mm for b , and 25 ° for β . A w = 10 mm × 1 m 1,000 mm × 80 mm × 1 m 1,000 mm cos β = 8 × 10 − 4 cos β m 2 (1) Show the cross-section of the welded plate as a cut section as in Figure 2. Find the shear force ( F s ) by resolving the shearing forces along the shear axis as follows. ∑ F s = 0 F s − 100 sin β = 0 F s = 100 sin β Find the angle β using the relation. τ w = F s A w Substitute 30 MPa for τ w , 100 sin β for F s , and 8 × 10 − 4 cos β m 2 for A w (b) To determine Find the normal stress perpendicular to the weld.

7th Edition

ISBN: 8220102804487

Author: BEER

Publisher: YUZU

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

Chapter 7, Problem 160RP

(a)

To determine

Find the angle β.

(b)

To determine

Find the normal stress perpendicular to the weld.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7, Problem 159RP

Chapter 7, Problem 161RP

Students have asked these similar questions

4. Two links made of heat treated 6061 aluminum (Sy = 276 MPa, Sys = 160 MPa) are pinned together using a steel dowel pin (Sy = 1398 MPa, Sys = 806 MPa) as shown below. The links are to support a load P with a factor of safety of at least 2.0. Determine if the link will fail first by tearout, direct shear of the pin, bearing stress on the link, or tensile stress at section AA. (Hint: find the load P for each case and choose the case that gives the smallest load.) P 8 mm P 8 mm ¡+A 3 mm →A 10 mm P

1. For a feature other than a sphere, circularity is where: A. The axis is a straight line B. The modifier is specified with a size dimension C. All points of the surface intersected by any plane perpendicular to an axis or spine (curved line) are equidistant from that axis or spine D. All points of the surface intersected by any plane passing through a common center are equidistant from that center 2. What type of variation is limited by a circularity toler- ance zone? A. Ovality B. Tapering C. Bending D. Warping 3. How does the Rule #1 boundary affect the application of a circularity tolerance? A. The modifier must be used. B. The feature control frame must be placed next to the size dimension. C. The circularity tolerance value must be less than the limits of size tolerance. D. Circularity cannot be applied where a Rule #1 boundary exists. 4. A circularity tolerance may use a modifier. A. Ø B. F C. M D. ℗ 5. A real-world application for a circularity tolerance is: A. Assembly (i.e.,…

3. A steel bar is pinned to a vertical support column by a 10 mm diameter hardened dowel pin, Figure 1. For P = 7500 N, find: a. the shear stress in the pin, b. the direct bearing stress on the hole in the bar, c. the minimum value of d to prevent tearout failure if the steel bar has a shear strength of 175 MPa. support column pin bar thickness of bar = 8 mm h d 150 mm

Chapter 7 Solutions

EBK MECHANICS OF MATERIALS

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Find the angle β .

Solution Summary: The author shows the cross-section of the welded steel plates as a cut section as in Figure 1.